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(B) For a large separation $L$,a charged capacitor can be treated as an electric dipole.The dipole moment $p$ of a parallel plate capacitor is given b

Vedclass · Accepted Answer

$d^2/L^4$

Vedclass · Answer

(B) For a large separation $L$,a charged capacitor can be treated as an electric dipole.The dipole moment $p$ of a parallel plate capacitor is given by $p = qd$,where $q$ is the charge on the plates and $d$ is the separation between them. Thus,$p \propto d$.The force of interaction $F$ between two dipoles separated by a distance $L$ is given by $F \propto \frac{p_1 p_2}{L^4}$.Substituting $p \propto d$,we get $F \propto \frac{d \cdot d}{L^4} = \frac{d^2}{L^4}$.Therefore,the force of interaction is proportional to $d^2/L^4$.

Two charged parallel plate capacitors,each with separation between plates equal to $d$,are separated by a large distance $L >> d$. Then the force of interaction between them is proportional to

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