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Show that the force on each plate of a parallel plate capacitor has a magnitude equal

Question:

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $(1 / 2) Q E$, where $Q$ is the charge on the capacitor, and $E$ is the magnitude of electric field between the plates. Explain the origin of the factor $1 / 2 .$

Solution:

Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx

As a result, the potential energy of the capacitor increases by an amount given as uAΔx.

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.,

$F \Delta x=u A \Delta x$

$F=u A=\left(\frac{1}{2} \in_{0} E^{2}\right) A$

Electric intensity is given by,

$E=\frac{V}{d}$

$\therefore F=\frac{1}{2} \in_{0}\left(\frac{V}{d}\right) E A=\frac{1}{2}\left(\epsilon_{0} A \frac{V}{d}\right) E$

However, capacitance, $C=\frac{\in_{0} A}{d}$

$\therefore F=\frac{1}{2}(C V) E$

Charge on the capacitor is given by,

Q = CV

$\therefore F=\frac{1}{2} Q E$

The physical origin of the factor, $\frac{1}{2}$, in the force formula lies in the fact that just outside the conductor, field is $E$ and inside it is zero. Hence, it is the average value, $\frac{E}{2}$, of the field that contributes to the force.

 

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