Derive an expression for the potential energy of an electric dipole of dipole moment $\vec{P}$ placed in a uniform electric field. **OR** Show that energy stored in a parallel plate capacitor is $\frac{1}{2}QV$.

> [!TLDR] > The potential energy is found as the external work done in rotating the dipole from a reference orientation ($90°$ to $\vec E$) to angle $\theta$, giving $U=-pE\cos\theta$. > …

Derive an expression for the potential energy of a

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\vec{P}

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The potential energy of a dipole in a u \dots

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Question of 54

\vec{P}

Nagaland NbseNagaland Board of School Education 2022Subjective· 3mImportance★★★★★est

0% · 0/54 Questions

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

✓Final answer

The potential energy of a dipole in a u \dots

Q.Derive an expression for the potential energy of an electric dipole of dipole moment $\vec{P}$ placed in a uniform electric field. OR Show that energy stored in a parallel plate capacitor is $\frac{1}{2}QV$ .

Q.Derive an expression for the potential energy of an electric dipole of dipole moment $\vec{P}$ placed in a uniform electric field. OR Show that energy stored in a parallel plate capacitor is $\frac{1}{2}QV$ .

Q.Derive an expression for the potential energy of an electric dipole of dipole moment P⃗\vec{P}P placed in a uniform electric field. OR Show that energy stored in a parallel plate capacitor is 12QV\frac{1}{2}QV21​QV.

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Q.Derive an expression for the potential energy of an electric dipole of dipole moment P⃗\vec{P}P placed in a uniform electric field. OR Show that energy stored in a parallel plate capacitor is 12QV\frac{1}{2}QV21​QV.

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Q.Derive an expression for the potential energy of an electric dipole of dipole moment $\vec{P}$ placed in a uniform electric field. OR Show that energy stored in a parallel plate capacitor is $\frac{1}{2}QV$ .

Unlock the full solution

Q.Derive an expression for the potential energy of an electric dipole of dipole moment $\vec{P}$ placed in a uniform electric field. OR Show that energy stored in a parallel plate capacitor is $\frac{1}{2}QV$ .

Unlock the full solution