A voltage has been applied across a capacitor. If the dielectric is replaced with another dielectric constant eight times as great and the voltage is reduced to half of what it was, the ENERGY STORED in the capacitor is how many times the original stored energy?

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$, where $U$ is the energy, $C$ is the capacitance, and $V$ is the voltage. The capacitance $C$ is directly proportional to the dielectric constant ($\kappa$), so if the dielectric constant is increased by a factor of 8, the capacitance increases by a factor of 8. If the voltage is reduced to half, $V^2$ becomes $\frac{1}{4}$ of its original value. Therefore, the new energy stored is $U_{new} = \frac{1}{2} \cdot 8C \cdot \left(\frac{1}{2}V\right)^2 = 2 \cdot \frac{1}{2}CV^2 = 2U$, which is twice the original stored energy.

A Voltage Has Been Applied Across a Capacitor