# Homework Help: A Charged Sheet Between Grounded Plates

1. Feb 11, 2014

### Xeroxer

1. The problem statement, all variables and given/known data
Two infinite conducting planes are held at zero potential at z=-d and z=d. An infinite sheet with uniform charge per unit area σ is interposed between them at an arbitrary point.
a) Find the charge density induced on each grounded plane and the potential at the position of the sheet of charge.
b) Find the force per unit area which acts on the sheet of charge.

2. Relevant equations
C = $\frac{Q}{\varphi}$
∫$\rho$a$\varphi$bd3r = ∫$\rho$b$\varphi$ad3r
σ = $\frac{dQ}{dS}$

3. The attempt at a solution
There is a picture associated with the problem but all it shows is that the sheet of charge must be off-center. Because both plates are grounded $\varphi$=0 for both. Charge for both plates are non-zero. The electric field is the gradient of the potential, thus E at the plates and from z>|d| must also equal zero. I'm unclear what the electric field would be inside though. If the sheet was exactly in between the two plates the charge induced on both plates would be equal and have to balance the charge on the sheet so that E remains 0 outside. Offset, like it is I'm unclear how to solve for the induced charge density.

Potential of Plate further away from sheet (P=C-1):
0 = P11Q1+P12Q2+P13Q3

Potential of Plate closest to the sheet:
0 = P21Q1+P22Q2+P23Q3

Potential of the sheet:
$\varphi$ = P31Q1+P32Q2+P33Q3

Capacitance:
C = $\frac{Q}{\varphi}$
Where:
$\varphi$ = P31Q1 + P32Q2 + P33Q3 - (P11Q1 + P12Q2 + P13Q3) - (P21Q1 + P22Q2 + P23Q3)

From here I'm thinking there must be a way to solve for each individual Q in a nice manner, but I'm stuck. Any help would be much appreciated.

2. Feb 12, 2014

### Simon Bridge

I'd have started from the physics - what would the electric field due to an infinite sheet of charge normally be?
What is the effect of adding a grounded sheet of conductor? (the ground provides charges to the conductor to do what to the electric field?)

3. Feb 12, 2014

### Xeroxer

The electric field from the sheet would be E=$\frac{σ}{2ε0}$. The grounded sheets means that the potential at each plate will be zero.

E = -∇$\varphi$
$\varphi$ = -∫E dl
0 = -∫E dl (at each plate)

4. Feb 12, 2014

### Simon Bridge

How does the grounded sheet ensure that the potential at each sheet is zero?
What physically happens?

Presumably you can sketch the electric field for this situation using your existing knowledge of electrostatics?
What does it look like?

5. Feb 12, 2014

### Xeroxer

I got it. Thanks for the help. In case anyone else stumbles upon it, here is my solution:

The electric field for an infinite plane is
E = $\frac{σ}{2*ε_0}$

The generalized electric potential is then
$\varphi$ = -∫$\frac{σ}{2*ε_0}$dx = - $\frac{σ*x}{2*ε_0}$ (where x is distance from plate)

The charge distribution for each plate can be found by determining the potential at each plate
σ1 = -$\frac{σ(d-z)}{2d}$
σ2 = -$\frac{σ(d+z)}{2d}$
$\varphi$Sheet = $\frac{σ(d^2-z^2)}{2dε_0}$

To find the force just take the negative gradient of $\varphi$Sheet and times by q.

6. Feb 12, 2014

### Simon Bridge

Well done.
Aside: you'll find equations less painful using LaTeX :)

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