Question

32. A charge is distributed uniformly over a
ring of radius a. Obtain an expression for
the electric field 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. Delhi 2016​

Answer 1

Proof :

Let's consider an elemental part of ring containing charge dq. The radius of ring is "a" and the centre of ring is "x" units from the concerned point.

Refer to the diagram to understand better.

Let Electrostatic Field Intensity due to elemental charge be dE ;

At that concerned point , all the sine components will get cancelled and the cos components will be added :

$dE = \dfrac{k(dq)}{ { \bigg( \sqrt{ {a}^{2} + {x}^{2} } \bigg) }^{2} }$

$= > dE = \dfrac{k(dq)}{ {a}^{2} + {x}^{2} }$

Integrating on both sides with respect to cos component :

$\displaystyle = > \int dE = \int\dfrac{k(dq)}{ {a}^{2} + {x}^{2} } \cos( \theta)$

$\displaystyle = > \int dE = \int \dfrac{kq}{ {a}^{2} + {x}^{2} } \times \frac{x}{ \sqrt{ {a}^{2} + {x}^{2} } }$

$\displaystyle = > \int dE = \dfrac{k}{ { \bigg \{ {a}^{2} + {x}^{2} \bigg \}}^{ \frac{3}{2} } } \int q$

$= > E = \dfrac{kqx}{ { \bigg \{ {a}^{2} + {x}^{2} \bigg \}}^{ \frac{3}{2} } }$

Putting value of Coulomb's Constant :

$= > E = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{qx}{ { \bigg \{ {a}^{2} + {x}^{2} \bigg \}}^{ \frac{3}{2} } }$

So final answer is:

$\boxed{ \red{ \bold{ E = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{qx}{ { \bigg \{ {a}^{2} + {x}^{2} \bigg \}}^{ \frac{3}{2} } } }}}$