A) 0.03kV
B) 0.9kV
C) 1.8kV
D) 3.6kV
Correct Answer: C
Solution :
Here, \[C=4\pi {{\varepsilon }_{0}}R\] \[V=\frac{q}{C}=\frac{2\times {{10}^{-8}}}{4\pi {{\varepsilon }_{0}}R}\] \[=\frac{9\times {{10}^{9}}\times 2\times {{10}^{-3}}}{10\times {{10}^{-2}}}\] \[=1.8\times {{10}^{3}}V=1.8\,kV\]You need to login to perform this action.
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