Potential due to a Line Charge

Potential due to a Line Charge

AU : May-05, 09

• Consider a line charge having density ρL C/m, as shown in the Fig. 4.6.1.

• Consider differential length dL' at a distance r'. Then the differential charge on the length dL' is given by,

dQ = ρL(r ')dL'     ...(4.6.1)

where ρL(r') = Line charge density at r'

Let the potential at A is to be determined. Then

• The R = |r - r '| indicates the distance from the differential charge.

• The dV_A is a differential potential at A. Hence the potential V_A can be obtained by integrating dV_A over the length over which line charge is distributed.

Key Point : Note that R is the distance and not the vector and for uniform line charge density ρL(r') = ρL.

Ex. 4.6.1 Find the potential V on z-axis at a distance z from origin when uniform line charge ρL in the form of a ring of radius a is placed in the z = 0 plane.

Sol. : The arrangement is shown in the Fig. 4.6.2.

The point A (0, 0, z) is on z-axis, at a distance z from the origin while radius of the ring is a.

Consider differential length dL' at point P on the ring. The ring is in z = 0 plane hence dL' in cylindrical system is,

dL = r’d ϕ = ad ϕ

The distance of point A from the differential charge is R = l (PA).

R = √a² + z² ... From the Fig. 4.6.2

Hence the potential of A is to be obtained by integrating dVA over the circular ring i.e. path with radius r' = a and 0 varies from 0 to 2 π

Important

Key Point : Note that the potential at a point can be obtained by two ways.

1. If  is known then use,

C = 0 if reference is infinity.

2. If  is not known, then find differential charge dQ considering differential length dL' and

The integration depends on the charge distribution.

Ex. 4.6.2 A uniform line charge of 1 nCIm is situated along x-axis between the points (-500, 0) and (500, 0) mm. Find the electric scalar potential at (0,1000) mm.

AU : May-05, Marks 10

Sol . :

ρL = 1 nC/m along x-axis and find V at A (0, 1000) mm.

Sol . :

Consider elementary charge dQ on length dx at distance x from orign.

dQ = ρL dx

Distance of point A from charg dQ is

Ex. 4.6.3 A total charge of 10 ^-8 C is distributed uniformly along a ring of radius of 5 m. Calculate the potential on the axis of the ring at a point 5 m from the centre of the ring.

Sol. : Q = 10^-8 C, r = 5 m, h = 5 m. The ring is shown in the Fig. 4.6.4.

Consider the differential length dL on the ring.

Example for Practice

Ex. 4.6.4 A uniform line charge density pL C/m is existing from - L to + L on y-axis. Find potential at A (a, 0, 0).

Ex. 4.6.5 A total charge of 40/3 nC is uniformly distributed over a circular ring of radius 2 m placed in z = 0 plane, with center as origin. Find the electric potential at A (0, 0, 5).

[Ans.: 22.252 V]