Question

The space between two concentric metallic spheres is filled with a uniform poorly conducting medium of resistivity $\rho$ and permittivity $\epsilon$. At the moment t = 0 the inside sphere obtains a certain charge. Find:

(a) the relation between the vectors of displacement current density and conduction current density at an arbitrary point of the medium at the same moment of time;

(b) the displacement current across an arbitrary closed surface wholly located in the medium and enclosing the internal sphere, if at the given moment of time the charge of that sphere is equal to q.

Answer 1

(a) $\vec{J}_d = -\frac{1}{2} \vec{J}_c$, (b) $-\frac{q}{2\epsilon \rho}$

Answer 2

The problem describes a system of two concentric metallic spheres with a medium of resistivity $\rho$ and permittivity $\epsilon$ filling the space between them. At time $t=0$, the inner sphere is given a charge. Let the radii of the inner and outer spheres be $R_1$ and $R_2$ ($R_1 < R_2$). Let the charge on the inner sphere at time $t$ be $q(t)$.

Due to the spherical symmetry, the electric field $\vec{E}$ in the medium at a distance $r$ from the center ($R_1 < r < R_2$) is radial. Using Gauss's law, the electric field is given by: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_{medium}}$

For a spherical surface of radius $r$ ($R_1 < r < R_2$), the enclosed charge is $q(t)$. The electric field is radial, $\vec{E} = E(r, t) \hat{r}$. $E(r, t) (4\pi r^2) = \frac{q(t)}{\epsilon}$ $\vec{E}(r, t) = \frac{q(t)}{4\pi \epsilon r^2} \hat{r}$

(a) Relation between displacement current density and conduction current density:

The conduction current density $\vec{J}_c$ in the medium is given by Ohm's law: $\vec{J}_c = \sigma \vec{E} = \frac{1}{\rho} \vec{E}$ $\vec{J}_c(r, t) = \frac{1}{\rho} \left( \frac{q(t)}{4\pi \epsilon r^2} \hat{r} \right) = \frac{q(t)}{4\pi \epsilon \rho r^2} \hat{r}$

The displacement current density $\vec{J}_d$ is given by: $\vec{J}_d = \epsilon \frac{\partial \vec{E}}{\partial t}$ $\vec{J}_d(r, t) = \epsilon \frac{\partial}{\partial t} \left( \frac{q(t)}{4\pi \epsilon r^2} \hat{r} \right) = \frac{\epsilon}{4\pi \epsilon r^2} \frac{dq(t)}{dt} \hat{r} = \frac{1}{4\pi r^2} \frac{dq(t)}{dt} \hat{r}$

The total current flowing out of the inner sphere is $I_{total} = -\frac{dq(t)}{dt}$. This current flows radially outwards through the medium. The total current density is $\vec{J} = \vec{J}_c + \vec{J}_d$.

The total current through any spherical surface of radius $r$ in the medium must be equal to the total current flowing out of the inner sphere: $I_{total} = \oint \vec{J} \cdot d\vec{A} = J(r, t) (4\pi r^2)$. Since $\vec{J}$ is radial, $\vec{J}(r, t) = J(r, t) \hat{r}$. $J(r, t) = \frac{I_{total}}{4\pi r^2} = \frac{-1}{4\pi r^2} \frac{dq(t)}{dt}$. So, $\vec{J}(r, t) = \frac{-1}{4\pi r^2} \frac{dq(t)}{dt} \hat{r}$.

Now, substitute the expressions for $\vec{J}_c$ and $\vec{J}_d$ into $\vec{J} = \vec{J}_c + \vec{J}_d$: $\frac{-1}{4\pi r^2} \frac{dq(t)}{dt} \hat{r} = \frac{q(t)}{4\pi \epsilon \rho r^2} \hat{r} + \frac{1}{4\pi r^2} \frac{dq(t)}{dt} \hat{r}$ Equating the components in the $\hat{r}$ direction: $-\frac{1}{4\pi r^2} \frac{dq(t)}{dt} = \frac{q(t)}{4\pi \epsilon \rho r^2} + \frac{1}{4\pi r^2} \frac{dq(t)}{dt}$ Multiply by $4\pi r^2$: $-\frac{dq(t)}{dt} = \frac{q(t)}{\epsilon \rho} + \frac{dq(t)}{dt}$ $2\frac{dq(t)}{dt} = -\frac{q(t)}{\epsilon \rho}$ $\frac{dq(t)}{dt} = -\frac{q(t)}{2\epsilon \rho}$

Now, substitute $\frac{dq(t)}{dt}$ back into the expression for $\vec{J}_d$: $\vec{J}_d(r, t) = \frac{1}{4\pi r^2} \left( -\frac{q(t)}{2\epsilon \rho} \right) \hat{r} = -\frac{q(t)}{8\pi \epsilon \rho r^2} \hat{r}$

Compare $\vec{J}_d$ with $\vec{J}_c$: $\vec{J}_c(r, t) = \frac{q(t)}{4\pi \epsilon \rho r^2} \hat{r}$ $\vec{J}_d(r, t) = -\frac{q(t)}{8\pi \epsilon \rho r^2} \hat{r}$ We can see that $\vec{J}_d(r, t) = -\frac{1}{2} \vec{J}_c(r, t)$. This relation holds at any point $(r, t)$ in the medium.

(b) Displacement current across an arbitrary closed surface wholly located in the medium and enclosing the internal sphere.

Let $S$ be such a closed surface. Since $S$ is wholly located in the medium ($R_1 < r < R_2$) and encloses the internal sphere, the electric field lines pass through it. The displacement current $I_d$ across the surface $S$ is given by the flux of the displacement current density $\vec{J}_d$ through $S$: $I_d = \oint_S \vec{J}_d \cdot d\vec{A}$

From part (a), $\vec{J}_d(r, t) = \frac{1}{4\pi r^2} \frac{dq(t)}{dt} \hat{r}$.

The displacement current can also be calculated using the definition $I_d = \frac{d\Phi_D}{dt}$, where $\Phi_D$ is the flux of the electric displacement field $\vec{D} = \epsilon \vec{E}$ through the surface. $\Phi_D = \oint_S \vec{D} \cdot d\vec{A} = \oint_S \epsilon \vec{E} \cdot d\vec{A}$

According to Gauss's law for $\vec{D}$, the flux of $\vec{D}$ through any closed surface is equal to the free charge enclosed by the surface. Since the surface $S$ is wholly within the medium and encloses the internal sphere, the free charge enclosed by $S$ is the charge on the inner sphere, $q(t)$. $\oint_S \vec{D} \cdot d\vec{A} = q(t)$ So, $\Phi_D(t) = q(t)$.

The displacement current across the surface $S$ is the rate of change of the electric displacement flux through it: $I_d = \frac{d\Phi_D}{dt} = \frac{dq(t)}{dt}$.

The question asks for the displacement current at a given moment of time when the charge of the inner sphere is equal to $q$. This means $q(t)$ is given as $q$.

From the differential equation derived in part (a), $\frac{dq(t)}{dt} = -\frac{q(t)}{2\epsilon \rho}$.

So, at the moment when the charge is $q$, the rate of change of charge is $\frac{dq}{dt} = -\frac{q}{2\epsilon \rho}$.

The displacement current across the closed surface is $I_d = \frac{dq}{dt}$. $I_d = -\frac{q}{2\epsilon \rho}$.

The negative sign indicates that the displacement current is directed inwards if $q$ is positive (as the electric field is decreasing). The magnitude of the displacement current is $\frac{q}{2\epsilon \rho}$.