Question

Consider a capacitor placed in free space, consisting of two concentric circular parallel plates of radii r. The separation z between the plates oscillates with a constant frequency $\omega$, i.e., $z(t) = z_0 + z_1 cos \omega t$. Here $z_0$ and $z_1 (<z_0)$ are constants. The separation $z(t) (<< r)$ is varied in such a way that the voltage $V_0$ across the capacitor remains constant.

(a) Calculate the displacement current density and the displacement current between the plates through a concentric circle of radius r/2.

(b) Calculate the magnetic field vector between the plates at a distance r/2 from the axis of the capacitor.

Answer 1

The displacement current density is $\vec{J}_d = \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} \hat{k}$, the displacement current through a circle of radius r/2 is $I_d = \frac{\epsilon_0 \pi r^2 V_0 z_1 \omega \sin \omega t}{4 (z_0 + z_1 \cos \omega t)^2}$, and the magnetic field vector at distance r/2 is $\vec{B} = \frac{\mu_0 r}{4} \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} \hat{\phi}$.

Answer 2

The separation between the plates is $z(t) = z_0 + z_1 \cos \omega t$. The voltage across the capacitor is constant, $V_0$. The capacitance is $C(t) = \frac{\epsilon_0 A}{z(t)} = \frac{\epsilon_0 \pi r^2}{z(t)}$. The charge on the plates is $Q(t) = C(t) V_0 = \frac{\epsilon_0 \pi r^2 V_0}{z(t)}$.

The electric field between the plates is approximately uniform and perpendicular to the plates, given by $E(t) = \frac{V_0}{z(t)}$. Assuming the plates are in the xy-plane and the separation is along the z-axis, $\vec{E}(t) = \frac{V_0}{z(t)} \hat{k}$.

(a) The displacement current density is $\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$. $\vec{J}_d = \epsilon_0 \frac{\partial}{\partial t} \left( \frac{V_0}{z(t)} \hat{k} \right) = \epsilon_0 V_0 \frac{d}{dt} \left( \frac{1}{z(t)} \right) \hat{k}$. We have $z(t) = z_0 + z_1 \cos \omega t$. $\frac{d}{dt} \left( \frac{1}{z(t)} \right) = -\frac{1}{z(t)^2} \frac{dz}{dt} = -\frac{1}{(z_0 + z_1 \cos \omega t)^2} (-z_1 \omega \sin \omega t) = \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2}$. So, the displacement current density is $\vec{J}_d = \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} \hat{k}$.

The displacement current through a concentric circle of radius r/2 is given by $I_d = \int_S \vec{J}_d \cdot d\vec{A}$, where S is the area of the circle. The area vector $d\vec{A}$ is parallel to $\vec{J}_d$ (i.e., in the $\hat{k}$ direction). The area of the circle is $A' = \pi (r/2)^2 = \frac{\pi r^2}{4}$. Assuming $\vec{J}_d$ is uniform over this area (valid for $r/2 < r$), the integral becomes $I_d = J_d \cdot A'$. $I_d = \left( \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} \right) \left( \frac{\pi r^2}{4} \right) = \frac{\epsilon_0 \pi r^2 V_0 z_1 \omega \sin \omega t}{4 (z_0 + z_1 \cos \omega t)^2}$.

(b) To find the magnetic field vector between the plates at a distance $\rho = r/2$ from the axis, we use Ampere-Maxwell's law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_{enc} + I_{d, enc})$. Between the plates, there is no conduction current, so $I_{enc} = 0$. The law becomes $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d, enc}$. Due to the cylindrical symmetry, the magnetic field lines are concentric circles around the axis, and the magnitude of $\vec{B}$ depends only on the distance $\rho$ from the axis. We choose a circular Amperian loop of radius $\rho = r/2$ centered on the axis, lying in a plane between the plates. $\oint \vec{B} \cdot d\vec{l} = B(2\pi \rho)$. The enclosed displacement current $I_{d, enc}$ is the displacement current passing through the area enclosed by the loop, which is a circle of radius $\rho$. $I_{d, enc} = \int_{Area} \vec{J}_d \cdot d\vec{A} = J_d \cdot (\pi \rho^2)$, assuming $J_d$ is uniform. $I_{d, enc} = \left( \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} \right) (\pi \rho^2)$. Applying Ampere-Maxwell's law: $B(2\pi \rho) = \mu_0 \left( \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} \right) (\pi \rho^2)$. $B = \frac{\mu_0 \rho}{2} \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2}$. At a distance $\rho = r/2$: $B(r/2) = \frac{\mu_0 (r/2)}{2} \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} = \frac{\mu_0 r}{4} \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2}$. The direction of the magnetic field is azimuthal ($\hat{\phi}$) according to the right-hand rule, curling around the direction of $\vec{J}_d$. $\vec{B}(\rho=r/2, t) = \frac{\mu_0 r}{4} \epsilon_0 V_0 \frac{z_1 \omega \sin \omega t}{(z_0 + z_1 \cos \omega t)^2} \hat{\phi}$.