Question

consider a charged parallel - plate capacitor. how can its capacitance be halved? check all that apply. double the charge. double the plate area. double the plate separation. halve the charge. halve the plate area. halve the plate separation.

Explanation:

Step1: Recall capacitance formula

The capacitance of a parallel - plate capacitor is given by $C = \frac{\epsilon_0A}{d}$, where $\epsilon_0$ is the permittivity of free space, $A$ is the plate area and $d$ is the plate separation.

Step2: Analyze effect of changing plate area

If we want to halve $C$, from $C = \frac{\epsilon_0A}{d}$, when we halve the plate area $A$ (new $A'=\frac{A}{2}$), the new capacitance $C'=\frac{\epsilon_0A'}{d}=\frac{\epsilon_0\frac{A}{2}}{d}=\frac{1}{2}\frac{\epsilon_0A}{d}=\frac{C}{2}$.

Step3: Analyze effect of changing plate separation

If we double the plate separation $d$ (new $d' = 2d$), then the new capacitance $C'=\frac{\epsilon_0A}{d'}=\frac{\epsilon_0A}{2d}=\frac{1}{2}\frac{\epsilon_0A}{d}=\frac{C}{2}$.

Step4: Analyze effect of changing charge

The capacitance of a capacitor is a property of its geometry and the dielectric between the plates, and is independent of the charge $Q$ on the capacitor. So changing the charge (doubling or halving it) will not change the capacitance.

Answer:

C. Double the plate separation
E. Halve the plate area