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 .
- A 1.4
- B -90.7
- C 49.3
- D 24.3
Answer & Solution
Correct Answer
(C) 49.3
Step-by-step Solution
Detailed explanation
\(dV =-\overrightarrow{ E } \cdot \overrightarrow{ dr }\) \(\int_0^{ v } dV =-\int_{\infty}^{ r } \frac{ A }{ r ^2} dr\) \(V =-\left[\frac{- A }{ r ^2}\right]_{\infty}^{ r } \Rightarrow V =\frac{ A }{ r }\) \(U = U _{\text {self }}+ U _{\text {interaction }}\)…
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