Physics for Class XI

T 3.2 ELECTRIC POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL We know that a free positive charge, in an electric field, tends to move along the direction of field, so when a positive test

T 3.2 ELECTRIC POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL

We know that a free positive charge, in an electric field, tends to move along the direction of field,

so when a positive test charge is brought opposite to the direction of electric field, work is done against

the Coulomb’s force of repulsion The wwork done by external force in ———————————>

carrying per unit positive test charge from one point to another in an

electric field, always keeping the test charge in dynamic equilibrium,

is called the electric potential difference between those two points If me

Wis the work done in carrying positive test charge 4g from point B to x 8

A against the direction of electric field, the potential difference ——————— —»

‘As work Waa and charge qp are scalars, therefore, potential difference is also a scalar quantity

Unit of Potential difference : The unit of work W is joule and the unit of charge 4 is coulomb,

therefore, the unit of potential difference is joule/coulomb or volt

From equation (1)

1 volt=

1

That is, 1 volt is the potential difference between two given points in an electric field if the work

done in carrying 1 coulomb of charge from one point to another is 1 joule

1 volt = —tioule_ Teoulomb

Thatis, 1 volt is theelectric potential at a point in an electric field if the work done in bringing one coulomb of charge from infinity to that point is 1 joule, provided the charge of 1 coulomb does not affect the original electric field

The C.GS unit of electrostatic potential is erg/statcoulomb It is called statvolt (stv) Its relation with SI unit of potential is

Physical Significance of Electric Potential : We observe thata liquid always flows from higher

pressure to lower pressure (or from higher level to lower level), heat always flows from higher

temperature to lower temperature In a similar manner positive charge always flows from higher

potential to lower potential Just as there is no relation of direction of flow of liquid with the

quantity of liquid; and there is no relation of direction of flow of heat with the quantity of heat; in a

similar manner there is no relation of direction of flow of charge with the quantity of charge

The Earth as a Reference of Potential

It must be kept in mind that the potential difference

between two ee is more fundamental than the value of | _ 1: The reference level for zero potential is

absolute potential at individual point The value of absolute | ‘Ker (i)at infinity o (i) earth itself:

potential can be determined only if we arbitrarily choose the | 2 Jf @ conductor is earthed, its potential value of potential at some point to be zero In the definition | is taken as zero,

of potential, this reference point is taken as infinity Any

other point may also be chosen as reference point for zero potential For many purposes, the earth is

taken as reference point for zero potential The reason is that the earth is a good conductor and its size

is so large that any charge given to or taken from it does not at all alter its electrical conditions, so that the potential of earth always remains the same This may be compared with sea level which remains unaffected due to loss of water by evaporation or addition of water by rains

When earth is taken as reference for zero potential, a

body will be said to be at positive potential if there is flow

of positive charge from body to earth (or infact flow of

electrons from earth to body, on making the electrical

contact between the earth and the body), while a body will

be said at negative potential if there is flow of positive

charge from earth to the body (or infact the flow of

electrons from body to earth when electrical contact is

made between the two)

In both cases the flow of electrons continues until the potentials of body and earth are equalised, the common potential will be that of earth ie,, zero potential thus earthing a conductor makes the potential

of conductor zero

i 3.3 ELECTRIC POTENTIAL AS LINE INTEGRAL OF ELECTRIC FIELD

‘The electric potential may be defined as the negative of line integral of electric field This may be shown as follows :

Consider a region of electric field The field strength at any point is specified by E Let a positive test charge qa be displaced from point A to B, opposite to direction of electric field then external force on test

where integral extends along the path from A to B

Potential difference between two points A and B will be 8

If the point A is taken at infinity, the reference level for zero potential, i., Vq = 0, then potential at B,

Vp=-[2 Ear

Thus, the electric potential at any point in an electric field is defined as the negative of line integral of

electric field from infinity to given point

R 3.4 ELECTRIC FIELD IS CONSERVATIVE

{A conservative field is one in which work done is independent of pal th, cut Aaya corevate id defined th din whch werk dane

unchanged incomplete round tip, Consider the comy việc tp ACPDA T

electric ld in a complete round trip or closed eyele the inital'and final points A

are same, so the work done by electric field in carrying a charge q along closed 8

or Jf E-dr= f E-dr ~-@)

along ACB slong ADB

This implies that the line integral of electric field is same along two paths ACB and ADB between given initial and final points Thus, line integral of electric field is independent of path followed

between two given points

As potential difference between two points is defined in terms of line integral of electric field and line integral is independent of path; so potential difference between two points is independent of the path

Tiss catcutation oF ELECTRIC POTENTIAL

(i) Potential at a Point due to a Point Charge : Suppose a point charge + qis placed at point O The

electric potential, at point P at distance r from O is to be found Suppose an infinitesimal positive test

charge 4p is brought from infinity to point P When test charge is at point A at distance x from O, the electric force on test charge

4p _L 8b

a Rg _—==—=-

The work done by this force in displacing charge qy a small Fig 3.6(a)

distance dx closer to q will be

dW = external force x displacement

The work done is bringing charge q from infinity to point P will be the v„”

‘sum of all such small works; which is fund by integrating equation (1) from (4

Fig 3.6(b) shows the variation of electric potential and electric field with distance r from the point

coarge( Vt, B= ) -Coviouty, varie faster than V with distance r a 8 P đi) The electric potential due to a system of point charges : The

electric potential, being a scalar quantity, has no direction; therefore, in ⁄

order to find the electric potential at a given point due to a point charge, 5

the charge is substituted in the formula with its sign [fan electric field is /

due to a system of ‘n’ charges 4» 42 1 43 +» dy and their distances from % at given point Pare rrp , 13 , fy then net electric potential at P,

v-1%, 1h, #mgpn 4nepry 48t0 7; 1 8,

1 [8

olny (ii) The electric potential due to a continuos

charge distribution : Consider a continuous charge SIunit of potentiaL is volt Potential is a

distribution having charge q The charge distribution may | atrquantity and wo algebraically adit

be supposed to be formed of a large number of small grant aid sade eM charge elements Consider a charge element dq having

position vector r” relative to arbitrary origin O Let r” be the position vector of point P relative to O

‘Then the distance of point P from the charge element=|£ = r |

<The potential at P due to charge element dq

()_Ifthe charge distribution is linear charge of charge per unit

length (2), then charge on length element dl is dq = i dl

mei > 2

" Fig 3.8

where integration extends along the whole length

(ii) If the charge distribution is surface charge and charge per unit surface area is, then charge on

surface element dS is dq =6 dS, therefore

1 ods

any o>

integration extends over the whole surface

If the charge distribution is volume charge of volume charge densit

of volume dV is dq =p dV

1p then charge on element

1 pd Ane a>

ee er

where integration extends over whole volume.

Wms SS

Formulae Used: Electric potential, V "S

v =— ior a point charge

(i) Charge on nucleus isQ = + Ze

(ii) Electric potential is a scalar quantity, hence, algebraically additive

(iii) When two conductors are brought in contact, their electric potentials become equal

Ex 1 The work done in bringing a charge of 1-3 x 1077 coulomb from infinity to a point in am

electric field is 6-8 x 10” Š joule Find the electric potential at that point

Solution Key idea : Electric potential is work done in bringing unit charge from infinity to given point, ie,

electric potential, vet

Here W = 6-5 x 1075 joule, gg = 1-3 x 1077 coulomb

r5

+V= 6Š? — ~s00 volt

1:3x10

Ex 2 The electric field and electric potential at any point due to a point charge kept in air is

20 NC™ and 10 JC! respectively Compute the magnitude of this charge (CBSE 2006)

Solution Electric field, 14 - +4)

Amtg r

ụ 14 Electric potential, V=———4 oe

Ex 3 Calculate the potential at the centre of a square of q=~3I©

side {2 m which carries at its four corners charges of + 21 C,

+ lụC, — 3 Cand ~ 34 C respectively (CBSE 2000)

Solution Key idea : Electric potential is a scalar quantity Net

potential at a point is the algebraic sum of potentials due to

individual charges

Side of square = /2 m

Length of diagonal

‘AC = BD = v2)? + 3? = 2m

Distance of each charge from centre

r=OA =OB=OC =OD =1m

R 3.6 ELECTRIC POTENTIAL DUE TO AN ELECTRIC DIPOLE

Consider an electric dipole having charges =4 and +9 at P separation ‘2a’ The dipole moment of dipole is p = q(2),

directed from - qo + 9

‘The electric potential due to dipole is the algebraic sum of fe

potentials due to charges + qand ~ 4 wo

If n and rp are distances of point P from charge +4 and = c

respectively, then the potential due to electric dipole at point P,

Meg en âmSu|n rz As A®_ip

If(r,®) are polar coordinates of point P with respect to mid point “8 —*" a9"

and rR =r? +a? +2ar cosd (3) From (2) #«n|h->e9,s]

Substituting these values in (1), we get

Electric potential due to an electric dipole is

+ Electric potential due to an electric dipole is

This shows that the electric potential due to a dipole depends on distance r and also on the angle

between position vector 7 and dipole moment p The electrostatic potential at large distances falls off as

(Gi) When point P lies on the equatorial plane of the dipole, then

0=90

608 8 = cos 90° = 0 vx0

Ex 4 Two point charges 100 x 10” ° C and =100 x 10” ` € are placed at ä distanee 2-0 x 107 Ÿ m

and form an electric dipole Calculate the electric potential at a point distance 0-5 m from the centre of dipole if point is at (a) axial position and (b) broadside on position

Solution Here q = 100 x 1079 C,2a= 2-010" 3m

Similar charges repel and opposite charges attract Therefore, if charges are displaced farther or

nearer, some work is required This work is stored as the electric potential energy of the system of

charges The electric potential energy is assumed zero when charges are situated at infinite distance from one another Accordingly tie electric potential energy of a system of charges is defined as the work

that has to be done by external forces in bringing the charges of the system from infinity to their present positions without any acceleration

For Two charges : Let a system contain two charges q and qp at separation ‘r’ in air or vacuum

‘Suppose charge q; is at point A and charge 4p is at point B To find the electric potential energy it is assumed that initially charge qq is at infinite distance from charge 4, and q2 is brought from infinity to

point B

By definition of electric potential, the electric potential at point B due to charge 4,

= work doe in bringing unit charge from infinity to point B, distant fom

+ Work done in bringing charge ft from infinity to point B ——r——>

(charge q;) x (potential at B) Fig.3.11

If charges are of same sign, they repel each other Therefore to bring them closer, external work is done against the electric repulsive force, therefore potential energy of system increases, but if the charges are of opposite sign, they attract each other, therefore to bring them closer, the work is done by

the system itself, therefore, the potential energy of the system decreases

As electric potential energy is a scalar quantity, therefore to compute

the electric potential energy analytically, the charges q, and qy are

substituted with their sign

For a system containing more than two charges : If the system

contains more than two charges, then the potential energy of each pair of

charges is computed and added algebraically For example for a system of

three charges (qj , 42 , 43) shown in fig 3.12 is

Here in second summation we have used j >i, so that “4 terms appear orily once

(i) Electric potential energy is a scalar quantity and algebraically additive

(Gi) To find the potential and potential energy the sign of charge is retained in the formula

(iii) For n-charges, the total number of terms in potential energy will be” we 1,

(iv) Work done = gain in potential energy

(i) Let x be the distance of point P from A at which q electric potential is zero

x-070 Fig 3.14 (b)

Thus, electric.potential is zero at distance x = 0-50 m and x = 1-16m from point A

(Gi) Electrostatic potential energy of the system

strength E at separation Ar Let a positive test charge 4p be

brought by external force from A to B against the direction of

electric field By definition of potential difference, potential % E

difference between A and B is Vp = Vạ = workdoneinbringing "8 XÃ

per unit positive test charge from A to B

= Wap

Electric force on test charge, Fe = đụE Fig 3.17

External force on test charge Foxy = ~ đụ E

Work done in bringing test charge from A to B

(0 KINETIC ENERGY GAINED BY A CHARGED PARTICLE IN

UNIFORM ELECTRIC FIELD

Leta charged particle of charge q and mass mis situated in a uniform electric field of strength E

‘The force acting an charged particle due to electric field,

FeqE +)

If the charged particle is free to more, then acceleration,

mm

Clearly, the acceleration of charged particle in uniform electric field is constant Let initially the

particle be at rest and after traversing a distance s, the particle gains velocity 2, then from formula v? =u? + 2as , we have

+ The kinetic energy gained by particle in traversing the distance ’s’ is

2 m

If q is in coulomb, V in volts, the kinetic energy K will be in joule Thus it is clear that if a charge particle of charge q is acclerated through a potential difference V, the gain in kinetic energy of the particle will be qV

Small unit of energy : (Electron volt) : The small unit of energy is electron-volt (eV) :

1 Electron-volt is the energy gained by a particle of charge 1 elementary unit (like electron, proton)

when being accelerated through a potential difference of 1 volt

1 electron-volt= 1 eV = (1-6 x 10 9 coulomb) x (1 volt) = 1-6 x 10 !? joule

Illustrative Examples

Formulae Used: (i) Potential difference V = Ed

(ii) Work done = gain in KE = qV

Ex 7 Calculate the kinetic energy gained by a proton when accelerated through a potential

difference of 100 volt Charge of proton = 1-6 x 10”? coulomb"

Solution ‘The kinetic energy gained by particle = qV

Here q = ¢ = 1-6 x 10” ' coulomb, V = 100 volt

«+ Kinetic energy, K = (1-6 x 10~ 9 coulomb) x (100 volt)

=1-6 x10" !7 joule

As 1eV=1-6x 107 !9 joule

xi0

„16x10 =1 6X = 100eV 1:6 x 107 1

+ K=100eV = 1:6 x 107 "7 joule

Ex 8.A proton moves a distance of 0-1 m along a uniform electric field of 5000 NIC Calculate the :

(i) the potential difference between the points, between which the proton moves

(ii) change in kinetic energy of the proton after its indicated movement

Solution (i) Key idea : A proton (positive charge) gains kinetic energy when it moves along the

direction of electric field Ỷ

Potential difference,V = Ed Here E = 5000 N /C,d=0-1m

+:V =5000 x 0- Gi) Change (gain) in kinetic energy, Ex = eV

~ 1:6 x 10” ` x 800 = 8 x 107 17 joule

500 volt

I *3.11 ELECTRIC POTENTIAL DUE TO A UNIFORMLY CHARGED HOLLOW

CONDUCTING SPHERE

Consider a hollow conducting sphere of radius R It is given a charge Q The charge resides on the

‘outer surface of the sphere Let the charge distribution be uniform and o is the uniform surface charge

4RR?

(6) Atexternal point:Let P\ be an external point ata distance

(> R) from cenlre For external points, the electric field strength +

This is same as ifthe charge were placed at the centre O Hence for external points the entire charge may

be supposed to be concentrated at the centre Clearly at external points V; = : 5

(ii) At the surface of conductor, r = R

19

Electric potential at surface, Vj = ="—

0 (iii) At Internal Point : Let P; be an intemal point at distance

‘This potential is same as on the surface Hence within © Citence rom corte)

uniformly charged spherical surface, the electric potential is same ae