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. $\frac{3q}{r{ϵ}_{0}a}$
• B. $\frac{3q}{\sqrt{2}\pi {ϵ}_{0}a}$
• C. $\frac{q}{\pi {ϵ}_{0}a}$
• D. $\frac{3q}{2\sqrt{2}\pi {ϵ}_{0}a}$

Answer 1

Answer: (D)

$\frac{3q}{2\sqrt{2}\pi {ϵ}_{0}a}$

Potential at point $P$ due to charges $-q,3q$ and $+q$ is given by
Potential at $P$ due to $+q$ charge $+$ potential at $p$ due to $-q$ and $+$ potential at $p$ due to $+3q$ charge
i.e. $V_{Net}=V_A+V_B+V_c$
$[\text{As given figure}{V}_C+V_B=0,\because V_C=\frac{Kq}{\sqrt{2}\frac{a}{2}}\therefore V_B=\frac{-K_q}{\sqrt{2}\frac{a}{2}}]$
So, $V_{Net}=V_A=\frac{3Kq}{AP}=\frac{3Kq}{\frac{\sqrt{2}a}{2}}$
$[\because AP=\sqrt{a^2-{\left(\frac{\sqrt{2}a}{2}\right)}^2}=\sqrt{a^2-\frac{2a^2}{4}}=\frac{\sqrt{2a^2}}{4}=\frac{\sqrt{2}a}{2}]$
$V_{Net}=\frac{3q\times 2}{4\pi {ϵ}_0\sqrt{2}a}$
$\Rightarrow =\frac{3q}{2\sqrt{2}\pi {ϵ}_{0}a}$