# Is the electric field constant between two charged plates?

1. Feb 11, 2009

### n0083

Suppose there was a voltage difference between two charged plates, is the electric field constant at all points between these plates?

2. Feb 11, 2009

### jambaugh

Yes, in the idealized case where the plates are infinite then the field is constant. For finite plates there are edge effects. As long as the width of the plates is substantially larger than the separation you can assume the field is a constant. By field of course we are talking E-field. The potential varies linearly.

3. Feb 11, 2009

### n0083

Thanks, but i don't understand why the E-field is constant?

I imagine if the point in consideration was very near one of the plates then the E-field would be stronger there as the electric field is inversely proportional to distance.
E = kQ/r.

For instance, "Is the electric field constant at all points on the line between two positionally fixed charges?" The equation E = kQ/r would suggest to me the answer is no.

What is going on here? Why is the E-field constant?

4. Feb 11, 2009

Staff Emeritus
Before dealing with two charged plates, it helps to understand one charged plate. How does the field from a single charged plate behave?

5. Feb 16, 2009

### jambaugh

The standard exposition is to consider an infinite plane with a uniform charge density. You can then use a cylinder as a gaussian surface and argue that by symmetry all the flux must pass through the ends. Note that the total flux does not depend on how long the cylinder is.

But if it helps to think in terms of component effects consider this. Imagine you are a test charge suspended over a flat level (infinite) sheet with a uniform charge distribution.

Yes the close you are to the plane the more the charge directly below you contributes to the E field. However since the charges off to the side are pushing at smaller opposite angles the more their contributions cancel out. Only their components perpendicular to the plane contribute. This decrease offsets the other increase exactly.

Ultimately remember there is a charge density and not a finite point charge at each point. The closer you get the less charge is within a thin cone below you. As you almost touch the charge within a cone below you approaches zero.

It is a tricky integral but you can (I think) add up the contributions for each infinitesimal area of the plane and see that it is independent of the distance.

Hmmm... let's see using cylindrical coordinates with surface charge density $\rho$ on the plane $z = 0$
$$(z,r,\theta)$$
figure the electrostatic potential at a point on the z-axis.

The charge of a particular differential area is $dQ = \rho r\cdot dr \cdot d\theta$.
The distance to the point on the z-axis is $R=\sqrt(r^2 + z^2)$
The contributed electrostatic force is:
$$|dE| = \frac{kdQ}{R^2}$$
But the only contribution in the vertical direction will be the z-component which is
$$dE_z = \frac{kdQ}{R^2}\cos(\phi)$$
where $\phi$ is the angle the hypotenuse makes with the z-axis.
This angle will be
$$\cos(\phi)=\frac{z}{R}$$.

Now we are ready to set up the integral:

$$E_z = \int dQ\frac{kz}{R^3} = k \rho \int \frac{z r dr d\theta}{(z^2 + r^2)^{3/2}}$$

$$= 2\pi k\rho \int_{r=0}^{\infty} \frac{z r dr}{(z^2 + r^2)^{3/2}}$$

Now use the trig substitution $r = z\tan(\alpha)$ and $dr = z\sec^2(\alpha)d\alpha$
You get:

$$E_z=2\pi k\rho \int_{\alpha=0}^{\pi/2} \frac{z^3 \tan(\alpha)\sec^2(\alpha) d\alpha}{z^3\sec^3(\alpha)}$$

$$E_z=2\pi k\rho \int_{\alpha=0}^{\pi/2}\sin(\alpha)d\alpha$$

$$E_z=2\pi k\rho[\cos(0)-\cos(\pi/2)]=2\pi k\rho$$

The z dependence exactly cancels. This because you get that extra z/R contribution from the angle off the z-axis. In other word the higher up you are the more charge is almost below you instead of way off to the sides and canceling out. Yet the farther you are from it so the E field gets neither stronger nor weaker.

Hope that helps.

Last edited: Feb 16, 2009