Question:
Two electric charges \(+3.2\times10^{-19}\,\text{C}\) and \(-3.2\times10^{-19}\,\text{C}\) are placed \(2.4\,\text{\AA}\) apart to form an electric dipole. It is placed in a uniform electric field of intensity \(4\times10^5\,\text{V m}^{-1}\). The work done to rotate the electric dipole from equilibrium position by \(180^\circ\) is:
Two electric charges \(+3.2\times10^{-19}\,\text{C}\) and \(-3.2\times10^{-19}\,\text{C}\) are placed \(2.4\,\text{\AA}\) apart to form an electric dipole. It is placed in a uniform electric field of intensity \(4\times10^5\,\text{V m}^{-1}\). The work done to rotate the electric dipole from equilibrium position by \(180^\circ\) is:
Show Hint
Potential energy of dipole:
\[
U=-pE\cos\theta
\]
Hence rotating dipole from stable equilibrium \((0^\circ)\) to unstable equilibrium \((180^\circ)\):
\[
W=2pE
\]
Updated On: Jun 17, 2026
- \(3\times10^{-25}\,\text{J}\)
- \(6\times10^{-23}\,\text{J}\)
- \(12\times10^{-23}\,\text{J}\)
- Zero
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The Correct Option is B
Solution and Explanation
Dipole moment:
\[
p=qd
\]
Given:
\[
q=3.2\times10^{-19}\,\text{C}
\]
\[
d=2.4\times10^{-10}\,\text{m}
\]
Thus:
\[
p=(3.2\times10^{-19})(2.4\times10^{-10})
\]
\[
p=7.68\times10^{-29}\,\text{C m}
\]
Electric field:
\[
E=4\times10^5\,\text{V m}^{-1}
\]
Work done in rotating dipole from \(0^\circ\) to \(180^\circ\):
\[
W=2pE
\]
\[
W=2(7.68\times10^{-29})(4\times10^5)
\]
\[
W=6.14\times10^{-23}\,\text{J}
\]
\[
\boxed{6\times10^{-23}\,\text{J}}
\]
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