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Physics · 2026

JEE Main · 23 January 2026, Shift 2 · Q39

Two charges 7 μ C and -2 μ C are placed at ( -9,0,0 ) cm and ( 9,0,0 ) cm respectively in an external field E=A/r^2 r, where A=9 × 10^5 N / C. m^2.…

Two charges $\displaystyle 7 \mu \mathrm{C}$ and $\displaystyle -2 \mu \mathrm{C}$ are placed at ( $\displaystyle -9,0,0$ ) cm and ( $\displaystyle 9,0,0$ ) cm respectively in an external field $\displaystyle E=\frac{\mathrm{A}}{r^2} \widehat{r}$, where $\displaystyle A=9 \times 10^5 \mathrm{~N} / \mathrm{C} . \mathrm{m}^2$. Considering the potential at infinity is $\displaystyle 0$ , the electrostatic energy of the configuration is $\displaystyle \_\_\_\_$J.
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.