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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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 .

  1. A 1.4
  2. B -90.7
  3. C 49.3
  4. D 24.3
Verified Solution

Answer & Solution

Correct Answer

(C) 49.3

Step-by-step Solution

Detailed explanation

\(dV =-\overrightarrow{ E } \cdot \overrightarrow{ dr }\) \(\int_0^{ v } dV =-\int_{\infty}^{ r } \frac{ A }{ r ^2} dr\) \(V =-\left[\frac{- A }{ r ^2}\right]_{\infty}^{ r } \Rightarrow V =\frac{ A }{ r }\) \(U = U _{\text {self }}+ U _{\text {interaction }}\)…