question_answer

Charges \[2q,\] \[-q\] and \[-q\] lie at the vertices of a triangle. The value of E and Vat the centroid of equilateral triangle will be

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