Question

From ring of charge q and radius R is small length x removed determine the electric field at centre due to the remaining

Answer 1

Answer:

Given:

From a whole ring , a small length x has been removed.

To find:

Electric field at centre of ring due to remaining part of ring.

Concept:

We know that in the centre of a whole ring , the Electric Field intensity remains zero due to Vector cancellation of intensity due to diametrically opposite segments.

Now since a small part has been removed the diametrically opposite part of the ring will provide Electric field intensity at centre of ring.

The field intensity will be equal to the magnitude of intensity provided by the removed part of the ring.

Calculation:

Charge of that small part :

$\boxed{ \large{ \sf{dq = \dfrac{Q}{2\pi r} x}}}$

Field Intensity due to that removed part :

$\sf{E = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{dq}{ {r}^{2} }}$

$\sf{ \implies E = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{ (\dfrac{Qx}{2\pi r} )}{ {r}^{2} }}$

$\sf{ \implies \: E = \dfrac{1}{8 {\pi }^{2} \epsilon_{0} } \dfrac{Qx}{ {r}^{3} }}$