question_answer

Charges 2q, -q and - q lie at the vertices of a triangle. The value of E and V at the centroid of equilateral triangle will be :

A) \[E\ne 0andV\ne 0\]                     

B) \[E=0andV=0\]

C) \[E\ne 0andV=0\]                           

D) \[E=0andV\ne 0\]

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                 \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{2q}{{{x}^{2}}}+\frac{2q}{{{x}^{2}}}\cos \theta  \right)\ne 0\]