Question

Three charges $1\mu C,2\mu C,$ and $3\mu C$ are kept at vertices of an equilateral triangle of side 1 m. If they are brought nearer, so that they now form an equilateral triangle of side $0.5m$, then work done is :

• A. $0.11J$
• B. $11J$
• C. $0.01J$
• D. $1.1J$

Answer 1

The correct option is C $0.01J$

Initial PE of the three charges,

$U_1=k\frac{q_1{q}_2+q_2{q}_3+q_1{q}_3}{r}$

$=9\times {10}^9\left[\begin{array}{c}\frac{1\times 2\times {10}^{-12}+2\times 3\times {10}^{-12}+1\times 3\times {10}^{-12}}{1}\end{array}\right]$

$=99\times {10}^{-3}J$

Final PE of the system

$U_f=9\times {10}^9\left[\begin{array}{c}\frac{1\times 2\times {10}^{-12}+2\times 3\times {10}^{-12}+1\times 3\times {10}^{-12}}{0.5}\end{array}\right]$

$=\frac{99\times {10}^{-3}}{0.5}=198\times {10}^{-3}J$

$W=U_f-U_i=(198-99)\times {10}^{-3}J$

$=99\times {10}^{-3}J$

$=0.099=0.01J$