Question

From the given figure below:

(a) Find potential of q at O.
(b) Find the potential of induced charge at O.
(c) Find the net potential at O.
(d) Find potential of induced charges at A.

Answer 1

Hint: All the subparts of the given question can be solved using just a single expression. As a first step, we could recall the expression for potential at a point due to a charge q. Then, you could accordingly substitute for each part in that expression. Also, remember that the charge q will induce an opposite charge of the same magnitude.
Formula used:
Potential due to charge q at distance r,
$P=k\dfrac{q}{r}$

Complete step-by-step solution
In the question, we are given a neutral sphere and a charge q is kept at $\sqrt{3}R$ distance from the center of the sphere O. Also, we have a point A inside the sphere at distance 5R from the point at which the charge q is kept. This set of information is followed by four subparts.
(a) We are asked to find the potential at O due to the charge q at distance$\sqrt{3}R$. For that, let us recall the standard expression for potential at a point due to a charge q at distance r given by,
$P=k\dfrac{q}{r}$ …………………………………………….. (1)
Where k is a constant given by,
$k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$
Using (1) the potential O due to q at $\sqrt{3}R$ distance is given by,
${{P}_{1}}=k\dfrac{q}{\sqrt{3}R}$ ………………………………….. (2)
Therefore, the potential at O due to charge q is found to be, $k\dfrac{q}{\sqrt{3}R}$
(b) Charge of the induced charge will be opposite to that causing it, so, the potential due to induced charge at O is given by,
${{P}_{2}}=k\dfrac{-q}{\sqrt{3}R}=-k\dfrac{q}{\sqrt{3}R}$ …………………………………………… (3)
Therefore, potential due to induced charges at point O is found to be, $-k\dfrac{q}{\sqrt{3}R}$
(c) The net potential at O could be given by the sum of that due to charge q and that due to induced charge, that is the sum of (2) and (3),
At O,
${{P}_{net}}={{P}_{1}}+{{P}_{2}}$
$\Rightarrow {{P}_{net}}=k\dfrac{q}{\sqrt{3}R}-k\dfrac{q}{\sqrt{3}R}$
$\therefore {{P}_{net}}=0$
Therefore, we found the net charge to be zero at O.
(d) Potential of induced charges at A,
The charges induced by q will be –q, so, the potential at A due to them will be,
${{P}_{3}}=-k\dfrac{q}{5R}$
Therefore, the potential at A due to induced charges is found to be $-k\dfrac{q}{5R}$

Note: No movement of electrons happens from the charge to the sphere but here the charge is just aiding the electron movement in the sphere. It is worthwhile to mention here that only the conductors can be charged by the process of induction. Also, the induced charge will be opposite to the charge that caused it.