A parallel plate capacitor has a paper dielectric having dielectric strength 8.00 kV/mm and dielectric constant 3.00.The plate area is 3000 cm² and the plate separation is 0.500 mm.If 10.0 V are applied to this capacitor,what is the charge on the capacitor?

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor with a dielectric can be calculated using the formula:

C = (ε*A)/d

where ε is the dielectric constant, A is the area of the plates, and d is the distance between them.

Plugging in the given values:

C = (3.00*8.85*10^-12*3000)/(0.500*10^-3) = 63.27 nF

The charge on a capacitor can be calculated using the formula:

Q = C*V

where V is the applied voltage.

Plugging in the given values:

Q = (63.27*10^-9)*10.0 = 632.7 nC

Converting to nanoCoulombs:

632.7 nC = 0.6327 μC

Therefore, the best choice is C) 160 nC.

A Parallel Plate Capacitor Has a Paper Dielectric Having Dielectric μ\muμ

A Parallel Plate Capacitor Has a Paper Dielectric Having Dielectric $\mu$