Magnetic field for parallel plate capacitors

[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.

 

[qbert]

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

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)

How do I determine the 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.

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

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

 

[Reshma]

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

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 .