Question

consider a charged parallel - plate capacitor. how can its capacitance be halved? check all that apply. double the charge. halve the charge. double the plate area. halve the plate area. double the plate separation. halve the plate separation. consider a charged parallel - plate capacitor. which combination of changes would quadruple its capacitance? double the charge and double the plate area. halve the charge and double the plate separation. double the charge and double the plate separation. halve the charge and double the plate area. double the plate separation and halve the plate area. halve the plate separation and double the plate area.

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 quadrupling capacitance

We want to find how to quadruple $C$. If we double the plate area $A$ and halve the plate separation $d$, the new capacitance $C'=\frac{\epsilon_0(2A)}{\frac{d}{2}} = 4\frac{\epsilon_0A}{d}=4C$.

Step3: Analyze halving capacitance

To halve the capacitance, we can either double the plate separation $d$ (since $C\propto\frac{1}{d}$) or halve the plate area $A$ (since $C\propto A$).

Answer:

To quadruple the capacitance: Double the charge and double the plate area.
To halve the capacitance: Double the plate separation, Halve the plate area.