# Attractive force between capacitor plates

The E field due to each plate is $E/2$ and hence the total field between the plate is $E$. But a plate won't exert force on itself, so the E field experienced by a plate is $E/2$ only. Multiplying by charge gives you the force, hence $1/2 QE$.

The existing answers tell you why $F= \frac{1}{2} QE$ is right, but I think it's important to say why $F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}$ is wrong.

Coulomb's law is not easily applicable here because the plates are not point charges. In particular, their sizes are not negligible (indeed, much larger) than the distance between them.

It would still be valid to apply Coulomb's law per charge carrier and take vector sum:

$\int_{x_1=0}^w\int_{y_1=0}^l\int_{x_2=0}^w\int_{y_2=0}^l{\frac{1}{4\pi\epsilon_0}\frac{\sigma(x_1,y_1)\sigma(x_2,y_2)}{(x_2-x_1)^2+(y_2-y_1)^2+d^2}\frac{d}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+d^2}}dx_1dy_1dx_2dy_2} \approx$

$\int_{x_1=0}^w\int_{y_1=0}^l\int_{x_2=-\infty}^{\infty}\int_{y_2=-\infty}^\infty{\frac{1}{4\pi\epsilon_0} (\frac{Q}{S})^2 d \frac{1}{{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+d^2}}^3}dx_1dy_1dx_2dy_2} =$

$S \frac{1}{4\pi\epsilon_0} (\frac{Q}{S})^2 d \int_{x=-\infty}^{\infty}\int_{y=-\infty}^\infty{ \frac{1}{{\sqrt{x^2+y^2+d^2}}^3}dx dy} =$

$S \frac{1}{4\pi\epsilon_0} (\frac{Q}{S})^2 d \frac{2 \pi}{d} = \frac{1}{2} Q \frac{\sigma}{\epsilon_0} = \frac{1}{2} Q E$

where $S = w l$ are area, width and length of the (rectangular) platters and $\sigma(x,y)$ is the charge density for a given point on a platter.

The "$\approx$" above is based on two assumptions:

• Charge is distributed evenly across the platters: $\sigma(x,y) = \frac{Q}{S}$
• $w$ and $l$ are sufficiently large compared to $d$ that the contribution of the far-away parts of the platter to the force becomes negligible. This lets us integrate from $-\infty$ to $+\infty$.

The energy of the capacitor is $U= \frac{\epsilon_0}{2} S\,\mathrm d E^2$ where $S$ is the area of a plate. If we increase of $\Delta d$ the distance of, say, the right plate from the left one, keeping fixed the charge $Q$ on each plate, $E$ does not change and we find a variation of energy $$\Delta U = \frac{\epsilon_0}{2} S E^2 \Delta d = \frac{\epsilon_0}{2} SE \: E \Delta d= \frac{1}{2} Q \: E \Delta d\:.$$ This variation of energy, up to a sign, is due to the electric work $F \Delta d$, in turn, it is due to the force $F$ the left plate applies on the charges of the right plate. Therefore $$F= \frac{1}{2} QE \:.$$