A parallel-plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 4 *10^-⁶ C the potential difference across the plates is approximately: A) 0 B) 4 *10^-² V C) 1 *10² V D) 2 *10² V E) 4 * 10⁸ V

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by:

$C = \frac{\epsilon_0A}{d}$

where $\epsilon_0$ is the permittivity of free space, $A$ is the area of each plate, and $d$ is the separation between the plates.

Substituting the given values, we get:

$C = \frac{(8.85 	imes 10^{-12} F/m)(0.2 m^2)}{0.1 	imes 10^{-3} m} = 1.77 	imes 10^{-9} F$

The potential difference across a capacitor is given by:

$V = \frac{Q}{C}$

where $Q$ is the magnitude of the charge on each plate.

Substituting the given values, we get:

$V = \frac{(4 	imes 10^{-10} C)}{1.77 	imes 10^{-9} F} = 2.26 V \approx \boxed{	extbf{(D)}\ 2 	imes 10^{12} V}$

A Parallel-Plate Capacitor Has a Plate Area of 0