A parallel plate capacitor has plates of area 2.0 × 10^-³ m² and plate separation 1.0 × 10^-⁴ m. Air fills the volume between the plates. What potential difference is required to establish a 3.0 $\mu$C charge on the plates?

Question

Quizplus · Accepted Answer

The capacitance of the parallel plate capacitor is given by:

C = εA/d

where ε is the permittivity of air, A is the area of the plates, and d is the separation between the plates.

Plugging in the given values, we get:

C = (8.85 × 10^-12 F/m)(2.0 × 10^-1 m^2)/(1.0 × 10^-4 m)
C = 1.77 × 10^-7 F

The potential difference required to establish a charge q on the plates is given by:

V = q/C

Plugging in the given values, we get:

V = (3.0 × 10^-6 C)/(1.77 × 10^-7 F)
V = 16.9 V

The closest answer choice to 16.9 V is C) 1.7 × 10^14 V (which should be 1.7 × 10^4 V).

A Parallel Plate Capacitor Has Plates of Area 2 μ\muμ C Charge on the Plates? A)9

A Parallel Plate Capacitor Has Plates of Area 2 $\mu$ C Charge on the Plates?
A)9