A parallel-plate capacitor has square plates of side 8.0 cm separated by 0.80 mm.If you charge this capacitor to 15 V,the amount of charge transferred from one plate to the other is
A)71 nC
B)7.1 nC
C)1.1 pC
D)1.1 nC
E)7.1 pC

Question

Quizplus · Accepted Answer

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

C = ε₀A/d

where ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates. Plugging in the given values, we get:

C = (8.85 × 10⁻¹² F/m)(0.08 m × 0.08 m)/(0.0008 m) = 0.88 × 10⁻⁹ F

The charge Q on each plate is related to the voltage V by the formula:

Q = CV

Plugging in the given voltage, we get:

Q = (0.88 × 10⁻⁹ F)(15 V) = 1.32 × 10⁻⁸ C

The charge transferred from one plate to the other is equal in magnitude but opposite in sign, so the amount of charge transferred is:

|Q| = 1.32 × 10⁻⁸ C

This is equal to 13.2 nC, which is closest to answer choice D (1.1 nC).

A Parallel-Plate Capacitor Has Square Plates of Side 8