Question

part d consider a charged parallel - plate capacitor. which combination of changes would quadruple its capacitance? double the charge and double the plate area. double the charge and double the plate separation. halve the charge and double the plate separation. halve the charge and double the plate area. halve the plate separation and double the plate area. double the plate separation and halve 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 each option

• Option 1: Doubling the charge does not affect capacitance ($C$ is independent of $Q$). Doubling $A$ gives $C_1=\frac{\epsilon_0(2A)}{d}=2C$.
• Option 2: Doubling the charge does not affect $C$. Doubling $d$ gives $C_2=\frac{\epsilon_0A}{2d}=\frac{1}{2}C$.
• Option 3: Halving the charge does not affect $C$. Doubling $d$ gives $C_3=\frac{\epsilon_0A}{2d}=\frac{1}{2}C$.
• Option 4: Halving the charge does not affect $C$. Doubling $A$ gives $C_4=\frac{\epsilon_0(2A)}{d}=2C$.
• Option 5: Halving $d$ gives $C_{d}=\frac{\epsilon_0A}{\frac{1}{2}d}=2C$, and doubling $A$ gives $C_{A}=\frac{\epsilon_0(2A)}{\frac{1}{2}d}=4C$.
• Option 6: Doubling $d$ gives $C_5=\frac{\epsilon_0A}{2d}$, and halving $A$ gives $C_6=\frac{\epsilon_0\frac{1}{2}A}{2d}=\frac{1}{4}C$.

Answer:

Halve the plate separation and double the plate area.