Initially (before the connection of points M and N) conductor B is neutral, while charge appeared on conductor A is q. After connecting the spheres, the potential of sphere B is
A) \[\frac{q}{4\pi {{\varepsilon }_{0}}(r+R)}\]
B) \[\frac{q}{4\pi {{\varepsilon }_{0}}R}\]
C) \[\frac{q}{4\pi {{\varepsilon }_{0}}(r+R)}\]
D) Zero
Correct Answer: B
Solution :
After connection, the total charge on inner conductor transfers to outer one and in this situation potential of sphere B is given as \[{{V}_{B}}=\frac{q}{4\pi {{\varepsilon }_{0}}R}\] Because both the spheres will attain an equal potential.
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