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

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2