Question

A charge - is uniformly distributed over a non-
conducting semi-circular rod of radius R. The potential
at the centre is:​

Answer 1

Answer:

Given:

Charge q has been distributed over a semi-circular rod of radius of R .

To find:

Potential at the centre of the semi-circle.

Calculation:

We shall use Integration to solve this kind of questions. Let a small portion of rod give dV amount of potential at centre.

$dV = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{\dfrac{dq}{r} \bigg \}$

Now, integration for the whole rod , we can say :

$= > \displaystyle \int dV = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \int \dfrac{dq}{r} \bigg \}$

Trying to experience charge in terms of circumference , we get :

$= > \displaystyle \int dV = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \int \dfrac{ \frac{q(dr)}{\pi r} }{r} \bigg \}$

$\displaystyle = > V = \dfrac{q}{4 {\pi}^{2} \epsilon_{0} } \int \frac{dr}{ {r}^{2} }$

$\displaystyle = > V = \dfrac{q}{4 {\pi}^{2} \epsilon_{0} } \times \dfrac{1}{r}$

$\displaystyle = > V = \dfrac{q}{4 {\pi}^{2} \epsilon_{0}r }$

So final answer :

$\boxed{ \red{ \huge{ \bold{ V = \dfrac{q}{4 {\pi}^{2} \epsilon_{0}r } }}}}$

Answer 2

$\tt{\huge{\pink{Hello!!! }}}$

Refer to the above attachment...

Now Charge At Point O :

$\tt{ \purple { \leadsto {dq \: = \dfrac{1}{4 \: \pi \: \e_0 } \times \dfrac{dq}{r} }}}$

Total Potential At O :

$\tt{ \purple{ \leadsto{ = \dfrac{1}{4 \: \pi \: e0} \\ \\ \displaystyle \int\limits_{0}^{ - q} \:dq \: = \tt{ \purple{ \leadsto{dq \: = \dfrac{ - 1}{4 \: \pi \: e0} . \dfrac{Q}{r} }}} }}}$