# Force between the plates of a capacitor

As we know that the electric field between two parallel plates of a capacitor is $$E=\frac{\sigma}{\epsilon_0}$$, so the magnitude of force exerted by one plate on the other should have been $$F=QE$$. But in reality it is$$F=\frac{QE}{2}$$. How is that possible? Where have i mistaken?

Also, I would like to refer to an answer by David Z to this question:

• The more realistic explanation is that essentially all of the charge on each plate migrates to the inside surface. This charge, of area density $\sigma$, is producing an electric field in only one direction, which will accordingly have strength $\frac{\sigma}{\epsilon_0}$. But when using this explanation, you do not also superpose the electric field produced by charge on the inside surface of the other plate. Those other charges are the terminators for the same electric field lines produced by the charges on this plate; they're not producing a separate contribution to the electric field of their own. So if they're not producing a separate contribution to the electric field of their own, then why do we consider individual fields of the plates while calculating force exerted by each plate?

Suppose the surface charge densities on the bottom plate is $\sigma$ and on the top plate $-\sigma$, then the electric field due to the bottom plate is $\frac{\sigma}{2 \epsilon_0}{\bf n}$ and that due to the top plate $-\frac{\sigma}{2 \epsilon_0}{\bf n}$, where ${\bf n}$ is a unit vector pointing from the bottom plate to the top plate. This gives the total electric field between the plates as $(\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0}){\bf n} = \frac{\sigma}{\epsilon_0} {\bf n}$, which is the electric field ${\bf E}$ as expected (the fields are acting in opposite directions).

The charge on the top plate only exerts a force on the charge on the bottom plate (and vice versa) and doesn't exert a charge on itself. This gives the force acting between the two plates as just $Q \times \frac{\sigma}{2 \epsilon_0} = Q\times \frac{{\bf E}}{2}$ where $Q$ is the charge on either plate.

• 'The charge on the top plate only exerts a force on the charge on the bottom plate (and vice versa) and doesn't exert a charge on itself.' I suppose that is the point. Right? – Aaron John Sabu Apr 3 '17 at 8:57

This sentence points out your mistake (May not be the exact words):

You cannot push a basket in which you sit. --David J.Griffith

(Which is actually Newton's third law in a new form.)

That electric field $\sigma\over\epsilon_0$ is for in between the plates and used to determine the force exerted by the capacitor to some other charge inside.

If you want to calculate the force on one of the plates, then, according to the rule above, you need to ignore the charges inside your system boundary (here, all charges on the plate).

Each element of charge $\mathrm dq$ on each plate exerts a force on all other elements of charge $\mathrm dq'$ on both plates. But $\mathrm dq$ does not exert a force on itself. Summing the forces from all charges on one plate on any given charge element on that plate gives zero net force. Therefore, the only force on a given charge element is the force from the charge elements on the opposite plate.

The same approach applies when calculating the force on a charge element $\mathrm dq$ at the surface of a spherical shell of charge, due to all other charge elements making up the shell. Setting up the integral explicitly and taking the limit as the contributions from the shell include all elements except the one excluded, we find a field at the boundary of magnitude $\sigma/(2\epsilon_0)$, or half what one would obtain if using the field just outside the shell, due to the entire shell.

The key point in both problems is the same: Namely, each charge element $\mathrm dq$ acts like a point charge, and point charges do not exert forces on themselves. The resulting factor of $1/2$ is counter-intuitive, but it is correct.

A different way to obtain the same results is to consider the change in the assembly energy of the distributions under virtual displacements of the excluded charge elements. See Jackson, 3rd edition, pp. 42-43, for a terse but accurate explanation.

If air is the medium between the plates of the parallel plate capacitor, then the electrical field at the position of the grounded plate will be E=σ/2ε; and the electrical field at that place for the grounded plate itself will be E"=0, as for the grounded plate itself there will be equal but opposite amount of field produced. So net will be zero. Now, at the place of that grounded plate, net electrical field will be, E=E+E"=σ/2ε. Again, the amount of negative charge on the inward surface of the plate is σA, where A is its area.Therefore, the attractive force between them will be, F=E(σA)Or, F=((σ^2)A/2ε)Or, F= q^2/2εA [as σ=q/A]