# Linear Charge Distribution

1. Jul 10, 2012

### E_M_C

The problem is stated:

The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centers of the plates (the plates are circular). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w « a. Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.

Find the electric field between the plates, as a function of t.

I understand that I have to use Gauss' Law to find the E-field, but first I have to find the charge distribution Q(t), this is where I'm having some difficulty. After a lot of frustration, I peeked at part of the solution, and I found that the charge distribution is Q(t) = It. I assume it was arrived at as follows:

$I = \frac{dQ(t)}{dt} → \int dQ = \int I dt → Q(t) = It$

I was surprised to see that the charge distribution is a linear function, as I was expecting an exponential expression. Maybe I'm not getting the concept of a "uniform surface charge", but I don't see how a linear charge distribution is a "realistic model." For example, what happens as t → ∞? The charge distribution blows up. How is that realistic?

Any help is appreciated.

2. Jul 10, 2012

### ehild

The charge of the capacitor is the time integral of the current. If the current is constant in time the charge is a linear function of time.
Charge distribution refers to the distribution of charge along the surface of the capacitor plate. It is assumed to be homogeneous: The surface charge density is σ=Q/A where A is the area of a plate, A=a22π. The problem asks the electric field as function of time. Apply Gauss' Law.
Of course, the constant current can not be maintained for infinity: but it is possible for some finite time. There are devices "current generators" which provide constant current.

ehild

3. Jul 10, 2012

### E_M_C

Thanks ehild.

I suppose that I was just over-thinking the problem. The idea of the capacitor being attached to a current source did cross my mind, but I couldn't think past the limitations of a linear solution.

After some more careful thought, I realized that the exponential solution I was expecting would only arise when the capacitor is in series with a resistor.

4. Jul 11, 2012

### ehild

You thought it correctly in case you charge the capacitor with a voltage source with constant emf through a resistor. The current will decrease exponentially with time constant RC.

ehild