Kvant Physics Problem 45
A spherical capacitor consists of two concentric conducting spheres of radii $a$ and $b$, with $a<b$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 5m52s
Source on kvant.digital
Problem
F45. A spherical capacitor filled with a dielectric and charged to some potential difference discharges through its dielectric. What will be the magnetic field of the discharge currents in the space between the spheres?
Setup and Assumptions
A spherical capacitor consists of two concentric conducting spheres of radii $a$ and $b$, with $a<b$. The space between them is completely filled with a dielectric of conductivity $\sigma$ and permittivity $\varepsilon$.
At some initial moment the capacitor is charged, creating a potential difference $U(t)$ between the spheres. Because the dielectric is not perfectly insulating, a conduction current flows through the dielectric and the capacitor gradually discharges.
The required quantity is the magnetic field $\mathbf B$ in the region
$$a<r<b .$$
The dielectric is assumed homogeneous and isotropic, the discharge is quasistatic, and the geometry remains perfectly spherically symmetric. Radiation effects and inductive distortions of the electric field are neglected.
Physical Principles
The solution rests on Maxwell's equations.
The conduction current density in the dielectric satisfies Ohm's law,
$$\mathbf j=\sigma \mathbf E .$$
The electric displacement is
$$\mathbf D=\varepsilon \mathbf E .$$
The Maxwell-Ampère equation is
$$\nabla\times\mathbf B = \mu_0 \left( \mathbf j+ \frac{\partial\mathbf D}{\partial t} \right).$$
In integral form,
$$\oint \mathbf B\cdot d\mathbf l = \mu_0 \left( I_{\rm cond}+I_{\rm disp} \right),$$
where $I_{\rm cond}$ is the conduction current through the chosen surface and $I_{\rm disp}$ is the displacement current through the same surface.
The spherical symmetry of the system implies that all physical quantities depend only on the distance $r$ from the common center. The electric field is radial,
$$\mathbf E=E(r,t),\mathbf e_r.$$
The current density is also radial.
Derivation
Let $Q(t)$ be the charge on the inner sphere. The electric field in the dielectric has the same form as in an ordinary spherical capacitor,
$$E(r,t)=\frac{Q(t)}{4\pi\varepsilon r^2}.$$
The conduction current density is
$$j(r,t) = \sigma E(r,t) = \frac{\sigma Q(t)}{4\pi\varepsilon r^2}.$$
The total conduction current through any concentric spherical surface of radius $r$ is
$$I_{\rm cond} = j(r,t),4\pi r^2 = \frac{\sigma}{\varepsilon}Q(t).$$
This quantity is independent of $r$, as required by charge conservation.
Next, calculate the displacement current. Since
$$D(r,t)=\varepsilon E(r,t) = \frac{Q(t)}{4\pi r^2},$$
the electric flux through a concentric spherical surface is
$$\Phi_D = \int \mathbf D\cdot d\mathbf S = Q(t).$$
Hence
$$I_{\rm disp} = \frac{d\Phi_D}{dt} = \frac{dQ}{dt}.$$
The charge on the capacitor decreases because of the conduction current. Charge conservation for the inner sphere gives
$$\frac{dQ}{dt} = -,I_{\rm cond} = -\frac{\sigma}{\varepsilon}Q(t).$$
Consequently,
$$I_{\rm disp} = -\frac{\sigma}{\varepsilon}Q(t).$$
Comparing this with the conduction current,
$$I_{\rm disp} = -,I_{\rm cond}.$$
The sum entering Maxwell's equation is therefore
$$I_{\rm cond}+I_{\rm disp}=0.$$
Substituting into the integral form of the Maxwell-Ampère equation yields
$$\oint \mathbf B\cdot d\mathbf l=0$$
for every closed contour.
Since the system possesses complete spherical symmetry, there is no preferred direction along which a magnetic field could exist. The only field compatible with both the symmetry and the vanishing circulation is
$$\mathbf B=0.$$
Thus the conduction current and the displacement current cancel exactly as sources of magnetic field.
Result
The conduction current through any spherical surface is
$$I_{\rm cond} = \frac{\sigma}{\varepsilon}Q(t),$$
while the displacement current is
$$I_{\rm disp} = -\frac{\sigma}{\varepsilon}Q(t).$$
Their sum is identically zero,
$$I_{\rm cond}+I_{\rm disp}=0.$$
Therefore Maxwell's equation gives
$$\boxed{\mathbf B=0} \qquad (a<r<b).$$
No numerical data are given in the problem. The magnetic field is exactly
$$\boxed{B=0\ {\rm T}}.$$
Sanity Checks
The result has the correct dimensions. The magnetic field is predicted to vanish identically, so the unit is tesla and the numerical value is zero.
If the conductivity tends to zero, the conduction current disappears. Then the charge becomes constant, the displacement current also vanishes, and the magnetic field remains zero. The result is unchanged.
If the conductivity is finite, the conduction current is nonzero, but the charge decreases at precisely the rate required by charge conservation. The displacement current is equal in magnitude and opposite in sign to the conduction current, producing exact cancellation in Maxwell's equation.
The most common mistake is to calculate only the conduction current and apply Ampère's law to it. That procedure would incorrectly predict a nonzero magnetic field. The crucial step is the inclusion of the displacement current term, which is equal and opposite to the conduction current throughout the discharge.