Worked examples · Sheet 1

Displacement currents in capacitors

Theory used: Displacement current Displacement current in a parallel-plate capacitor Ampère-Maxwell law Maxwell's equations in local form

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A parallel-plate capacitor in vacuum has plates of area $A = 2.0 \cdot 1 0^{- 2} m^{2}$ . During charging the uniform electric field between the plates grows at the constant rate $\frac{d E}{d t} = 3.0 \cdot 1 0^{12} V/ (m \cdot s)$ . Neglecting edge effects, compute the rate of change of the electric flux through a surface parallel to the plates and lying between them, the displacement current, and the current in the connecting wires.

Φ_{E} = E A, \frac{d Φ _{E}}{d t} = A \frac{d E}{d t}, i_{s} = ε_{0} \frac{d Φ _{E}}{d t} = ε_{0} A \frac{d E}{d t}, i = i_{s}

\frac{d Φ _{E}}{d t} i_{s} i = 2.0 \cdot 1 0^{- 2} \cdot 3.0 \cdot 1 0^{12} V \cdot m/s = 6.0 \cdot 1 0^{10} V \cdot m/s = 8.85 \cdot 1 0^{- 12} \cdot 6.0 \cdot 1 0^{10} A \approx 0.53 A = i_{s} \approx 0.53 A

Edge effects are negligible, so between the plates the field is uniform and perpendicular to the surface, while outside the plates it is zero; hence the flux is the product $E A$ , and since the area is constant its derivative is $A \frac{d E}{d t}$ . By the definition of displacement current this derivative is multiplied by $ε_{0}$ , which gives about $0.53 A$ . By the theorem on the displacement current in a parallel-plate capacitor, this current is equal to the conduction current that charges the plates, so a current of 0.53 A flows in the wires.

Unit check: $C^{2} / (N \cdot m^{2}) \cdot V \cdot m/s = C^{2} \cdot V/ (N \cdot m \cdot s)$ , and since $1 V = 1 N \cdot m/C$ the result reduces to $C/s = A$ .

Exercise 2

A parallel-plate capacitor in air, with plates of area $A = 0.050 m^{2}$ placed at a distance $d = 2.0 mm$ , is connected to a generator that makes the potential difference between the plates grow at the rate $\frac{d V}{d t} = 1.5 \cdot 1 0^{6} V/s$ . Treating air as equivalent to vacuum and neglecting edge effects, compute the capacitance of the capacitor, the rate of change of the electric field between the plates and the displacement current, and verify that it coincides with the charging current.

C = \frac{ε _{0} A}{d}, E = \frac{V}{d} ⟹ \frac{d E}{d t} = \frac{1}{d} \frac{d V}{d t}

i_{s} = ε_{0} A \frac{d E}{d t} = \frac{ε _{0} A}{d} \frac{d V}{d t} = C \frac{d V}{d t}, i = \frac{d q}{d t} = C \frac{d V}{d t}

C \frac{d E}{d t} i_{s} = \frac{8.85 \cdot 1 0 ^{- 12} \cdot 0.050}{2.0 \cdot 1 0 ^{- 3}} F \approx 2.2 \cdot 1 0^{- 10} F = \frac{1.5 \cdot 1 0 ^{6}}{2.0 \cdot 1 0 ^{- 3}} V/ (m \cdot s) = 7.5 \cdot 1 0^{8} V/ (m \cdot s) = 2.21 \cdot 1 0^{- 10} \cdot 1.5 \cdot 1 0^{6} A \approx 3.3 \cdot 1 0^{- 4} A

The capacitance of the parallel-plate capacitor is $\frac{ε _{0} A}{d}$ , about $220 pF$ . Between the plates the field is uniform, so it is related to the potential difference by $E = \frac{V}{d}$ ; since the distance is constant, the rate of change of the field is that of the voltage divided by $d$ .

By the definition of displacement current, with the flux $E A$ , we have $i_{s} = ε_{0} A \frac{d E}{d t}$ , and substituting $\frac{d E}{d t}$ the product $\frac{ε _{0} A}{d} \frac{d V}{d t} = C \frac{d V}{d t}$ appears. On the other hand the charge on the positive plate is $q = C V$ , so the charging current is $\frac{d q}{d t} = C \frac{d V}{d t}$ . The two currents are equal, about $0.33 mA$ , as stated by the theorem on the displacement current in a parallel-plate capacitor.

Exercise 3

A parallel-plate capacitor in vacuum has circular plates of radius $R = 5.0 cm$ at a distance $d = 1.0 mm$ , and the alternating voltage $v (t) = V_{0} sin (ω t)$ , with $V_{0} = 50 V$ and frequency $f = 1.0 kHz$ , is applied between the plates. Neglecting edge effects, compute the amplitude of the displacement current, the amplitude of the displacement current density $ε_{0} \frac{\partial E}{\partial t}$ , and the phase shift of the displacement current with respect to the voltage.

E (t) = \frac{v ( t )}{d} = \frac{V _{0}}{d} sin (ω t) ⟹ \frac{d E}{d t} = \frac{V _{0} ω}{d} cos (ω t)

i_{s} (t) = ε_{0} π R^{2} \frac{d E}{d t} = \frac{ε _{0} π R ^{2} V _{0} ω}{d} cos (ω t) = \frac{ε _{0} π R ^{2} V _{0} ω}{d} sin (ω t + \frac{π}{2})

i_{s, 0} = \frac{ε _{0} π R ^{2} V _{0} ω}{d}, J_{s, 0} = \frac{i _{s, 0}}{π R ^{2}} = \frac{ε _{0} V _{0} ω}{d}

ω J_{s, 0} i_{s, 0} = 2 π \cdot 1.0 \cdot 1 0^{3} rad/s \approx 6.28 \cdot 1 0^{3} rad/s = \frac{8.85 \cdot 1 0 ^{- 12} \cdot 50 \cdot 6.28 \cdot 1 0 ^{3}}{1.0 \cdot 1 0 ^{- 3}} A/ m^{2} \approx 2.8 \cdot 1 0^{- 3} A/ m^{2} = 2.78 \cdot 1 0^{- 3} \cdot π \cdot 0.05 0^{2} A \approx 2.2 \cdot 1 0^{- 5} A

v(t)

i_s(t)

Voltage

v (t)

and displacement current

i_{s} (t)

, each in units of its own amplitude, as functions of

ω t

: the displacement current reaches its maximum a quarter of a period before the voltage

The frequency is $1 kHz$ , so the corresponding wavelength, $\frac{c}{f} = 300 km$ , is enormously larger than the dimensions of the capacitor; hence at every instant the field between the plates can be computed as in electrostatics, $E = \frac{v}{d}$ , uniform. The derivative of $sin (ω t)$ is $ω cos (ω t)$ , and by the definition of displacement current, with the flux $E π R^{2}$ , we obtain a current proportional to $cos (ω t)$ , whose amplitude is the coefficient of the cosine, about $22 μ A$ .

The displacement current density is uniform between the plates, because the field is, so its amplitude is the amplitude of the current divided by the area $π R^{2}$ ; the result does not depend on the radius of the plates. Since $cos (ω t) = sin (ω t + \frac{π}{2})$ , the displacement current leads the voltage by a quarter of a period, that is, by $0.25 ms$ , as the graph shows. This is the same phase shift as the conduction current in a capacitor driven by an alternating voltage, consistently with the fact that the two currents are equal.