Question

A parallel-plate vacuum capacitor has 7.64 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.30 mm,
a) what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?
b) What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

Answer 1

Concept and reason

The concepts that are used to solve this problem are energy stored in the capacitor. Use the formula for the capacitance of a parallel plate capacitor and the energy stored in a capacitor. The relation between the separations between the two plates is obtained by using the formula for the capacitance of a parallel plate capacitor. The final potential energy is calculated using the formula for energy stored in a capacitor.

Fundamentals

Potential energy is defined as the energy possessed by an object by its position. Capacitance is defined as the ability of a capacitor to store energy. The capacitance is defined as the ratio of charge to potential difference and is given by, $C=\frac{Q}{V}$

Here, $Q$ is the charge, and $V$ is the voltage. The capacitance of a parallel plate capacitor is, $C=\frac{\varepsilon_{0} A}{d}$

Here, $\varepsilon_{0}$ is the permittivity, $\mathrm{A}$ is the area, and $\mathrm{d}$ is the separation between the two plates. The potential energy stored in the capacitor is, $U=\frac{Q^{2}}{2 C}$

Here, $Q$ is the charge, and $C$ is the capacitance. The potential energy stored in the capacitor is, $U=\frac{1}{2} C V^{2}$

Here, $V$ is the potential difference, and $C$ is the capacitance.

(a) The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates. So, the ratio of initial capacitance to final capacitance is, $\frac{C_{1}}{C_{2}}=\frac{d_{2}}{d_{1}}$

Explanation | Hint for next step

The initial capacitance of the capacitor is,

$C_{1}=\frac{\varepsilon_{0} A_{1}}{d_{1}}$

Here, $\varepsilon_{0}$ is the permittivity, $A_{1}$ is the plate area, and $d_{1}$ is the initial separation between the two plates. The final capacitance of the capacitor is,

$C_{2}=\frac{\varepsilon_{0} A_{2}}{d_{2}}$

Here, $\varepsilon_{0}$ is the permittivity, $A_{2}$ is the plate area, and $d_{2}$ is the final separation between the two plates. The ratio of the capacitance of the first capacitor to the capacitance of the second capacitor is $\frac{C_{1}}{C_{2}}=\frac{\left(\frac{\varepsilon_{0} A_{1}}{d_{1}}\right)}{\left(\frac{\varepsilon_{0} A_{2}}{d_{2}}\right)}$

$=\left(\frac{A_{1}}{A_{2}}\right)\left(\frac{d_{2}}{d_{1}}\right)$

The distance between the plates is decreased. The plates of the capacitors are not changed. So, the area of the plates remains the same.

Substitute A for $A_{1}$ and $A$ for $A_{2}$. $\begin{aligned} \frac{C_{1}}{C_{2}} &=\left(\frac{A_{1}}{A_{2}}\right)\left(\frac{d_{2}}{d_{1}}\right) \\ &=\left(\frac{A}{A}\right)\left(\frac{d_{2}}{d_{1}}\right) \\ &=\frac{d_{2}}{d_{1}} \end{aligned}$

The potential energy stored in the first capacitor is, $U_{1}=\frac{Q_{1}^{2}}{2 C_{1}}$

Here, $Q_{1}$ is the charge, and $C_{1}$ is the capacitance. The potential energy stored in the second capacitor is, $U_{2}=\frac{Q_{2}^{2}}{2 C_{2}}$

Here, $Q_{2}$ is the charge, and $C_{2}$ is the capacitance. The ratio of potential energies is. $\frac{U_{2}}{U_{1}}=\frac{\left(\frac{Q_{2}^{2}}{2 C_{2}}\right)}{\left(\frac{Q_{1}^{2}}{2 C_{1}}\right)}$

$=\left(\frac{Q_{2}}{Q_{1}}\right)^{2}\left(\frac{C_{1}}{C_{2}}\right)$

The battery is disconnected, and the distance between the plates is decreased. Hence, the charge remains the same.

Substitute $Q$ for $Q_{1}$ and $Q$ for $Q_{2}$ $\begin{aligned} \frac{U_{2}}{U_{1}} &=\left(\frac{Q}{Q}\right)^{2}\left(\frac{C_{1}}{C_{2}}\right) \\ &=\frac{C_{1}}{C_{2}} \\ \text { Substitute } \frac{d_{2}}{d_{1}} & \frac{C_{1}}{C_{2}} \end{aligned}$

$\frac{U_{2}}{U_{1}}=\frac{d_{2}}{d_{1}}$

Rearrange the equation for the final potential energy. $U_{2}=\left(\frac{d_{2}}{d_{1}}\right) U_{1}$

Substitute $1.30 \mathrm{~mm}$ for $d_{2}, 2.30 \mathrm{~mm}$ for $d_{1}$, and $7.64 \mathrm{~J}$ for $U_{1}$.

$\begin{aligned} U_{2} &=\left(\frac{1.30 \mathrm{~mm}}{2.30 \mathrm{~mm}}\right)(7.64 \mathrm{~J}) \\ &=4.318 \mathrm{~J} \end{aligned}$

Round off to three significant figures. So, the final potential energy is $4.32 \mathrm{~J}$.

(b) The potential energy stored in the first capacitor is,

$U_{1}=\frac{1}{2} C_{1} V_{1}^{2}$

Here, $V_{1}$ is the electric potential, and $C_{1}$ is the capacitance. The potential energy stored in the second capacitor is,

$U_{2}=\frac{1}{2} C_{2} V_{2}^{2}$

Here, $V_{2}$ is the electric potential, and $C_{2}$ is the capacitance. The ratio of potential energies is $\frac{U_{2}}{U_{1}}=\frac{\left(\frac{1}{2} C_{2} V_{2}^{2}\right)}{\left(\frac{1}{2} C_{1} V_{1}^{2}\right)}$

The battery is connected. Hence, the electric potential remains the same.

Substitute $V$ for $V_{1}$ and $V$ for $V_{2}$.

$\begin{aligned} \frac{U_{2}}{U_{1}} &=\frac{\left(\frac{1}{2} C_{2} V^{2}\right)}{\left(\frac{1}{2} C_{1} V^{2}\right)} \\ &=\frac{C_{2}}{C_{1}} \\ &=\frac{1}{\left(\frac{C_{1}}{C_{2}}\right)} \\ \text { Substitute } \frac{d_{2}}{d_{1}} \text { for } \frac{C_{1}}{C_{2}} . \\ \frac{U_{2}}{U_{1}} &=\frac{1}{\left(\frac{d_{2}}{d_{1}}\right)} \\ &=\frac{d_{1}}{d_{2}} \end{aligned}$

Rearrange the equation for the final potential energy. $U_{2}=\left(\frac{d_{1}}{d_{2}}\right) U_{1}$

Substitute $1.30 \mathrm{~mm}$ for $d_{2}, 2.30 \mathrm{~mm}$ for $d_{1}$, and $7.64 \mathrm{~J}$ for $U_{1}$.

$\begin{aligned} U_{2} &=\left(\frac{2.30 \mathrm{~mm}}{1.30 \mathrm{~mm}}\right)(7.64 \mathrm{~J}) \\ &=13.516 \mathrm{~J} \end{aligned}$

Round off to three significant figures. So, the final potential energy is $13.5 \mathrm{~J}$.

The capacitor is connected to the source of potential, and the separation between the plates is decreased. Since the capacitor is connected to the potential source, the electric potential does not change. Hence, the initial and final potentials of the capacitor are the same.

Part a

The potential energy of the capacitor when the battery is disconnected is $4.32 \mathrm{~J}$.

Part b

The potential energy of the capacitor when the battery is disconnected is $13.5 \mathrm{~J}$.