Question

A capacitor stores 9mC of charge when connected to a 12-V battery. 1. How much charge will it store when connected to a 10-V battery? 2. What is the capacitance of this capacitor?

Answer 1

By definition, the capacitance is:

$$C = \dfrac{q_1}{U_1}$$

where $q_1 = 9mC = 9\times 10^{-3}C$ is the stored charge, and $U_1 = 12V$ is the allpied voltage. Thus, obtain:

$$C = \dfrac{9\times 10^{-3}C}{12V}=7.5\times10^{-4}F = 750mF$$

The amount of charge in the second case can be expressed from the formula above:

$$q_2 = CU_2$$

where $U_2 = 10V$ is the voltage of the second battery. Thus, obtain:

$$q_2 = 7.5\times10^{-4}F\cdot 10V = 7.5\times 10^{-3}C = 7.5mC$$

Answer. 1) 7.5 mC, 2) 750 mF.