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 :
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, \[\therefore \] \[E=\frac{k{{Q}_{c}}}{{{r}^{2}}}+\frac{k{{Q}_{b}}}{{{r}^{2}}}=\frac{k}{{{r}^{2}}}\] \[(\sigma \times 4\pi {{c}^{2}}+(-\,\sigma )4\pi {{b}^{2}})=\frac{\sigma }{{{\varepsilon }_{0}}{{r}^{2}}}({{c}^{2}}-{{b}^{2}})\]You need to login to perform this action.
You will be redirected in
3 sec