Question

Three charges each $20\mu C$ are placed at the corners of an equilateral triangle of side of $0.4m$. The potential energy of the system is :

• A. $18\times {10}^{-6}J$
• B. $9J$
• C. $9\times {10}^{-6}J$
• D. $27J$

Answer 1

Answer: (D)

$27J$

$P.E,u=\frac{k{q}_1{q}_2}{r}$

Now total $P.E.=P.E_{AB}+P.E_{BC}+P.E_{AC}$

$=\frac{k\left(20\mu C\right)\left(20\mu C\right)}{\left(0.4\right)}\times 3$

$=\frac{9\times {10}^9\times 20\times 20\times {10}^{-12}\times 3\times 10}{4}$

$u=27J$