A) \[a=\frac{3R}{{{2}^{{}^{1}/{}_{4}}}}\]
B) \[a=R/\sqrt{3}\]
C) \[a={{8}^{-1/4}}R\]
D) \[a={{2}^{-1/4}}R\]
Correct Answer: C
Solution :
\[E4\pi {{a}^{2}}=\frac{\int_{0}^{a}{kr4\pi {{r}^{2}}dr}}{{{\varepsilon }_{0}}}\] \[E=\frac{k4\pi {{a}^{4}}}{4\times 4\pi {{\varepsilon }_{0}}}\] \[2Q=\int_{0}^{R}{kr4\pi {{r}^{2}}dr}\] \[k=\frac{2Q}{\pi {{R}^{4}}}\] \[QE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{(2a)}^{2}}}\] \[R=a{{8}^{1/4}}\]You need to login to perform this action.
You will be redirected in
3 sec