A) \[\frac{2Q}{4\pi {{\varepsilon }_{0}}R}\]
B) \[\frac{2Q}{4\pi {{\varepsilon }_{0}}R}-\frac{2q}{4\pi {{\varepsilon }_{0}}R}\]
C) \[\frac{2Q}{4\pi {{\varepsilon }_{0}}R}+\frac{q}{4\pi {{\varepsilon }_{0}}R}\]
D) \[\frac{(q+Q)}{4\pi {{\varepsilon }_{0}}R}\frac{2}{R}\]
Correct Answer: C
Solution :
At P, potential due to shell \[{{V}_{1}}=\frac{q}{4\pi \,{{\varepsilon }_{0}}R}\] At P, due to charge \[{{V}_{2}}=\frac{Q}{4\pi {{\varepsilon }_{0}}\frac{R}{2}}\] \[=\frac{2Q}{4\pi {{\varepsilon }_{0}}R}\] \[\therefore \] Net potential at P, \[V={{V}_{1}}+{{V}_{2}}=\frac{q}{4\pi {{\varepsilon }_{0}}R}+\frac{2Q}{4\pi {{\varepsilon }_{0}}R}\] (\[\because \] v is scalar quantity)You need to login to perform this action.
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