A) \[\frac{q}{4\pi {{\in }_{0}}2R}\]
B) \[\frac{4q}{12\pi {{\in }_{0}}R}\]
C) \[\frac{5q}{24\pi {{\in }_{0}}R}\]
D) Can't be determined
Correct Answer: B
Solution :
Due to induction - q charge will come at the inner surface of shell and +q will come on outer surface of shell |
Now |
\[{{V}_{at0}}={{V}_{due\,\,to\,\,q}}+{{V}_{due\,to-q\,on\,inner\,surface}}+{{V}_{due\,to+q\,on\,outer\,surface}}\] |
at \[0=\frac{kq}{R}+\frac{k\,(-\,q)}{2R}+\frac{kq}{3R}=\left( k=\frac{6q-3q+2q}{6R} \right)\] |
\[=\frac{5q}{24\pi {{\in }_{0}}R}\,\,\,\,\,\,\,\,\left( k=\frac{1}{4\pi {{\in }_{0}}} \right)\] |
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