Question

A parallel plate capacitor has a plate area of 0.2 square metre and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 4 microC, the potential difference across the plates is app

Answer 1

Answer:

A parallel-plate capacitor has a plate area of 0.2 m^2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 10^6 V/m between the plates, what is the magnitude of the charge on each plate?

Empower people to live healthier lives by sharing what you know.

The voltage across the capacitor terminals is

V = E . d

where E is the electric field, and d the distance between the plates.

So,

V = 2 . 10^6 . V/m . 0.1 . 10^(-3) m = 2 . 10^2 V = 200 V

Capacitor charge is, by definition

q = C . V

where C is the capacitance

Capacitance is given by

C = epsilon A / d

where epsilon is the permittivity, A is the plate area