Question

Charges $+q$ and $-q$ are placed at the vertices $\text{B}$ and $\text{C}$ of an isosceles triangle. The potential at the vertex $\text{A}$ is

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

Answer 1

The correct option is C zero

Potential at a point due to point charge is given by

$V=\frac{1}{4\pi {ϵ}_0}\frac{q}{r}$

So, here net potential at point $\text{A}$ due to $+q$ and $-q$ will be

$V_{net}=V_{\text{due to B}}+V_{\text{due to C}}$

$\Rightarrow V_{net}=\frac{k.\left(q\right)}{r_{AB}}-\frac{k\left(q\right)}{r_{AC}}$

$\because r_{AB}=r_{AC}=a$

$\therefore V_{net}=\frac{kq}{a}-\frac{kq}{a}=0$

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