Question

Three charges $+q,-qand-q$ are kept at the vertices of an equilateral triangle of $10cm$ side. The potential at the mid point in between $-q,-q,$ if $q=5\mu C$ is :

• A. $-6.4\times {10}^{5}V$
• B. $-12.8\times {10}^{4}V$
• C. $-6.4\times {10}^{4}V$
• D. $-12.8\times {10}^{5}V$

Answer 1

Answer: (D)

$-12.8\times {10}^{5}V$

We know $V=\frac{kq}{d}$
now potential at A$V_A=\frac{k(-q)}{5\times {10}^{-2}}+\frac{k(-q)}{5\times {10}^{-2}}+\frac{k\left(q\right)}{5\sqrt{3}\times {10}^{-2}}$
$=\frac{kq}{5\times {10}^{-2}}(-1-1+\frac{1}{\sqrt{3}})$

$=\frac{9\times {10}^9\times 5\times {10}^{-6}}{5\times {10}^{-2}}(-2+\frac{1}{\sqrt{3}})$

$=9\times {10}^5\times (-1.422)$
$V_A=-12.80\times {10}^{5}V$