Question

A right isosceles triangle of side a has charges $q, + 3q$ and $- q$ arranged on its vertices as shown in the figure . What is the electric potential at point $P$ midway between the line connecting the $+ q$ and $- q$ charges ?

• A. $\frac{q}{\pi\varepsilon_{0}a}$
• B. $\frac{3q}{2\sqrt{2}\pi\varepsilon_{0}a}$
• C. $\frac{3q}{\pi\varepsilon_{0}a}$
• D. $\frac{3q}{\sqrt{2}\pi\varepsilon_{0}a}$

Answer 1

Answer: (B)

Potential at point $P$ due to charges $-q$, $3 q$ and $+q$ is given by Potential at $P$ due to $+q$ charge $+$ potential at $p$ due to $-q$ charge $+$ potential at $p$ due to $+3 q$ charge
i.e. $V_{\text {Net }}=V_{A}+V_{B}+V_{C}$
$\left[\right.$ As given figure $\left. V_{C}+V_{B}=0, \because V_{C}=\frac{K q}{\sqrt{2} \frac{a}{2}} \therefore V_{B}=\frac{-K_{q}}{\sqrt{2} \frac{a}{2}}\right]$
So, $V_{\text {Net }}=V_{A}=\frac{3 K q}{A P}=\frac{3 K q}{\frac{\sqrt{2} a}{2}}$
$\left[\because A P=\sqrt{a^{2}-\left(\frac{\sqrt{2} a}{2}\right)^{2}}=\sqrt{a^{2}-\frac{2 a^{2}}{4}}=\frac{\sqrt{2 a^{2}}}{4}=\frac{\sqrt{2} a}{2}\right]$
$V_{\text {Net }} =\frac{3 q \times 2}{4 \pi \varepsilon_{0} \sqrt{2} a}$
$\Rightarrow = \frac{3 q}{2 \sqrt{2} \pi \varepsilon_{0} a}$