A parallel plate capacitor above has square plates 10 cm across on each side. The plates are 50 μm apart, and the electric field strength inside the capacitor is 4.0 x 10⁵ V/m.
A. What is the electric potential difference (voltage) across the capacitor?
B. What is the magnitude of the charge on either plate of the capacitor?
C. What is the capacitance of this capacitor?

a) ΔV = 20 V , b) Q = 35.4 10⁻⁹ C, c)  C = 1.77 10⁻⁹ F

Explanation:

a) The electric potential is

                 ΔV = E x

                 ΔV = 4.0 10⁵ 50 10⁻⁶

                  ΔV = 20 V

c) The capacity of a capacitor is

         C = ε₀  A / d

The area of a square plate is

       A = L²

       A = 0.10²

       A = 0.01 m²

       L = 50mm = 50 10⁻⁶ m

Let's calculate

        C = 8.85 10⁻¹²  0.01 /50 10⁻⁶

        C = 1.77 10⁻⁹ F

b)  the charge is

             Q = C ΔV

             Q = 1.77 10⁻⁹ 20

              Q = 35.4 10⁻⁹ C

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

be subject to a force andchange its motion