What charge was imparted to a flat capacitor with an electrical capacity of 20pF

What charge was imparted to a flat capacitor with an electrical capacity of 20pF, if the field strength between the plates is 50kV / m, and the distance between the plates is 5mm.

To determine what charge q was imparted to a flat capacitor with a given electrical capacity C and a distance between the plates d to create an electric field, strength E, it is necessary to take into account the relationship between the voltage U and the field strength U = E ∙ d and determine the capacitance of the capacitor C = q / U … Then:

q = С ∙ U or q = С ∙ Е ∙ d.

From the condition of the problem it is known that C = 20 pF = 20 ∙ 10 ^ (- 12) F, the field strength between the plates is E = 50 kV / m = 50,000 V / m, and the distance between the plates is d = 5 mm = 0.005 m. :

q = 20 ∙ 10 ^ (- 12) F ∙ 50,000 V / m ∙ 0.005 m;

q = 5 ∙ 10 ^ (- 6) C = 5 nC.

Answer: The flat capacitor was charged with a charge of 5 nC.