Question

each plate of a capacitor stores a charge of magnitude 6 mc when a 200 - v potential difference is applied. if the charge is doubled and the plates are moved so that the potential is halved, how will the new capacitance compare to the original? cnew = 1/2 ci cnew = ci cnew = 4ci cnew = 2ci none of the choices are correct.

Explanation:

Step1: Recall capacitance formula

The formula for capacitance is $C=\frac{Q}{V}$, where $C$ is capacitance, $Q$ is charge and $V$ is potential - difference. Initially, $C_i=\frac{Q_i}{V_i}$.

Step2: Analyze the changes

We are given that $Q_{new} = 2Q_i$ and $V_{new}=\frac{V_i}{2}$. Then, using the capacitance formula $C_{new}=\frac{Q_{new}}{V_{new}}$.

Step3: Substitute the values

Substitute $Q_{new} = 2Q_i$ and $V_{new}=\frac{V_i}{2}$ into the formula for $C_{new}$: $C_{new}=\frac{2Q_i}{\frac{V_i}{2}}$.

Step4: Simplify the expression

$C_{new}=4\frac{Q_i}{V_i}$. Since $C_i=\frac{Q_i}{V_i}$, we have $C_{new} = 4C_i$.

Answer:

$C_{new}=4C_i$