A) \[\frac{2Q}{\pi {{a}^{2}}}\]
B) \[\frac{Q}{2\pi {{a}^{2}}}\]
C) \[\frac{Q}{2\pi ({{b}^{2}}-{{a}^{2}})}\]
D) \[\frac{2Q}{\pi ({{a}^{2}}-{{b}^{2}})}\]
Correct Answer: B
Solution :
Gaussian surface at distance r from center \[\frac{Q+\int\limits_{a}^{r}{\frac{A}{r}4\pi {{r}^{2}}dr}}{{{\in }_{0}}}=E4\pi {{r}^{2}}\]2 \[E=\frac{Q+2\pi A{{r}^{2}}-2\pi A{{a}^{2}}}{4\pi {{r}^{2}}{{\in }_{0}}}\] make E independent of r then \[Q-2\pi {{a}^{2}}A=0\Rightarrow A=\frac{Q}{2\pi {{r}^{2}}}\]You need to login to perform this action.
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