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
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
[tex]C = \epsilon_o \frac{A}{d}[/tex]
[tex]Q = CV = \epsilon_o \frac{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[/tex]
Therefore, the charge on the capacitor is 1.77nC
