Displacement current and a charging capacitor

A gap in the circuit

A camera flash charges a capacitor from a small battery. A current flows in the wires. Between the two plates there is air, or nothing at all, and no charge can cross that gap. Yet the current in the wire runs up to a plate and stops there. Does the circuit have a break in it?

This lesson shows that something does cross the gap. It is not a charge. It is a changing electric field. The idea comes from James Clerk Maxwell, around 1861 to 1865, and it leads to the electromagnetic wave.

Ampere's law, and a puzzle

For steady currents, Ampere's law says

Draw a closed loop. Add up $B$ along it, step by step. The sum is $\mu_0$ times the current that passes through a surface bounded by the loop. The law holds for any surface with that rim. That word is the trouble.

Take a circular loop round the wire, on the left of the left plate. We choose two surfaces that have this loop as their rim.

1. S1 is a flat disc across the loop. The wire goes through it, so the current through S1 is $I$. Ampere's law gives $\mu_0 I$.
2. S2 is a bowl with the same rim. It bulges to the right, past the plate, and its bottom lies in the gap. No wire goes through it, so the current through S2 is zero. Ampere's law gives $0$.

The same loop and the same magnetic field give two different answers. Something is missing from the law.

Maxwell's remedy

Maxwell saw that something does go through S2: an electric field. While charge $Q$ gathers on the plates, the field between them grows. For two plates of area $A$ the field is $E = \dfrac{Q}{\varepsilon_0 A}$, so the electric flux through the gap is

Its rate of change is $\dfrac{d\Phi_E}{dt} = \dfrac{1}{\varepsilon_0}\dfrac{dQ}{dt} = \dfrac{I}{\varepsilon_0}$. So the quantity $\varepsilon_0\,d\Phi_E/dt$ equals the charging current $I$. Maxwell gave it a name, the displacement current:

He added it to Ampere's law. The new law is called the Ampere-Maxwell law:

Here $I_c$ is the conduction current, the current of moving charges. Now the two surfaces agree. Through S1, $I_c = I$ and $I_d = 0$. Through S2, $I_c = 0$ and $I_d = I$. Both give $\mu_0 I$.

A word of care. The displacement current is not a flow of charge. Nothing moves across the gap. The name only means that a changing electric flux makes a magnetic field, just as a current does. The sum $I_c + I_d$ is the same at every point of the circuit. In the wire it is all $I_c$. In the gap it is all $I_d$. In a copper wire $I_d$ is far too small to matter.

The field between circular plates

Let the plates be discs of radius $R$, charged by a steady current $I$. The field $E$ is the same everywhere between them, so the displacement current is spread evenly over the area $\pi R^2$. A circular loop of radius $r < R$, centred on the axis, takes in only the fraction $r^2/R^2$ of it. Ampere-Maxwell gives $B\cdot 2\pi r = \mu_0 I\,r^2/R^2$, so

Outside the plates, the loop takes in all of $I_d$, and the field is the same as for a long straight wire. At $r = R$ the two formulas agree, and $B$ is largest there.

Set the distance to $R$ and read the largest field. Then make the plates smaller. What happens to that field?

A worked example. A capacitor has circular plates of radius $5.0$ cm. A steady current of $0.80$ A charges it. Use $\mu_0/2\pi = 2 \times 10^{-7}\ \text{T m/A}$ and $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$.

1. The plate area. $A = \pi R^2 = \pi \times (0.050\ \text{m})^2 = 7.85 \times 10^{-3}\ \text{m}^2$.
2. How fast $E$ must change. We want $dE/dt$, so $\dfrac{dE}{dt} = \dfrac{I_d}{\varepsilon_0 A}$. First, $\varepsilon_0 A = 8.85 \times 10^{-12} \times 7.85 \times 10^{-3} = 6.95 \times 10^{-14}$ in SI units. Then $\dfrac{dE}{dt} = \dfrac{0.80\ \text{A}}{6.95 \times 10^{-14}} = 1.15 \times 10^{13}\ \text{V/(m s)}$. The number is huge because $\varepsilon_0$ is tiny.
3. The field at $r = 3.0$ cm. The loop takes in the fraction $\left(\dfrac{3.0}{5.0}\right)^2 = 0.36$ of $I_d$, which is $0.288$ A. So $B = \dfrac{\mu_0 I_d}{2\pi r} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.288\ \text{A}}{0.030\ \text{m}} = 1.92\ \mu\text{T}$.
4. At the edge, $r = R$. $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.80\ \text{A}}{0.050\ \text{m}} = 3.20\ \mu\text{T}$.
5. At $r = 10.0$ cm, outside. $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.80\ \text{A}}{0.100\ \text{m}} = 1.60\ \mu\text{T}$.

An alternating supply. If the voltage across a parallel-plate capacitor of plate distance $d$ is $V$, then $E = V/d$ and $I_d = \varepsilon_0 A\,\dfrac{1}{d}\dfrac{dV}{dt}$. Since $C = \varepsilon_0 A/d$,

This is the current that an AC source seems to push "through" a capacitor. It is a displacement current in the gap.

Other books. A-Level and IB Physics do not ask for the displacement current. It belongs to the Indian Class 12 syllabus and to first-year university physics.

Practise with new numbers

Try it in Python

The program charges a capacitor with circular plates by a steady current. It finds how fast the field must grow, checks that the displacement current equals the charging current, and lists the magnetic field at five distances.

Common mistakes

• Thinking charge crosses the gap. No charge does. The displacement current is a changing electric flux, and it makes a magnetic field like a current.
• Forgetting the fraction inside the plates. A loop with $r < R$ takes in only $r^2/R^2$ of the displacement current. The field there rises in a straight line from zero at the axis.
• Using the current of the wire for every loop. Outside the plates ($r > R$), the loop takes in all of $I_d$, and $B = \mu_0 I/2\pi r$. Inside, only a part.
• Mixing up the two parts of the law. The conduction current $I_c$ is moving charge. The displacement current $I_d$ is $\varepsilon_0\,d\Phi_E/dt$.

Remember these

• Ampere's law alone gives two answers for a charging capacitor. The surface through the gap has no conduction current.
• Displacement current: $I_d = \varepsilon_0\,d\Phi_E/dt$. For a charging capacitor, $I_d = I_c$.
• Ampere-Maxwell law: $\oint \vec{B}\cdot d\vec{l} = \mu_0(I_c + I_d)$.
• Between circular plates, $B = \mu_0 I r/2\pi R^2$ inside, and $B = \mu_0 I/2\pi r$ outside.
• For parallel plates, $I_d = C\,dV/dt$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why does Ampere's law, in its old form, fail for a charging capacitor?

2. A steady current of $3.0$ A charges a capacitor with circular plates of radius $4.0$ cm. What is the displacement current in the gap, and what is the magnetic field at the edge of the plates?

3. Does the displacement current mean that charge moves across the gap?

4. In the gap of a capacitor the electric field is steady. What is the displacement current?

References

• OpenStax, University Physics Volume 2, Chapter 16 (Electromagnetic Waves): the displacement current and Maxwell's equations.
• NCERT Physics Class 12, Chapter 8 (Electromagnetic Waves): the conduction and displacement currents, and the Ampere-Maxwell law.