Question

As shown in the figure, charges +q and -q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is

• A. $\frac{1}{4\pi {\epsilon }_0}.\frac{2q}{\sqrt{a^2+b^2}}$
• B. $\frac{1}{4\pi {\epsilon }_0}.\frac{q}{\sqrt{a^2+b^2}}$
• C. $\frac{1}{4\pi {\epsilon }_0}.\frac{(-q)}{\sqrt{a^2+b^2}}$
• D. zero

Answer 1

The correct option is D zero

$V_A=0$

$\therefore$ it is equidistance from positive and negative charge of equal magnitude$.$

Hence$,$ option $\left(D\right)$ is correct$.$