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

अवेशो \(- q , q\) तथा \(Q\) को चित्रानुसार रखा गया है। यदि \(D \gg a\) है तो विध्यूत स्थितिज ऊर्जा का मान होगा

  1. A \(\frac{1}{{4\pi {\varepsilon _0}}}\left[ { - \frac{{{q^2}}}{d} - \frac{{qQd}}{{{D^2}}}} \right]\)
  2. B \(\frac{1}{{4\pi {\varepsilon _0}}}\left[ { - \frac{{{q^2}}}{d} - \frac{{qQd}}{{{2D^2}}}} \right]\)
  3. C \(\frac{1}{{4\pi {\varepsilon _0}}}\left[ { - \frac{{{q^2}}}{d} + \frac{{2qQd}}{{{D^2}}}} \right]\)
  4. D \(\frac{1}{{4\pi {\varepsilon _0}}}\left[ { + \frac{{{q^2}}}{d} - \frac{{qQd}}{{{D^2}}}} \right]\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{{4\pi {\varepsilon _0}}}\left[ { + \frac{{{q^2}}}{d} - \frac{{qQd}}{{{D^2}}}} \right]\)

Step-by-step Solution

Detailed explanation

\(U_{total}\) \(=\) \(U_{self\,of\, dipole} + U_{interaction}\) \(=-\frac{k q^{2}}{d}-\left(\frac{k Q}{D^{2}}\right) q d\) \(=-k\left[\frac{q^{2}}{d}+\frac{q Q d}{D^{2}}\right]\)
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