Question

3.50 µf capacitor has a charge with a magnitude of 4.50 µC on each parallel plate. Find the potential difference between the plates.

a) 0.85 V
b) 1.29 V
c) 1.70 V
d) 2.00 V

Answer 1

b) 1.29 V

Step-by-step explanation:

Potential Difference: This is the work done when one coulomb of charge moves from one point to another in an electric field

The expression for the potential difference is

Q = CV .................... Equation 1

Where Q = amount of charge, C = Capacitance of the capacitor, V = potential difference.

Making V the subject of the equation,

V = Q/C................. Equation 2

Given: Q = 4.5 µC = 4.5×10⁻⁶ C, C = 3.5 µF = 3.5×10⁻⁶ F

Substituting into equation 2

V = 4.5×10⁻⁶ /3.5×10⁻⁶

V = 1.2857 V

V ≈ 1.29 V.

Hence the right option is b) 1.29 V