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