question_answer

Two small identical metal balls of radius rare at a distance a from each other and are charged, one with a potential \[{{V}_{1}}\]and the other with a potential \[{{V}_{2}}\]. The charges on the balls are:

A) \[{{q}_{1}}={{V}_{1}}a,{{q}_{2}}={{V}_{2}}a\]

B) \[{{q}_{1}}={{V}_{1}}r,\,\,{{q}_{2}}={{V}_{2}}r\]

C) \[{{q}_{1}}=\left( \frac{{{V}_{1}}+{{V}_{2}}}{2} \right)a,{{q}_{2}}=\left( \frac{{{V}_{1}}+{{V}_{2}}}{2} \right)r\]

D) \[{{q}_{1}}=-\frac{r}{a}(r{{V}_{2}}-a{{V}_{1}}),{{q}_{2}}=-\frac{r}{a}(r{{V}_{1}}-a{{V}_{2}})\]

Correct Answer: D

Solution :

\[{{V}_{1}}=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{{{q}_{1}}}{r}+\frac{{{q}_{2}}}{a} \right]\]

and \[{{V}_{2}}=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{{{q}_{2}}}{r}+\frac{{{q}_{1}}}{a} \right]\]

After solving above equations, and neglecting \[{{r}^{2}}\]in comparison to a, we get

\[{{q}_{1}}=-\frac{r}{a}(r{{V}_{2}}),\]and \[{{q}_{2}}=-\frac{r}{a}(r{{V}_{1}}-a{{V}_{2}}),\]