Physics 12th EMF and Electric Measurement

EMF and Electric Measurement

(1) Introduction

• In previous chapter we have already studied about electric current and resistance.
• We know that a force must be applied on free charges of a conductor in order to maintain a continous current in the conductor.
• Here a question arises how can we maintain this force in order to maintain a continous flow of current. You will find answer to this question while studying this chapter.
• In this chapter we will learn about ElectroMotive Force(emf) and sources of emf ( responsible for driving charge round the closed circuit). We'll also learn about electric circuits and measurements. 

(2) ElectroMotive Force(emf)

• Consider a conductor lying in presence of electric field as shown below in the figure such that an electric field exists inside the conductor.
  
   
• We know that when electric field exists in a conductur electric current begins to flow inside the conductor. Now a question arises what happens to the
  charge carriers when they reach the ends of the conductor and would this current remains constant with the passage of time.
• We can easily conclude that for an open ended conductor as shown in the figure , charges would accumulate at the ends of the conductor resulting a change in electric field with the passage of time. Due to this electric current would not remain constant and would flow only for a very short interval of time , digrametically shown below in the figure.
  
  
  
   
• Thus, in order to maintain a steady current throughout a conducting path the path must be in the form of a closed loop forming a complete circuit. Even this condition is not sufficient to maintain a steady current in the circuit.
• This is because charge always moves in the direction of decreasing potential and electric field always does a positive work on the charge.
• Now after travelling through a complete circuit when charge returns to a point where it has started, potential at that point must be same as the potential at that point in the begning of the journey but flow of current always involves loss of potential energy.
• Hence we need some external source in the circuit in which maintains a potential difference at its terminals by increasing the potential energyof the electric charge.
• Such a source make charge travel from lower potential to higher potential energy in direction opposit to the electrostatic force trying to push charge from higher potential to lower potential.
• This force that makes charge move from lower potential to higher potential is called electro mative force (EMF).
• The source or device which provide emf in a complete circuit is known as source of EMF and examples of such devices are generator, batteries, thermocuples etc.
• The source of EMF are basically energy converters that convert mechanical, thermal, chemical or any other form of energy into electrical potential energy and transform it into the circuit to which the source of emf is connected.
• Now we know that a source of emf or battery maintains a potential difference between its two terminals as shown in below figure
  
   
• Generally a battery consist of two terminals one positive and other is negative.
• Some internal force F_n generally non electric in nature is exerted on the charges of the material of the battery. This non-electric force depends on the nature of source of EMF.
• These Force(F_n) drives the positive charges of the material towards P and nigative charges of the material towards Q. This battery force
  F_n is directed Q to P.
• Positive charge accumulate on plate Pand negative charge accumulate on plate Q and a potential developes between plates P and Q. Thus an electric field would set up inside the battery from P to Q which excert an electric force on the charge of the material.
• When a steady state is reached, the the electric force and battery force F_n would become equal and opposit
  mathematically,
  qE=F_n                                (1)
  and after a steady state is reached no further accumulation of charhe takes place.
• Workdone by battery force F_n in taking poisitive charge from terminal Q to terminal P would be
  W=F_nd
  where d is distance between plates P and Q.
• Workdone by force F_n per unit charge is
  EMF=W/q=F_nd/q                                (2)
  where the quantity E is known as E.M.F. of the battery.
• For steady state
  ENF=qEd/q=ED=V                                (3)
  where V_eq=Ed is the potential difference across the terminals of the battery when nothing is connected externally between P and (i.e. when circuit is open)

(3)Internal Resistance of Battery (or cell)

• The resistance offered by medium in between plates of battery (electrolytes and electrodes of the cell) to the flow of current within the battery is called internal resistance of the battery.
• Internal resistance of a battery usually d branch containing batteryenoted by r and in electric circuit its representation is shown below in the figure
  
   
• Internal resistance of a battery depends on factors like seperation between plates, plate area, nature of material of plate etc. For an ideal cell r=0 , but real batteries or sourcesof emf always has same finite internal resistance.
• If P and Q are two terminals of the battery shown below in the figure
  
  
  then potential difference between terminals P and Q is
  V_P=(V_P-V_x) - (V_Q - V_x) = E-Ir
  let V_P-V_Q=V
  V=E-IR
  now for I=0 and V=EMF
  and this potential difference V is called the terminal difference of the cell or battery and defined as the emf of the batterywhen no current drawn from it.
• For real battery equation(4) which gives V=E-Ir whereI is the current in the branch containing battery.
• From figure(4) potential difference across the external resistance R of the circuit would be equal to terminal potential difference of the cell. Thus
  V=IR also V=E-Ir
  or, IR=E-Ir
  which gives
  I=E/(R+r) =Net EMF/Net resistance
• From equation(4) we can calculate that when current is drawn from the battery terminal potential difference is less than the EMF of the battery.

(4)Electric Energy and Power

• To understand the process of energy transfer in a simple circuit consider a simple circuit as shown in the figure given below
  
   
• Positive terminal of the battery as we all know is always at higher potential.
• Let ΔQ amount of charge begin to flow in the circuit from point E through the battery and resistor and then back to point E.
• When Charge ΔQ moves from point E to point F through the battery electric potential potential energy of the system increased by the amount
  ΔU=ΔQ V -(6) and the electric energy of the battery decreased by the same amount.
• When charge ΔQ moves from point G to S through resistoe R, there comes a decrease in electric potential energy.
• This loss in potential energy appears as the increased in thermal energy of the resistor.
• Thermal energy of the resistor increases because when the charge moves through the resistor they loss there electrical potential energy by colliding with the atom in the resistor. This way electrical energy is transformed internal energy crossesponding to increase in vibrational motion of the atom of the resistor and this cause increase in temprature of the resistor.
• The connecting wires are assumed to have negligible resistance and no energy transfer occur for the path FG and HE.
• In time Δt charge ΔQ moves through the resistor i.e. from G to H. The rate at which it loss potential energy
  ΔU/Δt=(ΔQ/Δt)ΔV=IΔV where I is thecurrent in the resistor and ΔV is the potential difference across it.
• This charge ΔQ regain its energy when it passes through the battery at the cost of conservation of chemical energy of electrolyte to the electrical energy.
• This loss of potential energy as stated earlier appears as increased thermal energy of the resistor. If P represents the rate at which energy is delivered to the resistor then
  P=IΔV
• We know that ΔV =IR for a resistor hence alternative forms of equation(8) are
  P=I²R=ΔV²/R where I is expressed in amperes, ΔV iv volts and resustance R in ohm(Ω)
• SI unites of power is watt such that
  1watt=1volt*1ampare
  Bigger unit of electric power are Kilowatt(KW) and Megawatt(MW).

(5) Kirchoff's Rules

• We have already analyzed simple circuit using ohm's laws and reducing these circuit to series and parallel combination of resistors
• But we also come across circuits containing sources of EMF and grouping of resistors can be far more complex and can not be easily reduced to a single equivalent resistors
• Such complex circuits can be anaylzed using two kirchoff's rules

(A) The junction Rule (or point rule)

• This law states that "The algebric sum of all the currents entering juntion or any point in a circuit must be equal to the sum of currents leaving the junction"
• Alternatively this rule can also be stated as " Algebric sum of the currents meeting at a point in a electric circuit is always zero i.e
  ΣI=0 at any point in a circuit
• This law is based on the law of conservation of charge
• Consider a point P in an electric circuit at which current I₁,I₂,I₃ and I₄ are flowing through conductors in
  the direction shown below in the figure below
   
• If we take current flowing towards the junction as positive and current away from the junction as negative,then from kirchoff's law
  I₁+I₂+(-I₃)+(-I₄)=0
  or,
  I₁+ I₂=I₃+ I₄
• From this law ,we conclude that netcharge coming towards a point must be equal to the net charge going away from this point in the same interval of time

(B) The Loop Rule (or Kirchoff's Voltage Law)

• The rule states that " the sum of potential difference across all the circuit elements along a closed loop in a circuit is zero
  ΣV=0 in a closed loop
• Kirchoff's loop rule is based on the law of conservation of energy becuase total amount of energy gained and losed by a charge round a trip in a closed loop is zero
• when applying this kirchoff's loop rule in any DC circuit,we first choose a closed loop in a circuit that we are analyzing
• Next thing we have to decide is that whether we will traverse the loop in a clockwise direction or in anticlockwise direction and the answer is that ,the choice of direction of travel is arbitrary to reach the same point again
• When traversing the loop ,we will be following convention to note down drop or rise in the voltage across the resistors or battery
  i) If the resistor is being traversed in the direction of the current then change in PD across it is negative i.e -IR
  ii)If the resistor is being traversed in the direction opposite to the current then change in PD across it is negative i.e IR
  iii) If a source of EMF is traversed in the direction from -ve terminal to its positive terminal then change in electric potential is positive i.e E
  iv)If a source of EMF is traversed in the direction from +ve terminal to its negative terminal then change in electric potential is negative i.e -E
• We would now demonstrate the use of kirchoff's loop law in finding equations in simple circuit
• Consider the circuit as shown below
   
• First consider loop ABDA.Lets traverse loop in anticlock wise direction.From kirchoff's loop law
  ΣV=0
• Neglecting internal resistance of the cell and using sign conventions stated previously we find
  -I₃R₃+E-I₁R₁-I₂R₂=0
  or
  I₁R₁+I₂R₂+I₃R₃=E
  And similarly if we traverse the loop ABCA in clock wise direction
  -I₂R₂+I₅R₅+I₄R₄=0
  or,
  I₅R₅+I₄R₄-I₂R₂=0

(6)Grouping of the cell's

• A limited ammount of current can be drawn from a single cell or battery
• There are situations where single cell fails to meet the current requirement in a circuits
• To overcome the problem cells can be grouped in series and in parallel combinations or mixed grouping of cells is done in order to obtain a large value
  of electric current

(A) Series combination

• Figure below shows the two cells of emf's E₁ and E₂ and internal resistance r₁ and r₂ respectively connected in series combination through external resistance
  
  
• Points A and B in the circuit acts as two terminals of the combination
• Applying kirchoff's loop rule to above closed circuit
  -Ir₂-Ir₁-IR+E₁+E₂=0
  or
  I=E₁+E₂/R+(r₁+r₂)
  Where I is the current flowing through the external resistance R
• Let total internal resistance of the combination by r=r₁+r₂ and also let E=E₁+E₂ is the total EMF of the two cells
• Thus this combination of two cells acts as a cell of emf E=E₁+E₂ having total internal resistance r=r₁+r₂ as shown above in the figure
  
   

(B) Parallel combinations of cells

• Figure below shows the two cells of emf E₁ and E₂ and internal resistance r₁ and r₂ respectively connected in parallel combination through external resistance
  
• Applying kirchoff's loop rule in loop containing E₁ ,r₁ and R,we find
  E₁-IR-I₁r₁=0 ------------------------(1)
  Similarly applying kirchoff's loop rule in loop containing E₂ ,r₂ and R,we find
  E₂-IR-(I-I₁)r₂=0 ------------------------(2)
• Now we have to solve equation 1 and 2 for the value of I,So multiplying 1 by r₂ and 2 by r₁ and then adding these equations results in following equation
  IR(r₁+r₂)+r₂r₁I-E₁r₂-E₂r₁=0
  which gives
  
  
  We can rewrite this as
  
  E is the resulting EMF due to parallel combination of cells and r is resulting internal resistance. 

(7) Wheat stone bridge

• Wheat stone bridge was designede by british physicist sir Charles F wheatstone in 1833
• It is a arrangement of four resistors used to determine resistance of one resistors in terms of other three resistors
• Consider the figure given below which is an arrangement of resistors and is knowns as wheat stone bridge
  
  
• Wheatstone bridge consists of four resistance P,Q,R and S with a battery of EMF E.Two keys K₁ and K₂ are connected across terminals A and C and B and D respectively
• ON pressing key K₁ fisrt and then pressing K₂ next if galvanometer does not show any deflection then wheatstone bridge is said to be balanced
• Galavanometer is not showing any deflection this means that no current is flowing through the galvanameter and terminal B and D are at the same
  potential .THus for a balanced bridge
  V_B=V_D
• Now we have to find the condition for the balanced wheatstone bridge .For this applying kirchoff's loop rule to the loop ABDA ,we find the relation
  -I₂R+I₁P=0
  or I₁P=I₂R --(a)
  Again applying kirchoff's rule to the loop BCDB
  I₁Q-I₂S=0
  or I₁Q=I₂S --(b)
  From equation a and b we get
  I₁/I₂=R/P=S/Q
  or
  P/Q=R/S                       (12)
• equation 12 gives the condition for the balanced wheatstone bridge
• Thus if the ratio of the resistance R is known then unknown resistance S can easily be calculated
• One important thing to note is that when bridge is balanced positions of cell and galvanometer can be exchanged without having any effect on the balance of the bridge
• Sensitivity of the bridge depends on the relative magnitudes of the resistance in the four arm of the bridge is maximum for same order of four resistance.

(8) Meterbridge (slide wire bridge)

• Meter bridge is based on the principle of wheatstone bridge and it is used to find the resistance of an unknown conductor or to compare two unknown
  resistance 
• Figure below shows a schematic diagram of a meter bridge
  
  
• In above figure AC is a 1m long wire made of maganin or constanan having uniform area of cross-section
• This wire is stretched along a scale one a wooden base
• Ends A and C of the wire are screwed to two L shaped copper strips as shown in figure
• A resistance box R and an unknown resistance S are connected as shown in figure
• One terminal of galvanometer is connected to point D and another terminal is joined to a jockey that can be slided on a bridge wire
• when we adjust the suitable resistance of value R in the resistance box and slide this jockey along the wire then a balance point is obtained sat at point B
• Since the circuit now is the same as that of wheatstone bridge ,so from the condition of balanced wheatstone bridge we have
  P/Q=R/S
  Here resistance P equals
  P=ρl₁/A
  And Q=ρl₂/A
  where ρ is the resistivity of the material of the wire and A is the area of cross-section of wire
  Now P/Q=(ρl₁/A)(A/ρl₂)=l₁/l₂

(9) Potentiometer

• Potentiometer is an accurate instruments used to compare emf's of a cells,Potential difference between two points of the electric wire
• Potentiometer is based on the principle that potential drop across any portion of th wire of uniform crossection is proportional to the length of that portion of thw wire when a constant current flows through the wire
• Figure below shows the construction of a potentiometer which consists of a number of segments of wire of uniform area of cross-section stretched on a wooden board between two copper strips .Meter scale is fixed parallel to the lenght of the wire
  
  
• A battery is connected across terminals A and B through a rehestat so that a constant currents flows through the wire
• Potentiometer is provided with a jockey J with the help of which contact can be made at any point on the wire
• Suppose A and ρ are the area of cross-section and resistivity of the material of the wire the resitance
  R=ρl/A ----------------------------(i)
  where l is the lenght of the wire
• If I is the current flowing through the wire then from Ohm's Law,
  V=IR ------------------------------(ii)
  Where V is the potential differene across the position of the wire of length l
  Thus ,from (i) and (ii)
  V=IR=I(ρl/A)=kl
  where K=ρI/A
  => V is proportional to l when current I is constant
• K=V/l is also known as potential gradient which is the fall of potential per unit length of wire
• Senstivity of a potentiometer depends on its potential gradient .If the potential gradient of a potentiometer is small then the potentiometer is more sensitive and hence more accurate

(A) Comparison of EMF's of two cells using potentiometer

• Consider the circuit arrangement of potentiometer given below used for comparison of emf's of two cells
  
  
• Positive terminals of two cells of emf's E₁ and E₂( whose emf are to be compared ) are connected to the terminals A and negative terminals are connected to jockey through a two way key K₂ and a galvanometer
• Now first key K₁ is closed to establish a potential difference between the terminals A and B then by closing key Ksub>2 introduce cell of EMF E₁ in the circuit and null point junction J₁ is dtermined with the help of jockey.If the null point on wire is at length
  l₁ from A then
  E₁=Kl₁
  Where K -> Potential gradient along the length of wire
• Similarly cell having emf E₂ is introduced in the circuit and again null point J₂ is determined .If length of this null point from
  A is l₂ then
  E₂=Kl₂
  Therefore
  E₁/E₂=l₁/l₂
  This simple relation allows us to find the ratio of E₁/E₂
• if the EMF of one cell is known then the EMF of other cell can be known easily

(B) Determination of internal resistance of the cell

• Potentiometer can also be used to determine the internal resistance of a cell
  
  
• For this a cell whose internal resistance is to be determined is connected to terminal A of the potentiometer across a resistance box through a key K₂
• First close the key K₁ and obtain the null point .Let l₁ be the length of this null point from terminal A then
  E=Kl₁
• When key K₂ is closed ,the cell sends current through resistance Box (R).If E₂ is the terminal
  Potential difference and null point is obtained at length l₂(AJ₂) then
  V=Kl₂
  Thus
  E/V=l₁/l₂
  But E=I(R+ r) and V=IR
  This gives
  E/V=(r+R)/R
  So (r+R)/R=l₁/l₂
  giving
  r=R(l₁/l₂-1)
• Using above equation we can find internal resistance of any given cell

Physics 12th Electric Current ,Resistance and Resistivity

Electric Current ,Resistance and Resistivity

(1) Introduction

• In our previous few chapters of electrostatics,we have discussed various terms and characterstics related to charge at rest
• Now in this chapter we will study about the moving charges,phenomenom related to them and various effects related to charge in motion
• Consider two metallic conducting balls charged at different potential are hanged using a non conducting insulating wires .Since air is an insulator ,no charge transfer takes place
• Now if we join both the metallic wire using a conducting metallic wire then charge will flow from metallic ball at higher potential to the one at lower potential.
• This flow of charge will stop when the two balls would be at the same potentials.
• If somehow we could maintian the potential between the metallic balls,we will get constant flow of the charge in metallic wire,connecting the two conducting balls
• This flow of charge in metallic wire due to the potential dicference between two conducters used is called electric current about which we would be dicussing in this chapter.

(2)Electric current and Current density

Electric Current

• We already had a brief idea about the electric current which wed defined as the state of motion of the electric charge .Now we are going to study about the electric current in details
• Quantitatively electric current is defined as the time rate of flow of the net charge of the area of crosssection of the conducter i.e
  Electric current = Total charge flowing / time taken
• if q is the amount of charge flowing through the conducter in t sec,The current through the conducter is given by
  I=q/t                          (1)
• SI unit of the current is Ampere(A) named so in the honour of french scientist Andee marie Ampere(1775-1836).Now,
  1 Ampere= 1 Coulumb/ 1 sec=1 Cs^-1
• Thus current through any conducter is said to be 1 ampere,if 1 C of charge is flowing through the conducter in 1 sec
• Small amount of currents are accordingly expressed in milliamperes (1mA=10^-3 A) or in micro ampere (1 mA=10^-6 A)
• Direction of electric current is in the direction of the flow of positive charged carriers and this current is known as conventional current.
• Direction of the flow of electron in conductor gives the direction of electronic current. Direction of conventional current is opposite to that of electronic current
• Electric current is a scalar quantity .Although electric current represent the direction of the flow of positive charged carrier in the conductor,still current is treated as scalar quantity as current in wires in a circuit does not follows the laws of vector addition

Current density

• The current density at a point in the conductor is defined as the current per unit cross-section area.Thus if the charge is flowing per unit time uniformaly over the area of crosss-section A of the conductor,then current density J at any point on that area is defined as
  J=I/A -(2)
• It is the characterstic property of point inside the conductor nor of the conductor as a whole
• Direction of current density is same as the direction of conventional current
• Note that current density is a vector quantity unlike electric current
• Unit of current density is Ampere/meter² (Am^-2)

(3) Drift Velocity

• Metallic conductors have large numbers of elctrons free to move about.These elctrons which are free to move are called conduction electrons
• Thus valence electrons of atom becomes the conduction electrons of the metals
• At room temperature,these conduction electrons moves randomly inside the conductor more or less like a gas molecule
• During motion,these conduction electrons collide with ions(remaining positive charged atom after the valence electrons move away) again and again and there direction of motion changes after each and every collision.
• As a results of these collisions atoms moves in a zig-zag path
• Since in a conductor there are large number of elctrons moving randomly inside the conductor.Hence they have not net motion in any particular direction.Since the number of electrons crossing an imaginary area ΔA from left to right inside the conductor very nearly equals the number of electron crossing the same area element from right to left in a given interval of time leaving flow of electric current through that area nearly equals to zero
• Now when we applied some P.D using a battery across the two ends of the conductor,then an electric field sets up inside the conductor
• As a result of this electric field setup inside the conductor,conduction electron expeirence a force in direction opposite to electric field and this force accelerates the motions of the electrons
• As a result of this accelerated motion electrons drifts slowly along the length of the conductor towards the end at higher potential
• Due to this acceleration velocity of electron's increases only for short interval of time as each acclerated electrons suffers frequent collision with positive ions and looses their Kinectic energy
• After each collision electrons starts fresh in random direction ,again get accelerated and loose their gained Kinetic energy in another collision
• This extra velocity gained by the electrons is lost in subsequent collison and the processes continued till the electron reach positive end of the conductor
• Under the effect of electric field inside the conductor ,free electrons have random thermal velocities due to the room tmeperature and small velocities with which they drift towards the positive end of the conductor.
• if τ is the average time between two successive collisions and E is the strength of applied electric field then force on electron due to applied electric field is
  F=eE
  Where e is the amount of charge on electron
• if m is the mass of electron ,then acceleration produced is given by
  a=eE/m
• Since electron is acclerated for an average time interval τ,additional velocity acquired by the electron is
  v_d=aτ
  or v_d=(eE/m)τ                    (3)
  This small velocity imposed on the random motion of electrons in a conductor on the application of electric field is known as drift velocity
• This drift velocity is defined as the velocity with which free electrons gets drifted towards the positive end of the conductor under the influence of externally applied electric field

(4) Relation between drift velocity and electric current

• Consider a conducting wire of lenghth L and having uniform cross-section area A in which electric field is present
  
  
  
• Consider in the wire that there are n free electrons per unit volume moving with the drift velocity v_d
• In the time interval Δt each electron advances by a distance v_dΔt and volume of this portion is Av_dΔt and no of free electron in this portion is nAv_dΔt and all these electrons crosses the area A in time Δt
• Hence charge crossing the area in time Δt is
  ΔQ=neAv_dΔt 
  or
  I=ΔQ/Δt =neAv_d                    (4)
  This is the relation between the electric current and drift velocity
• If the moving charge carriers are positive rather than negative then electric field force on charge carriers would be in a direction of electric fields direction and drift velocity would be in left to right direction opposite to what shown in fig-1
• In terms of drift velocity current density is given as
  j=I/A=nev_d                       (5)

(5) Ohm's Law and Resistance

• Ohm's law is the relation between the potential difference applied to the ends of the conductor and current flowing through the conductor.This law was expressed by George Simon Ohm in 1826
• Statement of Ohm's Law
  'if the physical state of the conductor (Temperature and mechanical strain etc) remains unchanged ,then current flowing through a conductor is always ditectly proportional to the potential difference across the two ends of the conductor
  Mathematically
  V α I
  or
  V=IR                    (6)
  Where constant of proportionallity R is called the electric resistance or simply resistance of the conductor
• Value of resistance depends upon the nature ,dimension and physically dimensions of the conductor
• Ohm's Law can be deducted using drift velocity relation as given in equation -3 .Thus from the equation
  v_d=(eE/m)τ 
  but Now E=V/l
  Therfore
  v_d=(eV/ml)τ 
  Also I=neAv_d
  Substituting the value of v_d in I relation
  I=(ne²Aτ/ml) V                    (7)
  or V/I=(ml/ne²Aτ)=R a constant for a given conductor
  Thus
  V=IR
  Mathematical expression of Ohm's Law
  From Ohm's Law
  V=IR or R=V/I                    (8)
  Thus electric resitance is the ratio of potential difference across the two ends of conductor and amount of current flowing through the conductor
• electric resistance of a conductor is the obstraction offered by the conductor to the flow of the current through it.
• SI unit of resistance is ohm (Ω) where
  1 Ohm=1 volt/1 Ampere
  or 1Ω=1VA^-1
• Dimension of resistance is [ML²T^-3A^-2]

(6) Resistivity and conductivity

• In terms of drift velocity ,electric current flowing through a conducting wire of length L and uniform area of cross-section A
  is
  I=dQ/dt =neAv_d=(ne²Aτ/ml) V
  The above can be rearranged to give the ohm's law i.e,
  V=IR
  where R=(ml/ne²Aτ) Now R=ρl/A                    (9)
  Where ρ is called the specific resistance or resistivity of the conductor
  And ρ=m/ne²τ                    (10)
• From equation (9) ,we can see that resistance of the wire is proportional to its length and inversly proportional to its cross-sectional area.
• Thus resistance of a long and thin wire will greater then the resistance of short and thick wire of the same material
• Now from equation (9)
  R=ρl/A                    (11)
  And from ohm law R=V/I
  Therefore
  ρ=(V/I)(A/L)
     =(V/L) / (I/A)
    =E/J                    (12)
  Where E=V/L is the electric field at any point inside the wire and J=I/A is current density at any point in the wire. Unit of resistivity is ohm-meter.
• Thus from equation (12) ,electric resistivity can also be defined as the ratio of electric field intensity at any point in the conductor and the current density at that point.
• The greater the resistivity of the material ,greater would be the field needed to establish a given current densisty
• Perfect conductor have zero resistivities and for perfect insulators resistivity would be infinite
• Metals and alloys have lowest resistivities and insulators have high resistivities and exceeds those of metals by a factor of 10²²
• The reciprocal of resistivity is called conductivity and is represented by σ
• Unit of conductivity is ohm^-1meter^-1(Ω^-1m^-1) and
  σ is defined as
  σ=1/ρ
  Since ρ=E/J
  or σ=J/E
  or J=σE                    (13a)
• The above relation can also be written in vector form as both J and E are vector quatities where vector Jbeing directed towards E
  J=σE                    (13b)

(7) variation of resistivity with temperature

• Resistance and hence resistivity of conductor depends on mubers of factors
• One of the most important factors is dependence of resistance of metals on temperature
• Resistivity of the metallic conductor increases with increase on temperature 
• when we increase the temperature of the metallic conductor,its constituent atoms vibrate with greater amplitudes then usual.This results
  to the more frequent collison between ions and electrons
• As a result average time between the two successive collision decreases resulting the decrease in drift velocity
• Thus increase collison with the increase in tempearture results in increase resistivity
• For small temperature variations ,resistivity of the most of the metals varies according to the following relations
  ρ(T)=ρ(T₀)[1 + α(T-T₀)]                    (14)
  Where ρ(T) and ρ(T₀) are the resistivies of the material at temperature T and T₀ respectively and α is the constant for given materail and is known as coefficient of resistivity.
• Since resistance of a given conductors depends on the length and crosssectional area of the conductor through the relation
  R=ρl/A
  Hence temperature variation of the resistance can be given as
  R=R(T₀)[1 + α(T-T₀)]                    (15)
• Resistivity of alloys also increase with temperature but this increase is much small as compared to metals
• Resistivies of the non-metals decreases with increase in temperature .This is because at high temperature more electrons becomes avialable for conduction as they set themselves loose from atoms and hence temperature coefficient of resistivity is negative for non-metals
• A similar behavior occurs in case of semi-conductors .temperature coefficient of resistivity is negative for semi-conductors and its value is often large for a semi-conductor materials

(8) Current Voltage relations

• We know that current through any electrical device such as resistors depends on potential defference between the terminals
• Devices obeying ohm's law follow a linear relationship between current following and potential applied where current is directly proportional to voltage applied .Graphical relation between V and I is shown below in figure
  
  
  
• Graph for a resistor obeying ohm's law is a straight line through the origin having some finite slope
• There are many electrical devices that does not obey the ohm's law and current may depends on voltage in more complicated ways.Such devices are called non-ohmic devices for examples vaccum tubes,semiconductor diodes ,transistors etc
• Consider the case of a semi conductor junction diode which are used to convert alternating current to direct current and are used to perform variety of logic functions is a non=ohmic device
• Graphical voltage relation for a diode is shown below in the figure
  
  
• Figure clearly shows a non linear depeence of current on voltage and diode clearly does not follow the ohm's w
• When a device does not follow obey ohm's law,it has non linear voltage -current relation and the quantity V/I is no longer a constant however ratio is still known as resistance which now varies with current
• In such cases we define a quantity dV/dI known as dynamic resistance which expresses the relation between smaal change in current and resulting change in voltage
• Thus for non-ohmic electrical devices resistance is not constant for different values of V and I

(9) Colour code of carbon resistors

• Commercially resistors of different type and values are avilable in the market but in electronic circuits carbon resistors are more frequently used
• In carbon resistors value of resistance is indicated by four coloured bands marked on its surface as shown below in figure
  
  
• The first three bands a,b.c determine the value of the resistance and fourth band d gives the tolerance of the resistance
• The colour of the first and second band respectively gives the first and second significant figure of the resistance and third band c gives the power of the ten by which two significant digits are multiplied for obtainng the value of the resistance
• value of different colurs for making bands in carbon resistors are given below in the table
  
  

  Colur	Figure(first and second band)	Multiplier(for third band)	tolerance
  Black	0	1	-
  Brown	1	10	-
  Red	2	102	-
  Orange	3	103	-
  Yellow	4	104	-
  Green	5	105	-
  Blue	6	106	-
  Violet	7	107	-
  Gray	8	108	-
  white	9	109	-
  Gold	-	10-1	5%
  Silver	-	10-2	10%
  no Colour	-	-	20%

• For example in a given resistor let first strip be brown ,second strip be red and third be orange and fourth be gold then resitance of the resitor would be
  12X 10³ +/- 5% 

(10)Combination of Resistors

• We have earlier studied that sevral capacitors can be connected in series or parallel combination to form a network. In same way sevral resistor may be combined to form a network.
• Just like capacitors resistors can be grouped in series and parallel.
• Equivalent resistance of the combination of any number of resistors is a single resistance which draw same current as the combination of different resistances draw when the same potential difference is applied across it.
  
  (A) Resistors in Series
• Resisors are said to be connected in series combinaton. If same current flows through each resistor when same potential difference is applied across the combination.
• Consider the figure given below
  
  
  
• In figure given above three resistors if resistance R₁, R₂ and R₃ are connected ibn series combination.
• If battery is connected across the series combination so as to mintain potential difference V between points A and B, the current I would pass through each resistor.
• If V₁, V₂ andV₃ is the potential difference across each resistor R₁, R₂ and R₃ respectively, then according to Ohm's Law,
  V₁=IR₁
  V₂=IR₂
  V₃=IR₃
  Since in series combination current remains same but potential is divieded so,
  V=V₁+V₂+V₃
  or, V=I(R₁+R₂+R₃)
  If R_eqis the resistance equivalent to the series combination of R₁, R₂ and R₃ then ,
  V=IR_eq
  where, R_eq=R₁+R₂+R₃
• Thus when the resistors are connected in series, equivalent resistance of the series combination is equal to the sum of individual resistances.
• Value of esistance of the series combination is always greater then the value of largest individual resisnces.
• For n numbers of resistors connected in series equivalent resistance would be
  R_eq=R₁+R₂+R₃+...........................+R_n
  
  (B) Resistors in parallel
• Resistors are said to be connected in parallel combination if potential difference across each resistors is same.
• Thus , in parallel combination of resistors potential remains the same but current is divided.
• Consider the figure given below
  
  
• Battery B is connected across parallel combination of resistors so as to maintain potential difference V across each resistors.Then total current in the circuit would be
  I=I₁+I₂+I₃                           (16)
• Since potential difference across each resistors is V. Therefore, on applying Ohm's Law
  V=I₁R₁=I₂R₂=I₃R₃
  or,
  
  
  From equation (16)
  
   
• If R <sub>is the equivalent resistance of parallel combination of three resistors heaving resistances R1, R2 and R3 then from Ohm's Law
  V=IReq
  or,
  
  Comparing equation (16) and (17) we get</sub>
  
   
• For resistors connected in parallel combination reciprocal of equivalent resistance is equal to the sum of reciprocal of individual resistances.
• Value of equivalent resistances for capacitors connected in parallel combination is always less then the value of the smallest resistance in circuit.
• If there are n number of resistances connected in parallel combination, then quivalent resistance would be reciprocal of
  

Physics 12th Capacitance

Capacitance

1. Introduction

• A capacitor (formerly known as condenser) is a device that can store electronic charge and energy.
• All capacitors consists of a combination of two conductors separated by an insulator.
• The insulator is called dielectric which could be oil, air or paper and many more such materials are there wich can act as a dielectric medium between conducting plates of a capacitor.
• Figure 1 below shows the symbol used to represent a capacitor.
  
  
• Now plates of the capacitor are connected to the terminals of a battery, shown below in figure 2, in order to charge it's conducting plates.
  
  
• As soon as capacitor is connected to the battery , charge is transferred from one conductor to another.
• Plate connected to positive terminal of the battery becomes positively charged with charge +Q in it and plate connected to negative terminal of the battery becomes negatively charged with cahrge -Q on it i.e. both plates have equal amount of opposite charge .
• Once the capacitor is fully charged potential difference between the conductors due to their equal and opposite charges becomes equal to the potential difference between the battery terminals.
• For a given capacitor Q∝V and the ratio Q/V is constant for a capacitor.
  Thus,
            Q=CV                              (1)
  where the proportionality constant C is called the capacitance of the capacitor.
• Capacitance of any capacitor depends on shape , size and geometrical arrangement of the conductors.
• When Q is in coulumbs (C) and V is in volts(V) then the S.I. unit of capacitance is in farads(F) where
            1F=1 coulumb/volt
• One farad is the capacitance of very large capacitor and it's submultiples such as microfarad(1μF=10^-6) or picofarad(1pF=10^-12) are generally used for practical applications.

2. Calculation of capacitance

• For calculating capacitance of a capacitor first we need to find the potential difference between it's two conducting plates having charge +Q and -Q.
• For simple arrangements of conductors like two equivalent parallel plates kept at distance d apart or two concentric conducting spheres etc., potential difference can be found first by calculating electric field from gauss's law or by Coulumb's law.                      
• After calculating electric field , potential difference can be found by integrating ellectric field using the relation
            V_a-V_b=∫E.dr
  where the limits of integration goes from a to b.
• Once we know the potential difference between two conductors of the capacitor , it's capacitance can be calculated from the relation
            C=Q/V                              (2)
• Calculation of capacitance of some simple arrangements would be illustrated in following few articles.

3. Parallel plate capacitor

• A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance shown below in the figure 3.
  
• Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of each plate then surface charge density on each plate is
            σ=Q/A
• We have already calculated field between two oppositely charged plates using gauss's law which is
            E=σ/ε₀=Q/ε₀A
  and in this result effects near the edges of the plates have been neglected.
• Since electric field between the plates is uniform the potential difference between the plates is
            V=Ed=Qd/ε₀A 
  where , d is the separation between the plates.
• Thus, capacitance of parallel plate capacitor in vacuum is
            C=Q/V=ε₀A/d               (3)
• From equation 3 we see that quantities on which capacitance of parallel plate capacitor depends i.e.,ε₀ , A and d are all constants for a capacitor.
• Thus we see that in this case capacitance is independent of charge on the capacitor but depends on area of it's plates and separation distance between the plates.

4.Cylinderical capacitor

• A cylinderical capacitor is made up of a conducting cylinder or wire of radius a surrounded by another concentric cylinderical shel of radius b (b>a).
• Let L be the length of both the cylinders and charge on inner cylender is +Q and charge on outer cylinder is -Q.
• For calculate electric field between the conductors using Gauss's law consider a gaussian consider a gaussian surface of radius r and length L¹ as shown in figure 4.
  
  
• According to Gauss's law flux through this surface is q/ε₀ where q is net charge inside this surface.
• We know that electric flux is given by
            φ=E.A
              =EAcosθ  
              =EA
  since electric field is constant in magnitude on the gaussian surface and is perpandicular to this surface. Thus,
             φ=E(2πrL)
  since            φ=q/ε₀
  =>           E(2πrL)=(λL)/ε₀
  where λ = Q/L = charge per unit length
  =>           

  E =	λ
  	2πε0r

  or, 

  E =	λ
  	2πε0r

                                (4)
• If potential at inner cylinder is V_a and V_b is potential of outer cylinder then potential difference between both the cylinders is
            V=V_a and           V_b=∫Edr
  where limits of integration goes from a to b.
• Potential of inner conductor is greater then that of outer conductor because inner cylinder carries positive charge. Thus potential difference is

  V =	Qln(b/a)
  	2πε0L

• Thus capacitance of cylinderical capacitor is
            C=Q/V
  or, 

  C =	2πε0L Qln(b/a)
  	ln(b/a)

                                     (5)
• From equation 5 it can easily be concluded that capacitance of a cylinderical capacitor depends on length of cylinders.
• More is the length of cylinders , more charge could be stored on the capacitor for a given potential difference.

5. Spherical capacitor

• A spherical capacitor consists of a solid or hollow spherical conductor of radius a , surrounded by another hollow concentric spherical of radius b shown below in figure 5
  
• Let +Q be the charge given to the inner sphere and -Q be the charge given to the outer sphere.
• The field at any point between conductors is same as that of point charge Q at the origin and charge on outer shell does not contribute to the field inside it.
• Thus electric field between conductors is 

  E =	Q
  	4πε0r2

            
• Potential difference between two conductors is
       V=V_a-V_b
       =-∫E.dr
  where limits of integration goes from a to b.
  On integrating we get potential difference between to conductors as
  

  V =	Q(b-a)
  	4πε0ab

• Now , capacitance of spherical conductor is
       C=Q/V
  or, 

  C =	4πε0ab
  	(b-a)

                                (6)
• again if radius of outer conductor aproaches to infinity then from equation 6 we have
       C=4πε₀a          (7)
• Equation 7 gives the capacitance of single isolated sphere of radius a.
• Thus capacitance of isolated spherical conductor is proportional to its radius.

6. Capacitors in series and parallel combinations

For prectical applications , two or more capacitors are often used in combination and their total capacitance C must be known.To find total capacitance of the arrangement of capacitor we would use equation
     Q=CV
(i) Parallel combination of capacitors

• Figure below shows two capacitors connected in parallel between two points A and B
  
• Right hand side plate of capacitors would be at same common potential V_A. Similarly left hand side plates of capacitors would also be at same common potential V_B.
• Thus in this case potential difference V_AB=V_A-V_B would be same for both the capacitors, and charges Q₁ and Q₂ on both the capacitors are not necessarily equal. So,
       Q₁=C₁V and Q₂=C₂V
• Thus charge stored is divided amongst both the capacitors in direct proportion to their capacitance.
• Total charge on both the capacitors is,
       Q=Q₁+Q₂
        =V(C₁+C₂)  
  and
       Q/V=C₁+C₂                              (8)
  So system is equivalent to a single capacitor of capacitance
       C=Q/V
  where,
• When capacitors are connected in parallel their resultant capacitance C is the sum of their individual capacitances.
• The value of equivalent capacitance of system is greater then the greatest individual one.
• If there are number of capacitors connected in parallel then their equivalent capacitance would be
       C=C₁+C₂+ C₃...........               (10)
  (ii) Series combination of capacitors
• Figure 7 below shows two capacitors connected in series combination between points A and B.
  
• Both the points A and B are maintained at constant potential difference V_AB.
• In series combination of capacitors right hand plate of first capacitor is connected to left hand plate of next capacitor and combination may be extended foe any number of capacitors.
• In series combination of capacitors all the capacitors would have same charge.
• Now potential difference across individual capacitors are given by
       V_AR=Q/C₁
  and,
       V_RB=Q/C₂
• Sum of V_AR and V_RB would be equal to applied potential difference V so,
       V=V_AB=V_AR+V_RB
        =Q(1/C₁ + 1/C₂)
  or,
  
  where
  
  i.e., resultant capacitance of series combination C=Q/V, is the ratio of charge to total potential difference across the two capacitors connected in series.
• So, from equation 12 we say that to find resultant capacitance of capacitors connected in series, we need to add reciprocals of their individual capacitances and C is always less then the smallest individual capacitance.
• Result in equation 12 can be summarized for any number of capacitors i.e.,
  

7. Energy stored in a capacitor

• Consider a capacitor of capacitance C, completely uncharged in the begning.
• Charhing process of capacitor requires expanditure of energy because while charging a capacitor charge is transferred from plate at lower potential to plate at higher potential.
• Now if we start charging capacitor by transporting a charge dQ from negative plate ti the positive plate then work is done against the potential difference across the plate.
• If q is the amount of charge on the capacitor at any stage of charging process and φ is the potential difference across the plates of capacitor then magnitude of potential difference is φ=q/C.
• Now work dW required to transfer dq is
  dW=φdq=qdq/C
• To charge the capacitor starting from the uncharged state to some final charge Q work required is
  Integrating from 0 to Q
       W=(1/C)∫qdq
       =(Q²)/2C      (14a)
       =(CV²)/2
       =QV/2 
  
  Which is the energy stored in the capacitor and can also be written as
       U=(CV²)/2 ---(15)
• From equation 14c,we see that the total work done is equal to the average potential V/2 during the charging process ,multiplied by the total charge transferred
• If C is measured in Farads ,Q in coulumbs and V in volts the energy stored would in Joules
• A parallel plate capacitor of area A and seperation d has capacitance
  
  C=ε₀A/d
• electric field in the space between the plates is
  E=V/d or V=Ed
  
  Putting above values of V and C in equation 14b we find
  W=U=(1/2)(ε₀A/d)(Ed)²
  =(1/2)ε₀E²(Ad)
  =(1/2)ε₀E².V ---(16)
• If u denotes the energy per unit volume or energy density then
  u=(1/2)ε₀E² x volume
• The result for above equation is generally valid even for electrostatic field that is not constant in space.

8. Effect of Dielectric

• Dielectric are non conducting materials for ex- Glass,mica,wood etc.
• What happened when space between the two plates of the capacitor is filled by a dielectric was first discovered by faraday.
• Faraday discovered that if the space between conductors of the capacitor is occupied by the dielectric,the capacitance of capacitor is increased.
• If the dielectric completely fills the space between the conductors of the capacitor ,the capacitance is increased by an factor K which is characterstics of the dielectric and This factor is known as the dielectric constan.
• Dielectric constant of vaccum is unity.
• Consider a capacitor of capacitance C₀ is being charged by the connecting it to a battery.
• If Q₀ is the amount of charged on the capacitor at the end of the charging and V₀ is potental diffrence across the plates of the capacitor then
  C₀=Q₀ /V₀ ----(17)
  
  Thus charge being placed on the capacitor is
  Q₀=C₀V₀
• If the battery is diconnected and space between the capacitor is filled by a dielectric the P.D decrease to a new value
  V=V₀/K.
• Since the original charge is still on the capacitor,the new capacitance will be
  C=Q₀/V=KQ₀/V₀=KC₀----(19)
• From equation 19 it follows that C is greater then C₀.
• Again if the dielectric is inserted while the battery is still connected then battery would have to supply some amount of charge to maintain the P.D between the plates and then total charge on the plates would be Q=KQ₀.
• In either of the cases ,capacitance of the capacitor is increase by the amount K.
• For a parallel plate capacitor with dielectric of dielectric constant K between its plates its capacitance becomes
  
  C=εA/D ----(20)
  where ε=Kε₀
• When a sufficiently strong electric field is applied to any dielectric material it becomes a conductor and this phenomenon is known as dielectric breakdown.
• The maximun electric field a material can withstand without the occurence of breakdown is called dielectric strength of that material.
• Thus field across the capacitor should never exceed breakdown limits in order to store charge on capacitor without leaking.

Physics 12th Electric Potential

Electric Potential

1. Introduction

• We already have an introduction of work and energy while studying mechanics.
• We know that central forces are conservative in nature i.e., work done on any particle moving under the influence of consrevative forces does not depend on path taken by the particle but depends on initial and final positions of the particle.
• Electrostatic force given by Coulumb's law is also a central force like gravitational force and is conservative in nature.
• For conservative forces, work done on particle undergoing displacement can be expressed in terms of potential energy function.
• In this chapter we will apply work and energy considerations to the electric field and would develop the concept of electric potential energy and electric potential.

2. Electric potential energy

• Conside a system of two point charges in which positive test charge q' moves in the field produced by stationary point charge q shown below in the figure.
  
  
  
• Charge q is fixed at point P and is displaced from point R to S along a radial line PRS shown in the figure.
• Let r₁ be the distance between points P and R and r₂ be the distance between P and S.
• Magnitude of force on positive test charge as given by Coulumb's law is
  
   
• If q' moves towards S through a small displacement dr then work done by this force in making the small displacement dr is
  dW=F·dr
  
   
• Total work done by this force as test charge moves from point R to S i.e., from r₁ to r₂ is,
  
  
  or
  
   
• Thus for this particular path work done on test charge q' depends on end points not on the path taken.
• Work done W in moving the test charge q' from point R to S is equal to the change in potential energy in moving the test charge q' from point R to S. Thus,
  W=U(r₁)-U(r₂)                                          (4)
  where
  
  is the potential energy of test charge q' when it is at point R and
  
  
  is the potential energy of test charge q' when it is at point S.
• Thus potential energy of test charge q' at any distance r from charge q is given by
  
  
  Equation 5 gives the electric potential energy of a pair of charges which depends on the separation between the charges not on the location of charged particles.
• If we bring the test charge q' from a very large distance such that r₂=∞ to some distance r₁=r then we must do work against electric forces which is equal to increase in potential energy as given by equation 5.
• Thus potential energy of a test charge at any point in the electric field is the work done against the electric forces to bring the charge from infinity to point under consideration.

3. Electric Potential

• We now move towards the electric potential which is potential energy per unit charge.
• Thus electrostatic potential at any point of an electric field is defined as potential energy per unit charge at that point.
• Electric potential is represented by letter V.
  V=U/q' or U=q'V                                             (6)
• Electric potential is a scalar quantity since both charge and potential energy are scalar quantities.
• S.I. unit of electric potential is Volt which is equal to Joule per Coulumb. Thus,
  1 Volt = 1 JC^-1
• In equation 4 if we divide both sides by q' we have
  
  
  where V(r₁) is the potential energy per unit charge at point R and V2) is potential energy per unit charge at point S and are known as potential at points R and S respectively.
• Again consider figure 1. If point S in figure 1 would be at infinity then from equation 7
  
  
  Since potential energy at infinity is zero therefore V(∞)=0. Therefore
  
  
  hence electric potential at a point in an electric field is the ratio of work done in bringing test charge from infinity to that point to the magnitude of test charge.
• Dimensions of electric potential are [ML²T^-3A^-1] and can be calculated easily using the concepts of dimension analysis. 

4. Electric potential due to a point charge

• Consider a positive test charge +q is placed at point O shown below in the figure.
  
  
  
• We have to find the electric potential at point P at a distance r from point O.
• If we move a positive test charge q' from infinity to point P then change in electric potential energy would be
  
   
• Electric potential at point P is
  
   
• Potential V at any point due to arbitrary collection of point charges is given by
  
   
• here we see that like electric field potential at any point independent of test charge used to define it.
• For continous charge distributions summation in above expressin will be replaced by the integration
  
  
  where dq is the differential element of charge distribution and r is its distance from the point at which V is to be calculated.

5. Relation between electric fiels and electric potential

• Consider the electric field E due to a point charge +q at point O in a radially outward direction shown below in the figure.
  
  
  
• Suppose R and S are two points at a distance r and r+dr from point O where dr is vanishingly small distance and V is electric potential at point R.
• Now force on any test charge q' at point R in terms of electric field is
  F=q'E
• Work done by the force in displacing test charge from R to S in field of charge q is
  dW = F·dr = q'E·dr
  and, change in potential energy is
  dU = -dW = -q'E·dr
  Change in electric potential would be
  dV = dU/q
  or dV = -E·dr                                             (11)
• From equation 11 electric field is
  E=-(dV/dr)                                                    (12)
  the quantity dV/dr is the rate of change of potential with the distance and is known as potential gradient. Negative sign in equation 12 indicates the decrease in electric potential in the direction of electric field.
• For cartesian coordinate system
  E=E_xi+E_yj+E_zk
  and,
  dr=dxi+dyj+dzk
  from equation 11
  dV=-E⋅dr
  or, dV=-(E_xdx+E_ydy+E_zdz)          (13)
• Thus components of E are related to corresponding derivatives of V in the following manner
  E_x=dV/dx                                          (14a)
  E_y=dV/dy                                          (14b)
  E_z=dV/dz                                          (14c)
  In equation (14a) we see that V is differentiated with respect to coordinate x keeping other coordinates constant. Same is the case with equations (14b) and (14c) in case of y and z coordinates respectively.

6. Equipotential surfaces

• Surface over which the electric potential is same everywhere is called an equipotential surface.
• Equipotential surfaces are graphical way to represent potential distribution in an electric field.
• We can draw equipotential surfaces through a space having electric field.
• For a positive charge , electric field would be in radially outward direction and the equipotential surfaces would be concentric spheres with centers at the charge as shown below in the figure.
  
  
  
• Since electric potential remains same everywhere on an equipotential surface from this it follows that PE of a charged body is same at all points on this surface.This shows that work done in moving a charged body between two points on an equipotential surface would be zero.
• At every point on equipotential surface electric field lines are perpandicular to the surface. This is because potential gradient along any direction parallel to the surface is zero i.e.,
  E=-dV/dr=0
  so component electric parallel to equipotential surface is zero.

7. Potential due to an electric dipole

• We already know that electric dipole is an arrangement which consists of two equal and opposite charges +q and -q separated by a small distance 2a.
• Electric dipole moment is represented by a vector p of magnitude 2qa and this vector points in direction from -q to +q. 
• To find electric potential due to a dipole consider charge -q is placed at point P and charge +q is placed at point Q as shown below in the figure.
  
  
  
• Since electric potential obeys superposition principle so potential due to electric dipole as a whole would be sum of potential due to both the charges +q and -q. Thus
  
  
  where r₁ and r₂ respectively are distance of charge +q and -q from point R.
• Now draw line PC perpandicular to RO and line QD perpandicular to RO as shown in figure. From triangle POC
  cosθ=OC/OP = OC/a
  therefore OC=acosθ similarly OD=acosθ
  Now ,
  r₁ = QR≅RD = OR-OD = r-acosθ
  r₂ = PR≅RC = OR+OC = r+acosθ
  
  
  since magnitude of dipole is
  |p| = 2qa
  
   
• If we consider the case where r>>a then
  
  
  again since pcosθ= p·rˆ where, rˆ is the unit vector along the vector OR then electric potential of dipole is
  
  
  for r>>a
• From above equation we can see that potential due to electric dipole is inversly proportional to r² not ad 1/r which is the case for potential due to single charge.
• Potential due to electric dipole does not only depends on r but also depends on angle between position vector r and dipole moment p.

8. Work done in rotating an electric dipole in an electric field

• Consider a dipole placed in a uniform electric field and it is in equilibrium position. If we rotate this dipole from its equllibrium position , work has to be done.
• Suppose electric dipole of moment p is rotated in uniform electric field E through an angle θ from its equilibrium position. Due to this rotation couple acting on dipole changes.
• If at any instant dipole makes an angle φ with uniform electric field then torque acting on dipole is
  Γ=pEsinφ                                                    (19)
  again work done in rotating this dipole through an infitesimaly small angle dφ is
  dW=torque x angular displacement
  =pEsinφdφ 
• Total work done in rotating the dipole through an angle θfrom its equilibrium position is
  
  
  This is the required formula for work done in rotating an electric dipole placed in uniform electric field through an angle θ from its equilibrium position.

9.Potential energy of dipole placed in uniform electric field

• Again consider equation 20 which gives the work done in rotating electric dipole through an infinetesimly small angle dφ is
  dW=pEsinφdφ
  which is equal to the change in potential energy of the system
  dW=dU=pEsinφdφ                                                    (22)
• If angle dφ is changed from 90⁰ to θ then in potential energy would be
  
   
• We have choosen the value of φgoing from π/2 to θ because at π/2 we can take potential energy to be zero (axis of dipole is perpandicular to the field). Thus U(90⁰)=0 and above equation becomes