An ideal parallel-plate capacitor has a capacitance of $C$. If the area of the plates is doubled and the distance between the plates is halved, what is the new capacitance?

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by $C = \epsilon_0 \frac{A}{d}$, where $A$ is the area of the plates and $d$ is the distance between them. Doubling the area and halving the distance results in a new capacitance of $C' = \epsilon_0 \frac{2A}{d/2} = 4\epsilon_0 \frac{A}{d} = 4C$.

An Ideal Parallel-Plate Capacitor Has a Capacitance Of CCC If the Area of the Plates Is Doubled and If

An Ideal Parallel-Plate Capacitor Has a Capacitance Of $C$ If the Area of the Plates Is Doubled and If