Two charges 5 nC and -2 nC are placed at points $(5 \mathrm{~cm}, 0,0)$ and $(23 \mathrm{~cm}, 0,0)$ in a…
- $10 \times 10^{-7} \mathrm{~J}$
- $5 \times 10^{-7} \mathrm{~J}$
- $15 \times 10^{-7} \mathrm{~J}$
- $25 \times 10^{-7} \mathrm{~J}$
Solution

$\begin{aligned} & \mathrm{q}_1=5 \mathrm{nC}=5 \times 10^{-9} \mathrm{C}, \mathrm{q}_2=-2 \mathrm{nC}=-2 \times 10^{-9} \mathrm{C} \\ & \mathrm{r}=(23-5)=18 \mathrm{~cm}=18 \times 10^{-2} \mathrm{~m} \end{aligned}$ $\therefore \quad$ The electrostatic potential energy of charge system is $\begin{aligned} \mathrm{U} & =\frac{\mathrm{kq}_1 \mathrm{q}_2}{\mathrm{r}}=\frac{9 \times 10^9 \times 5 \times 10^{-9} \times 2 \times 10^{-9}}{18 \times 10^{-2}} \\ & =5 \times 10^{-7} \mathrm{~J} \end{aligned}$
Asked in: AP EAMCET 2024 (22 May Shift 2)