Question

What is the capacitance of a capacitor that has 7.9 x 107 C of charge and a potential

difference of 12.0 V?

Answer 1

6.58×10⁶ F

Step-by-step explanation:

Applying,

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

Where Q = charge of the capacitor, C = Capacitance of the capacitor, V = Volatage

Make C the subject of the equation

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

From the question,

Given: Q = 7.9×10⁷ C, V = 12.0 V

Susbtitute these values into equation 2

C = 7.9×10⁷/12

C = 6.58×10⁶ F