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A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2~ ?$$

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  • $\begingroup$ Suppose to increase the distance between the plates of $\Delta x$ when the capacitor is charged and isolated from the generator, moving a plate and keeping fixed the other. What is the increment of energy stored in the capacitor? This increment is due to the work $F \Delta x$ of the attractive force $F$ between the plates... $\endgroup$ – Valter Moretti Apr 17 '16 at 7:55
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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 $$F= \frac{1}{2} QE \:.$$

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    $\begingroup$ I like the way the equation half way through your answer ends in 1/2 QEd presumably because you're halfway done. :) $\endgroup$ – M. Enns Apr 17 '16 at 14:11
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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 $1/2 QE$.

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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 $+\infty$.
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You would expect that the force would not depend on whether or not the capacitor was connected to a constant voltage source $V$.

If the capacitor $C = \dfrac {\epsilon_o A}{d}$ is connected to a constant voltage source the the energy stored in the capacitor is $U = \frac 1 2 CV^2 = \dfrac 1 2 \dfrac{\epsilon_o A}{d}V^2$.

The energy stored in the capacitor changes by $\Delta U$ if the separation of the plates $d$ increases by an amount $\Delta d$.

$\Delta U = - \dfrac 1 2 \left( \dfrac {\epsilon_o A}{d}\right)V \dfrac V d \Delta d = - \dfrac 1 2 Q E \Delta d $

A decrease in stored energy is because the capacitance has decreased at constant voltage.

$Q=CV= \dfrac {\epsilon_o A}{d} V\Rightarrow \Delta Q = - \dfrac {\epsilon A}{d^2} V \Delta d$

Thus the charge on the capacitor decreases and for the change $\Delta Q$ to flow through the voltage source the amount of work which must be done is $\Delta Q V $

$\Delta Q V = \left( \dfrac {\epsilon_o A}{d}\right)V \dfrac V d \Delta d = Q E \Delta d $

The net amount of work done by the external force $F$ is $F \Delta d$ and this is the magnitude of the attractive force between the plates.

$F\Delta x = Q E \Delta d - \dfrac 1 2 Q E \Delta d \Rightarrow F = \frac 1 2 QE$

Note that in the constant voltage case there is a decrease in the energy stored in the capacitor but at the same time work must be done in moving charge through the constant voltage source.

In the constant charge case increasing the separation of the plates increases the potential difference across the plates so more energy is stored in the capacitor and work must be done to do this.
For the constant change case $Q=CV = \dfrac {\epsilon_o A}{d}V$ so in the derivation the electric field $E$ is constant.

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protected by Qmechanic Apr 17 '16 at 12:55

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