Question

Three identical metallic uncharged spheres $$A,B,andC$$ each of the radius a, kept at the corners of an equilateral triangle of side $$d(d\gg a)$$ as shown in the figure. The fourth sphere (of radius$$a$$), which has a change in$$q$$, touches A and then moves to a position far away. Bis earthed and then the earth connection is removed C is then earthed. The charge on C is:

(A) $${\displaystyle \frac{qa}{2d}}({\displaystyle \frac{2d-a}{2d}})$$
(B) $${\displaystyle \frac{qa}{2d}}({\displaystyle \frac{2d-a}{d}})$$
(C) $$-{\displaystyle \frac{qa}{2d}}({\displaystyle \frac{d-a}{d}})$$
(D) $${\displaystyle \frac{qa}{2d}}({\displaystyle \frac{d-a}{2d}})$$

Answer 1

Hint:If a charged metallic sphere contacts the same size of metallic sphere, then half of the charge is transferred to another one.If a metallic sphere is earthed, then the net potential on that sphere has become zero. There is a charge dislocation in the earthed sphere by the influence of other charges nearby.

Complete step by step answer:
Initially, there are no charges in A, B, C, spheres. So-net charge $$q_{net}=0$$
Then there is a fourth sphere comes in contact with sphere A
So the charge is transferred to A by an amount of $$q/2$$
$$q_A=q/2$$
Then removed the fourth sphere and earthed the B sphere. (Indicates figure B)
Because of earthed net Potential at B, $$V_{net}=0$$ and there is a charge $$q^{\prime }$$ on B.
$$V_{net}=V_A+V_B$$

$$V_{net}={\displaystyle \frac{k{q}_A}{r_A}}+{\displaystyle \frac{k{q}_B}{r_B}}$$
$$r_A$$ Distance between centre of the sphere A and centre of the sphere B, $$r_A=a+d+a=d+2a$$ but $$d\gg a$$ so $$r_A=d+2a=d$$.
$$V_A$$ Is the potential due to the sphere A.
$$V_B$$ Is the potential due to the charge dislocation in B.
$$k$$ is a constant.
$$V_{net}={\displaystyle \frac{k(q/2)}{d}}+{\displaystyle \frac{k{q}^{\prime }}{a}}=0$$
$${\displaystyle \frac{kq}{2d}}={\displaystyle \frac{-k{q}^{\prime }}{a}}$$, then $$q^{\prime }={\displaystyle \frac{-qa}{2d}}$$this is the charge on B.
Now, earth connection removed from B and earthed C
So, net potential on C =0 $$V{}_{net}^{\prime }=0$$
$$V{}_{net}^{\prime }=V_A+V_B+V_C=0$$
We already know $$V_A$$and $$V_B$$ and substitute values.
$$0={\displaystyle \frac{k{q}_{}}{2d}}+{\displaystyle \frac{-kqa}{2d^2}}+{\displaystyle \frac{k{q_C}^{\prime }}{a}}$$
Cancel all $$k$$
$${\displaystyle \frac{k{q}_{}}{2d}}+{\displaystyle \frac{-kqa}{2d^2}}=-{\displaystyle \frac{k{q_C}^{\prime }}{a}}$$
$${\displaystyle \frac{q_{}}{2d}}+{\displaystyle \frac{-qa}{2d^2}}=-{\displaystyle \frac{{q_C}^{\prime }}{a}}$$
$$({\displaystyle \frac{q_{}}{2d}}+{\displaystyle \frac{-qa}{2d^2}})a=-{q_C}^{\prime }$$
 The charge on C is
$${q_C}^{\prime }=-{\displaystyle \frac{qa}{2d}}[{\displaystyle \frac{d-a}{d}}]$$
So the answer is (C) $$-{\displaystyle \frac{qa}{2d}}({\displaystyle \frac{d-a}{d}})$$

Note:Earthed means is the grounding of the metallic sphere. We should count this term $$2a$$ in $$d+2a$$, if this $$d\gg a$$ is not given. Earthling is used in electrical appliances to prevent electric shock by providing a path of unwanted charge flow.