If charge on each plate of a parallel plate c | vedclass

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Question

Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of $	riangle x .$ Hence, work done by the force to do

Vedclass · Accepted Answer

$\frac{1}{2}\,QE$

Vedclass · Answer

Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of $	riangle x .$ Hence, work done by the force to do so $=F 	riangle x$

As a result, the potential energy of the capacitor increases by an amount given as $u A 	riangle x$

Where,

$u=$ Energy density

$A=$ Area of each plate

$d=$ Distance between the plates

$V=$ Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e...

$F 	riangle x=u A 	riangle x$

$F=u A=\left(\frac{1}{2} \in_{0} E^{2}ight) A$

Electric intensity is given by, $E=\frac{V}{d}$

$	herefore F=\frac{1}{2} \in_{0}\left(\frac{V}{d}ight) E A=\frac{1}{2}\left(\in_{0} A \frac{V}{d}ight) E$

However, capacitance, $C=\frac{\epsilon_{0} A}{d}$

$	herefore F=\frac{1}{2}(C V) E$

Charge on the capacitor is given by, $Q=C V$

$	herefore F=\frac{1}{2} Q E$

The physical origin of the factor, $\frac{1}{2},$ in the force formula lies in the fact that just outside the conductor, field is $\mathrm{E}$ and inside it is zero. Hence, it is the average value,

$\frac{E}{2},$ of the field that contributes to the force.

If charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of electric field between the plates is $E$ then force on each plate of a parallel plate capacitor will be

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