# Why is the force required to keep a parallel plate capacitor EQ/2?

In our electrostatics course we were talking about how if you increase the separation of the plates by a tiny amount, then as the field is still constant, the potential difference increases. This energy comes from the work you do to separate the plates. As the surface has a charge Q, and this charge is in a field E, the overall net force upwards on the bottom plate here:

Should just be EQ. Now I can follow the math and show that if you increase the seperation by dx, the potential increases by X and this is equal to the work done or the force applied times dx, but I cant see why the force we apply ends up being less than the net force on the bottom plate.

• Hi Vishal, could you mention where you're getting the $EQ/2$ formula from? May 18 '19 at 15:37

"Should just be EQ." No. the force exerted on one plate by the other is $$EQ/2.$$ A simple way of understanding the factor of 1/2 is to think of each plate as contributing half the electric field in the gap. The force on the bottom plate arises only from that part of the electric field due to the top plate.

It may help if we look more carefully at the origins of the electric fields... The surface charge (negative) on the top of the bottom plate gives rise to an electric field normal to the surface, of magnitude $$\sigma/2\epsilon_0$$ and pointing downwards in the gap above the surface and pointing upwards in the metal below the surface. This follows from Gauss's law and planar symmetry. In the vicinity of the bottom plate, the top plate gives rise to a downward field of magnitude $$\sigma/2\epsilon_0$$. This field, due to a non-local charge, is downwards both in the gap and inside the metal of the lower plate. Therefore the fields in the metal of the lower plate (and, by the same argument, the top plate) cancel to zero, as they must, or charge would be moving inside the plates. In the gap the fields add to $$\sigma/\epsilon_0$$.

• Ah, a charged particle does not experience a force due to its own electric field. So in this case the charges on the bottom plate, are in a field of effective magnitude E/2 as far as the force is concerned. May 18 '19 at 16:12
• Sorry, just posted the same answer with a delay of a minute or so without having seen yours. Will edit mine to make it look a bit different I guess! :P May 18 '19 at 16:13
• I wouldn't worry! May 18 '19 at 16:24

Bob D's answer is correct and comprehensive. I'll try to take a more formal approach to the problem.

The force is the negative gradient of the energy stored within the plates, but since we're only interested in the magnitude, $$F = \frac{dU_E}{dx}$$ where $$x$$ is the separation between the plates.

So the problem now comes down to expressing the energy of the electric field stored $$U_E$$ in terms of the separation $$x$$. Recalling that the energy density for electric fields is $$u_E = \frac{1}{2}\epsilon_0E^2,$$

and since $$E$$ is constant within the plates, which can be justified using Gauss' law, $$\oint_V\textbf{E}\cdot d\textbf{A} = \frac{q}{\epsilon_0} \\ \pi a^2E=\frac{\pi a^2\rho}{\epsilon_0}=\frac{\pi a^2Q}{\epsilon_0 A} \\ E=\frac{Q}{\epsilon_0 A},$$

where we have chosen $$V$$ to be a cylinder of radius $$a$$ across the $$+Q$$ plate, the total energy $$U_E$$ is simply the density times the volume, $$\frac{1}{2}\epsilon_0 E^2Ax.$$

And since $$E$$ is independent of $$x$$, the derivative, hence the force, is $$\frac{dU_E}{dx} = \frac{1}{2}\epsilon_0 E^2A = \frac{1}{2}EQ$$

• +1: I don't see how this is equivalent to Bob's answer tho. They seem to reach the conclusion $F=QE$ for example. May 18 '19 at 17:04

The field $$E$$ is the net field created due to the charge distribution on the $$+Q$$ sheet and the $$-Q$$ sheet. The electrostatic force on any one of the sheets would only be due to the electric field of the other sheet. Due to symmetry, it is easy to see that the field created due to either one of the sheets would be of the magnitude $$\frac{E}{2}$$ and it will be in the same direction as $$E$$. Thus, the magnitude of the electrostatic force on any one of the sheets would indeed be $$\frac{QE}{2}$$ and so will be the magnitude of the force needed to keep the sheet in equilibrium.

Edit

However, one can object that this is a cope out. We should just deal with whatever is the net electric field $$\vec{E}_{net}$$ at a point and calculate the force on a charged particle with charge $$q$$ situated at that point as $$q\vec{E}_{net}$$. This is true. So, is there a contradiction here? Well, as expected, no. The point is that even the net electric field isn't really $$E$$ on any one of the sheets. As we recalled, the field $$E$$ is the summation of an $$\frac{E}{2}$$ contribution from each sheet. But this is only true between the sheets. Precisely on the sheet, the field due to the sheet itself is indeed zero (and then on the other side of the sheet, the field due to the same sheet is again $$\frac{E}{2}$$ but in the opposite direction), so the net electric field on the sheet is indeed only due to the electric field of the other sheet. So it isn't the case that the field on the sheet is actually $$E$$ but we are ignoring the half of it via invoking the argument that the sheet shouldn't apply a force on itself. A careful calculation respects this expectation all by itself in the way we explained.

• @VishalJain I have added some more points as "Edit" to make the original heuristic argument relatively more rigorous. Have a look! :) May 18 '19 at 16:26