# Electrostatic Pressure Concept

There was a Question bothering me.

I tried solving it But couldn't So I finally went up to my teacher asked him for help . He told me that there was a formula for Electrostatic pressure $\rightarrow$

$$\mbox{Pressure}= \frac{\sigma^2}{2\epsilon_0}$$

And we had just to multiply it to the projected area = $\pi r^2$

So what is it actually.Can someone Derive it/Explain it please.

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I haven't seen the term electrostatic pressure used explicitly before, but I can explain how to think about the problem.

You need to consider the total force on each hemisphere, which is of course the integral over the sphere of the (vector) force per unit area. Take, then, a surface element $dA$, with charge $\sigma dA$. As is nicely explained by Purcell, the force on such a surface element is given by the average of the electric field inside and outside. Since the field inside vanishes, the total force on the surface element is then $$d\mathbf{F}=\frac{1}{2}\sigma dA\times\frac{4\pi R^2\sigma}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{R^2}=\frac{\sigma^2}{2\epsilon_0}\hat{\mathbf{r}}\,dA.$$ By symmetry, the total force on each hemisphere will be along the axis of the problem, which I take in the $z$ direction. This total force will then be $$\mathbf{F}=\int d\mathbf{F}=\hat{\mathbf{z}}\int\frac{\sigma^2}{2\epsilon_0}\hat{\mathbf{z}}\cdot\hat{\mathbf{r}}dA=\hat{\mathbf{z}}\frac{\sigma^2}{2\epsilon_0}R^2\int\cos(\theta)d\Omega=\frac{\sigma^2\pi R^2}{2\epsilon_0}\hat{\mathbf{z}}.$$

The effect is indeed like having a gas inside exerting an outward pressure $p=\frac{dF}{dA}=\frac{\sigma^2}{2\epsilon_0}$, but this is hardly general - it depends on the precise, global arrangement of charges of this particular problem, while giving the impression of being a purely local thing (since it depends only on the "local" charge density, which is of course also a global parameter). If you do accept this "pressure" then yes, the total force is this constant pressure times the area vector of the surface, which is $\pi R^2\hat{\mathbf{z}}$.

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Can you please explain it to me why is there a factor of 1/2 in calculation of $dF$, or how is that a consequence of everaging the electric field there. – Satwik Pasani Jul 22 '13 at 17:06
Can the field averaged over inside and outside not be a 'quadratic average' rather than the factor of 1/2 that appears when the field increases linearly from inside. – Satwik Pasani Jul 22 '13 at 17:30

I think you can do this by dimensional analysis. I'll do the calculation because your professor almost certainly won't accept it, so it's not cheating :-)

We have the three quantities 1/$\epsilon_0$, $\sigma^2$ and $R^n$, where we don't know $n$, and the product has to have the dimensions of force. The dimensions are:

$1/\epsilon_0 = L^3MT^{-4}A^{-2} = L^3MT^{-4}Q^{-2}T^2 = L^3MT^{-2}Q^{-2}$

$\sigma^2 = Q^2L^{-4}$

$R^n = L^n$

and of course force has dimensions $MLT^{-2}$. Multiply together $1/\epsilon_0$, $\sigma^2$ and $R^n$ and set the dimensions equal to $MLT^{-2}$ and you get:

$L^3MT^{-2}Q^{-2}Q^2L^{-4}L^n = MLT^{-2}$

which simplifies to:

$ML^{-1}L^nT^{-2} = MLT^{-2}$

and therefore $n = 2$ and the answer is A.

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If you charge up a body, you get either an excess or deficiency of electrons on the body. It must be obvious to you that if we keep a bunch of similar charges close each charge will experience a repulsion. This repulsion per unit area of the body can be termed electrostatic pressure. To solve this problem I can give you three hints to solve this problem for which you won't be needing any calculus.

1) A charge cannot interact with its own electrostatic field.

2) Whenever you move across the surface of a conductor there is a discontinuity in the field around it.

3) Force divided by the surface normal to the force will give you the pressure on the surface.

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