A) \[E\ne 0\]and \[V\ne 0\]
B) \[E=0\]and \[V=0\]
C) \[E\ne 0\]and \[V=0\]
D) \[E=0\] and \[V\ne 0\]
Correct Answer: C
Solution :
The potential due to charge q at a distance r is given by \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Since, potential is a scalar quantity, it can be added to find the sum due to individual charges. \[\Sigma V={{V}_{A}}+{{V}_{B}}+{{V}_{C}}\] \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2q}{x}\] \[{{V}_{B}}=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{x}\] \[{{V}_{C}}=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{x}\] \[\therefore \] \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{2q}{x}-\frac{q}{x}-\frac{q}{x} \right)=0\] Electric field is a vector quantity, hence component along OD is taken \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{2q}{{{x}^{2}}}+\frac{2q}{{{x}^{2}}}\,\cos \theta \right)\ne 0\]You need to login to perform this action.
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