Question

Charges $+q$, $-4q$ and $+2q$ are arranged at the corners of an equilateral triangle of side $0.15m$. If $q=1\mu C$, their mutual potential energy is roughly:

• A. $0.4J$
• B. $0.5J$
• C. $0.6J$
• D. $0.8J$

Answer 1

Answer: (C)

$0.6J$

We know potential energy between two points $=\frac{k{Q}_1{Q}_2}{r.}$
Now P. E between A and B$=\frac{k\left(q\right)(-4q)}{a}$
P. E between A and C$=\frac{k\left(q\right)(+2q)}{a}$
P. E between B and C$=\frac{k(-4q)\left(2q\right)}{a.}$
total P. E of arrangement $=\frac{-4k{q}^2}{a}+\frac{2k{q}^2}{a}-\frac{8k{q}^2}{a}=\frac{-10k{q}^2}{a}$
$=\frac{-10\times 9\times {10}^9\times 1\times {10}^{-6}\times 1\times {10}^{-6}}{0\cdot 15}=-0\cdot 6J.$