Explanation

Total electrostatic energy of the system is the sum of:

(i) Mutual interaction energy between the two charges
(ii) Potential energy of charges due to the external field

Distance between the two charges is

$$ r_{12}=18\,cm=0.18\,m $$

Interaction energy:

$$ U_{12}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} $$

$$ U_{12}=9\times10^9 \times \frac{(7\times10^{-6})(-2\times10^{-6})}{0.18} $$

$$ U_{12}=-0.7\,J $$

Potential due to external field

Given

$$ \vec{E}=\frac{A}{r^2}\hat{r} $$

Potential is obtained from

$$ E=-\frac{dV}{dr} $$

$$ V(r)=\frac{A}{r} $$

For both charges,

$$ r=9\,cm=0.09\,m $$

$$ V=\frac{9\times10^5}{0.09}=10^7\,V $$

External field energy:

$$ U_{ext}=q_1V+q_2V=(7-2)\times10^{-6}\times10^7 $$

$$ U_{ext}=50\,J $$

Total electrostatic energy:

$$ U=U_{12}+U_{ext} $$

$$ U=50-0.7=49.3\,J $$

Final Answer: $$ \boxed{49.3\;J} $$

Related JEE Main Questions