Question

A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?

Answer 1

Answer:

$1.88\cdot 10^{-5} A$

Explanation:

The capacitance of a parallel plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$ (1)

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

$Q=CV$ (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

$J=\frac{Q}{t}$ (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

$J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}$

where we have

$A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2$

$d = 0.58 mm = 5.8\cdot 10^{-4} m$

$\frac{V}{t}=500,000 V/s$

Substituting into the equation, we find

$J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A$