A capacitor stores 40 J of energy when charged to 20 C. What charge is required to store 20 J of energy?

Question

ScanSolve AI Tutor · Accepted Answer

The energy stored in a capacitor is given by the formula:

E = \frac{Q^2}{2C}

where $E$ is the energy stored, $Q$ is the charge on the plates, and $C$ is the capacitance.

Identify the initial conditions.
Initial charge $Q_1 = 20 \, \text{C}$
Initial energy $E_1 = 40 \, \text{J}$
Determine the target energy.
The question asks for the charge when the capacitor stores half the energy.
Target energy $E_2 = \frac{1}{2} E_1 = \frac{1}{2} (40 \, \text{J}) = 20 \, \text{J}$ .
Use the relationship between energy and charge.
Since the capacitance $C$ of the capacitor remains constant, the energy stored is directly proportional to the square of the charge ( $E \propto Q^2$ ). We can set up a ratio:
$\frac{E_2}{E_1} = \frac{Q_2^2}{Q_1^2}$
Substitute the known values into the ratio.
$\frac{20 \, \text{J}}{40 \, \text{J}} = \frac{Q_2^2}{(20 \, \text{C})^2}$ $\frac{1}{2} = \frac{Q_2^2}{400 \, \text{C}^2}$
Solve for $Q_2^2$ .
$Q_2^2 = \frac{1}{2} \times 400 \, \text{C}^2$ $Q_2^2 = 200 \, \text{C}^2$
Calculate $Q_2$ .
$Q_2 = \sqrt{200} \, \text{C}$ $Q_2 \approx 14.142 \, \text{C}$
Compare with the given options.
(a) 10.0C

(b) 5.0C

(c) 14.4C

(d) 20.0C

The calculated value $14.142 \, \text{C}$ is closest to $14.4 \, \text{C}$ .

The final answer is $\boxed{\text{14.4C}}$ .

Answer