question_answer

Two point charges q and -2q are kept at distance d apart. At what point, potential due to the charges is zero:

A) At a distance d/2 from charge q

B) At a distance d/2 from charge -2q

C) At a distance d/3 from charge q

D) At a distance d/3 from charge -2q

Correct Answer: C

Solution :

(c) At a distance d/3 from charge q Let P be the required point at distance x from charge q. \[\therefore \,\,\,\,\,\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{x}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( -2a \right)}{d-x}=0\] \[\frac{1}{x}=\frac{2}{d-x}\] \[d-x=2x\] \[\Rightarrow \,\,x=\frac{d}{3}\] \[\therefore \] Required point is at a distance d/3 from charge q.