Calculating displacement current and magnetic field.

[weezy]

1. The problem statement,

A capacitor is there in free space consisting of 2 circular plates of radius $r$ separated by a distance $z$ which is a function of time. $z(t) = z_0 + z_1 cos (\omega t)$; $z_0(<<r)$ and $z_1(<z_0)$ are constants. The separation $z(t)$ is varied in such a way that the potential difference $V_0$ between the plates remains constant.

• Calculate the displacement current density & displacement current between the plates through a concentric of radius $\frac{r}{2}$
• Calculate $\vec H$ between the plates at a distance of $\frac{r}{2}$ from the axis of the capacitor.

Now, $\oint_{\frac{r}{2}} \vec H \cdot d\vec l = \int_S \vec J \cdot d\vec a + \int_S \vec J_d \cdot d\vec a$ where $J_d$ is displacement current density. I hope to use this equation and equate $J=0$ and find out $J_d$ in the region between the plates. What I don't understand is how to use the condition $V_0 = constant$ to solve this problem. I calculated the capacitance $C(t) = \frac{A\epsilon}{z_0+z_1 cos(\omega t)}$ and that's it. I'm stuck.

EDIT 1: I think the way to go is to use $\vec E(t) = -\nabla V(t)$ where $V(t) = \frac{Q}{C(t)}$ but I'm not sure if $Q$ is also a function of time. Is it?

 

[vela]

How do you propose to find the displacement current?

 

[weezy]

How do you propose to find the displacement current?

I updated the question. See Edit 1. I think I have to find $\partial_t \vec E$ first but I'm not sure if $Q=Q(t)$.

 

[vela]

The V(t) in $\vec{E} = -\nabla V(t)$ and $V(t) = \frac{Q}{C(t)}$ aren't the same quantity. The V in the first expression is also a function of the spatial coordinates, otherwise you'll get 0 when you take the gradient. The V(t) in the second expression is the potential difference across the capacitor; you were told it's equal to a constant $V_0$. (That should answer your question about whether Q is a constant or not.)

You're on the right track though. You want to relate the electric field inside the capacitor to the potential difference across the plates.

 

[weezy]

The V(t) in $\vec{E} = -\nabla V(t)$ and $V(t) = \frac{Q}{C(t)}$ aren't the same quantity. The V in the first expression is also a function of the spatial coordinates, otherwise you'll get 0 when you take the gradient. The V(t) in the second expression is the potential difference across the capacitor; you were told it's equal to a constant $V_0$.

You're on the right track though. You want to relate the electric field inside the capacitor to the potential difference across the plates.

But the question seems contradictory to me. if $V_0$ is constant then how can a displacement current exist? Does the constant V_0 imply a fixed charge on the capacitor?

 

[vela]

The charge on the capacitor is equal to Q = C(t)V₀. Do you still think the charge is fixed?

 

[weezy]

The charge on the capacitor is equal to Q = C(t)V₀. Do you still think the charge is fixed?

Obviously no. That was my first thought! But I don't see another way to explain $V_0$ = constant. Moreover I don't even see why that is a necessary condition in this question.

 

[weezy]

SO I did this :
$- \int _{0}^{z_0+z_1cos(\omega t)} \vec E \cdot d\vec l = V_0 \implies E= -\frac{V_0}{z_0+z_1cos(\omega t)}$ or $\partial_t E = \frac{V_0z_1sin(\omega t)}{z_0+z_1cos(\omega t)}$. Is this the right way?

 

[vela]

Yes, but the denominator should be squared in your final result.

 

[weezy]

Yes, but the denominator should be squared in your final result.

Oops. Sorry for that!