Q.
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) $\frac{1}{4\pi\varepsilon_{0}}.\frac{2q}{\sqrt{a^{2} + b^{2}}}$
B) Zero
C) $\frac{1}{4\pi\varepsilon_{0}}.\frac{q}{\sqrt{a^{2} + b^{2}}}$
D) $\frac{1}{4\pi\varepsilon_{0}}.\frac{( - q)}{\sqrt{a^{2} + b^{2}}}$
Explanation
(b) Potential at $A =$ Potential due to $( + q)$ charge + Potential due to $( - q)$ charge
$= \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{\sqrt{a^{2} + b^{2}}} + \frac{1}{4\pi\varepsilon_{0}}\frac{( - q)}{\sqrt{a^{2} + b^{2}}} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.