Two charges 5 nC and -2 nC are placed at points $(5 \mathrm{~cm}, 0,0)$ and $(23 \mathrm{~cm}, 0,0)$ in a…

Two charges 5 nC and -2 nC are placed at points $(5 \mathrm{~cm}, 0,0)$ and $(23 \mathrm{~cm}, 0,0)$ in a region of space where there is no other external field. The electrostatic potential energy of this charge system is
  1. $10 \times 10^{-7} \mathrm{~J}$
  2. $5 \times 10^{-7} \mathrm{~J}$
  3. $15 \times 10^{-7} \mathrm{~J}$
  4. $25 \times 10^{-7} \mathrm{~J}$

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)

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