CBSE Class 12 · 2016 · Delhi · ElectrostaticsShort Answer

A charge is distributed uniformly over a ring of radius

a

. Obtain an expression for the electric intensity

E

at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.

Solution

Let P be an axial point at distance

x

from the centre O; the ring has radius

a

and charge

q

.
A charge element

d q

is at distance

x^{2} + a^{2}

from P and gives

d E = \frac{1}{4 π ε _{0}} \frac{d q}{x ^{2} + a ^{2}}

Resolve

d E

along the axis and perpendicular to it, with

cos θ = \frac{x}{x ^{2} + a ^{2}}

.
For every element there is a diametrically opposite one whose perpendicular component cancels it, so only the axial components survive:

E = \int d E cos θ = \frac{1}{4 π ε _{0}} \frac{x}{( x ^{2} + a ^{2} ) ^{3/2}} \int d q

E = \frac{1}{4 π ε _{0}} \frac{q x}{( x ^{2} + a ^{2} ) ^{3/2}}

For

x ≫ a

(x^{2} + a^{2})^{3/2} \approx x^{3}

E \approx \frac{1}{4 π ε _{0}} \frac{q}{x ^{2}}

This is the field of a point charge

q

, so at large distances the ring behaves like a point charge.

CBSE Class 12 Board · 2016 · Delhi · Subjective