A) - q from ground to conductor
B) - q from conductor to ground
C) + q from ground to conductor
D) Zero
Correct Answer: A
Solution :
Charge distribution after induction is shown in figure. V of inner shell \[=\left( \frac{2q}{4\pi {{\varepsilon }_{0}}2r} \right)+\left( \frac{q'}{4\pi {{\varepsilon }_{0}}r} \right)\] \[q'\to \]charge flown from earth to shell or from shell to earth. \[V=0=k\frac{2q}{2r}+k\frac{q'}{r}.\] \[\left( k=\frac{1}{4\pi {{\varepsilon }_{0}}} \right)\] \[\Rightarrow \]\[q'=-q\] Since, charge is negative (- ve) so charge \[({{e}^{-1}})\]are flowing from ground to shell. So, - q charge flows from earth to shell.You need to login to perform this action.
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