# Homework Help: Magnetic field for parallel plate capacitors

1. Feb 20, 2006

### Reshma

From Griffiths again!

A large parallel plate capacitor with uniform surface charge $\sigma$ on
upper plate and $-\sigma$ on lower is moving with a constant speed v.

Q1]Find the magnetic field between the plates and also above and below them.

My work:
For a surface charge distribution: $\vec K = \sigma \vec v$
Magnetic induction: $B = \frac{\mu_0 K}{2}$
Here both the top plate produces a field:
$$B = \frac{\mu_0 K}{2}$$
And the bottom plate produces a field:
$$B = -\frac{\mu_0 K}{2}$$

How do I take take into account the directions of these fields in order to calculate the field between them?

Q2] Find the magnetic force per unit area on the upper plate and its direction.

My work:
$$\vec F_{mag} = \int \left(\vec K \times \vec B\right)d\vec a$$

So force per unit area is:
$$\vec f = \vec K \times \vec B$$

Magnitude of the force would be:
$$F = \frac{\mu_0 K^2}{2}$$

How do I determine the direction?

Q3] At what speed 'v' would the magnetic force balance the electrical force?

I need complete assistance on this question.

2. Feb 20, 2006

### qbert

right-hand rule has always worked for me.

If the positive sheet is coming out of the monitor
++++++++++++++ (z-direction)
the B is ----------> (x-direction)

If the negative sheet is coming out of the monitor
the B is ------------>(x-direction)
------------------------ (z-direction)

right hand rule has always worked for me.

Take the field from the positive sheet. (Say
it's in the x-direction) the negative sheet is
moving in the z-direction. and the negative
sheet sitting below the positive one.

+++++++++++ (moving out of monitor z-hat)
------> B (x-hat)
---------------------- (moving out of monitor)

then
F = qv x B = - z-hat cross x-hat = - y-hat
thus Force from the top on the bottom
pushes it away.

similarly force on top due to bottom pushes it away.

The electrostatic force is just force due to attracting charges.
sigma^2/(2 epsilon_0)
and you want it to balance mu_0 K^2 /2 =
mu_0 sigma^2 v^2 /2

3. Feb 21, 2006

### Reshma

Thanks once again for the help!
So the magnetic force would be:
$$f_m = \frac{\mu_0 \sigma^2 v^2}{2}$$

Electrical force for the lower plate:
$$f_e = \frac{\sigma^2}{2\epsilon_0}$$

Balancing condition: $f_m = f_e$

$$\mu_0 v^2 = \frac{1}{\epsilon_0}$$
$$v = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = c$$
c = velocity of light .

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