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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2026
  • Medium
  • MCQ

Two charges \(7 \mu C\) and \(-2 \mu C\) are placed at \((-9,0,0) cm\) and \((9,0,0) cm\) respectively in an external field \(E=\frac{ A }{ r ^2} \hat{ r }\), where \(A=9 \times 10^5 N / C . m ^2\).
Considering the potential at infinity is 0 , the electrostatic energy of the configuration is ___________ J .

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Verified Solution

Answer & Solution

Correct Answer

(C) 49.3

Step-by-step Solution

Detailed explanation


image
\(dV =-\overrightarrow{ E } \cdot \overrightarrow{ dr }\)
\(\int_0^{ v } dV =-\int_{\infty}^{ r } \frac{ A }{ r ^2} dr\)
\(V =-\left[\frac{- A }{ r ^2}\right]_{\infty}^{ r } \Rightarrow V =\frac{ A }{ r }\)
\(U = U _{\text {self }}+ U _{\text {interaction }}\)
\(= q _1 v _1= q _2 v _2+\frac{ kq _1 q _2}{2 r }\)
\(=7 \times 10^{-6} \frac{A}{9 \times 10^{-2}}-2 \times 10^{-6} \frac{A}{9 \times 10^{-2}}\)
\(-\frac{9 \times 10^9 \times 14 \times 10^{-12}}{2 \times 9 \times 10^{-2}}\)
\(=\frac{5 \times 10^{-6} \times 9 \times 10^5}{9 \times 10^{-2}}-7 \times 10^{-1}\)
\(=50-0.7\)
\(=49.3 J\)
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