How Do Voltage Changes Affect Fields in a Quasi-Electrostatic Capacitor?

[NotHeisenburg]

Homework Statement

I am trying to find the electric and magnetic fields between two parallel circular plates, where one plate is grounded and the the other has a voltage that increases linearly with time. I need the E field between plates, and I can ignore fringing fields. The permativity is specified to be $\epsilon_0$.

Homework Equations

$V(t)=at$ where $a$ is a constant.
The radius of the plates is given to be $R$

The Attempt at a Solution

$E(\vec{r},t)=\frac{1}{4\pi\epsilon_0}\int\frac{(\vec{r}-\vec{r}')\rho(\vec{r}',t)d^3r'}{|\vec{r}-\vec{r}'|^3}$

$\rho(\vec{r},t)=$?

With the E field I would find the B field using

$\nabla \times B=\mu_0\vec{J}+\frac{1}{c^2}\frac{\partial E}{\partial t}$

 

[gneill]

It sounds like a simple parallel plate capacitor setup. You can find the plate area easily enough from the given radius. Do you have a value for plate separation?

 

[NotHeisenburg]

I would assume that I could just call the plate separation $d$

 

[gneill]

I would assume that I could just call the plate separation $d$

Sure. And it's well known that, ignoring edge effects, the electric field between the plates of a parallel capacitor is uniform.

 

[paranormal]

First of all, let's define what a quasi-electrostatic capacitor is. A quasi-electrostatic capacitor is a type of capacitor where the electric field is approximately constant between the two plates, meaning that the electric field does not vary significantly between the two plates. This is in contrast to a regular capacitor, where the electric field varies significantly between the two plates.

Now, in order to solve for the electric and magnetic fields between the two parallel circular plates, we need to determine the charge density, $\rho(\vec{r},t)$, which is the charge per unit volume at any given point in space and time. In this case, since one plate is grounded and the other plate has a voltage that increases linearly with time, we can assume that the charge on the plates will also increase linearly with time.

We can express the charge density as:

$\rho(\vec{r},t)= \sigma(t)\delta(\vec{r}-\vec{r}_0)$

where $\sigma(t)$ is the surface charge density on the plates and $\vec{r}_0$ is the position vector of a point on the plate.

Using this charge density, we can now solve for the electric field between the plates using the equation you provided:

$E(\vec{r},t)=\frac{1}{4\pi\epsilon_0}\int\frac{(\vec{r}-\vec{r}')\sigma(t)\delta(\vec{r}'-\vec{r}_0)d^3r'}{|\vec{r}-\vec{r}'|^3}$

Since we are ignoring fringing fields, we can assume that the charge distribution is uniform and the electric field is constant between the plates. This means that the charge density, $\sigma(t)$, is also constant and equal to the charge on the plates divided by the area of the plates.

Substituting this into the equation above, we get:

$E(\vec{r},t)=\frac{\sigma(t)}{4\pi\epsilon_0}\int\frac{(\vec{r}-\vec{r}')\delta(\vec{r}'-\vec{r}_0)d^3r'}{|\vec{r}-\vec{r}'|^3}$

Since the charge on the plates is increasing linearly with time, we can express the charge density as:

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