A) \[\frac{q}{4\pi r_{2}^{1}}and\frac{Q}{4\pi r_{2}^{2}}\]
B) \[\frac{-q}{4\pi r_{2}^{1}}and\frac{Q+q}{4\pi r_{2}^{2}}\]
C) \[\frac{q}{4\pi r_{1}^{2}}and\frac{Q-q}{4\pi r_{2}^{2}}\]
D) \[0\,and\,\frac{Q-q}{4\pi r_{2}^{2}}\]
Correct Answer: C
Solution :
Surface charge density \[(\sigma )=\frac{Ch\arg e}{Surface\text{ }area}\] \[\therefore \] \[{{\sigma }_{inner}}=\frac{-q}{4\pi r_{1}^{2}}\] and \[{{\sigma }_{outer}}=\frac{Q-q}{4\pi r_{2}^{2}}\]You need to login to perform this action.
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