How Is Linear Charge Distribution a Realistic Model in Capacitor Charging?

E_M_C
Messages
43
Reaction score
0
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.
 
Physics news on Phys.org
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
 
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.
 
E_M_C said:
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.

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
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top