A) 16 kV
B) 4 kV
C) 400V
D) 160V
E) 16V
Correct Answer: B
Solution :
\[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{r}_{1}}}+\frac{{{Q}_{2}}}{{{r}_{2}}}+\frac{{{Q}_{3}}}{{{r}_{3}}} \right)\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{33\times {{10}^{-9}}}{93\times {{10}^{-3}}}-\frac{51\times {{10}^{-9}}}{\sqrt{2}\times 93\times {{10}^{-3}}} \right.\]\[+\left. \frac{47\times {{10}^{-9}}}{93\times {{10}^{-3}}} \right)\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{10}^{-9}}}{93\times {{10}^{-3}}}\left( 33-\frac{51}{\sqrt{2}}+47 \right)\] \[\approx 4\times 1000\,V=4\,kV\]You need to login to perform this action.
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