Question

Two insulating rods of equal length, one of them of charge q and other has charge -q form a circle of radius R (one rod forms one - half of circle) . the electric field at centre of circle is

Answer 1

Given:

Two insulating rods of equal length, one of them of charge q and other has charge -q form a circle of radius R (one rod forms one - half of circle) .

To find:

Net Electrostatic Field at centre.

Calculation:

Let the linear charge density for the rods be $\lambda$.

So, field due to positive part of circle:

$\vec{E1} = \dfrac{2k \lambda}{r} \: \: ( - \hat{j})$

Again, field due to negative part of the circle:

$\vec{E2} = \dfrac{2k \lambda}{r} \: \: ( - \hat{j})$

So, net field intensity:

$\therefore \: \vec{E} = \vec{E1} + \vec{E2}$

$= > \: \vec{E} = \bigg \{ \dfrac{2k \lambda}{r} + \dfrac{2k \lambda}{r} \bigg \} \: \: ( - \hat{j})$

$= > \: \vec{E} = \bigg \{ \dfrac{4k \lambda}{r} \bigg \} \: \: ( - \hat{j})$

$= > \: \vec{E} = \bigg \{ \dfrac{4\lambda}{4\pi \epsilon_{0}r} \bigg \} \: \: ( - \hat{j})$

$= > \: \vec{E} = \bigg \{ \dfrac{\lambda}{\pi \epsilon_{0}r} \bigg \} \: \: ( - \hat{j})$

$= > \: \vec{E} = \bigg \{ \dfrac{ (\frac{q}{\pi r} )}{\pi \epsilon_{0}r} \bigg \} \: \: ( - \hat{j})$

$= > \: \vec{E} = \bigg \{ \dfrac{ q}{ { \pi}^{2} \epsilon_{0} {r}^{2} } \bigg \} \: \: ( - \hat{j})$

So, final answer is:

$\boxed{ \red{ \bold{ \: \vec{E} = \bigg \{ \dfrac{ q}{ { \pi}^{2} \epsilon_{0} {r}^{2} } \bigg \} \: \: ( - \hat{j})}}}$