Question

At the corners of an equilateral triangle of side $25cm$ charges $1\mu C,2\mu C$ and $3\mu C$ are placed. The electrostatic potential energy of the system is :

• A. $396\times {10}^{-3}J$
• B. $132\times {10}^{-3}J$
• C. $396\times {10}^{3}J$
• D. $132\times {10}^{3}J$

Answer 1

Answer: (A)

$396\times {10}^{-3}J$

We know $P.E=\frac{k{q}_1{q}_2}{r}$
$P.E{.}_{total}=P.E{.}_{AB}+P.E{.}_{BC}+P.E{.}_{AC}$

$\Rightarrow P.E{.}_{total}=\frac{k\left(1\mu c\right)\left(2\mu c\right)}{25\times {10}^{-2}}+\frac{k\left(2\mu c\right)\left(3\mu c\right)}{25\times {10}^{-2}}+\frac{k\left(1\mu c\right)\left(3\mu c\right)}{25\times {10}^{-2}}$

$=\frac{k\times {10}^{-12}}{25\times {10}^{-2}}(2+6+3)$

$=\frac{9\times {10}^9\times {10}^{-12}\times 11}{25\times {10}^{-2}}$

$P.E.=396\times {10}^{-3}J$