Question

he plates each have an area of 2 x 10-3 m2, and the spacing between the plates is 6 x 10-3 m. There is no dielectric between the plates. What is the charge on the capacitor

Answer 1

Complete question:

The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10⁻³ m², and the spacing between the plates is 6 x 10⁻³ m. There is no dielectric between the plates. What is the charge on the capacitor?

Answer:

The charge on the capacitor is 1.77nC

Step-by-step explanation:

Given;

magnitude of electric field between the plates, E = 100 kV/m

Area of each plate, A = 2 x 10⁻³ m²

Distance between the plates, d = 6 x 10⁻³ m

Charge on the capacitor is calculated as;

Q = CV

V = Ed

`C = \epsilon_o (A)/(d)`

`Q = CV = \epsilon_o (A)/(d)*Ed\\\\ Q = \epsilon_o AE\\\\Q = 8.85*10^(-12) *2*10^(-3)* 100,000\\\\Q = 1.77 *10^(-9) \ C\\Q = 1.77 \ nC`

Therefore, the charge on the capacitor is 1.77nC