Question

3. A series combination of two uncharged capacitors are connected to a 15 V battery, 300 μJ of energy is drawn from the battery. If one of the capacitors has a capacitance of 5.0 pF, what is the capacitance of the other?

Answer 1

The total capacitance $C$ of the series can be found from the formula:

$$E = \dfrac{CU^2}{2}$$

where $E = 300\times 10^{-6}J$ is the energy of the system, $U = 15V$ is the voltage. Thus, obtain:

$$C = \dfrac{2E}{U^2}$$

On the other hand, the capacitance of the two capacitors in cseries is given as follows:

$$\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$

where $C_1 = 5\times 10^{-12}F, C_2$ are individual capacitances. Expressing $C_2$ and substituting the formula for $C$, obtain:

$$C_2 = \dfrac{CC_1}{C_1-C} = \dfrac{C_1}{C_1/C-1} = \dfrac{C_1}{C_1\dfrac{U^2}{2E}-1}\\ C_2 = \dfrac{5\times 10^{-12}F}{5\times 10^{-12}F\cdot \dfrac{(15V)^2}{2\cdot 300\times 10^{-6}J}-1} \approx -5\times 10^{-12}F$$

The ansver is negative, thus, the given numbers are wrong.

Answer. $C_2 = \dfrac{C_1}{C_1\dfrac{U^2}{2E}-1}$