A) 0
B) \[\frac{\sigma }{{{\in }_{0}}{{r}^{2}}}({{b}^{2}}-{{c}^{2}})\]
C) \[\frac{\sigma }{{{\in }_{0}}{{r}^{2}}}({{a}^{2}}+{{b}^{2}})\]
D) none of these
Correct Answer: B
Solution :
[B]Electric field at a distance r (a > r > b) will be due to charges enclosed in r only, & since, a sphere acts as a point charge for points outside its surface, |
\[\therefore \,\,E=\frac{k{{Q}_{c}}}{{{r}^{2}}}+\frac{k{{Q}_{b}}}{{{r}^{2}}}=\frac{k}{{{r}^{2}}}\,\,(\sigma \times 4\pi {{c}^{2}}+(-\,\sigma )4\pi {{b}^{2}})\] |
\[=\frac{\sigma }{{{\varepsilon }_{0}}{{r}^{2}}}({{c}^{2}}-{{b}^{2}})\] |
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