Chapter 17, Problem 102QAP

To determine

(a)

Change in the capacitance

Answer to Problem 102QAP

The capacitor is change by factor $4$.

Explanation of Solution

Given:

Area is doubled, $A_2=2A$

Distance is halved, $d_2=\frac{d}{2}$

Formula used:

The capacitance of the parallel plate capacitor is given as

  $C=\frac{{\epsilon }_{o}A}{d}$

Here,

  ${\epsilon }_o$ is the permittivity

  $A$ is the area

  $d$ is the distance between the plates

Calculation:

The capacitance of the parallel plate capacitor is given as

  $C=\frac{{\epsilon }_{o}A}{d}$

When the distance between the plates is halved and area is doubled, the capacitance become

  $\begin{array}{c}{C}_2=\frac{{\epsilon }_o\times 2A}{\frac{d}{2}}\\ =\frac{4{\epsilon }_{o}A}{d}\\ =4C\end{array}$

Conclusion:

The capacitor is change by factor $4$.

To determine

(b)

Charge on the positive plate

Answer to Problem 102QAP

Charge on the positive plate is changed by factor $4$.

Explanation of Solution

Given:

Area is doubled, $A_2=2A$

Distance is halved, $d_2=\frac{d}{2}$

Formula used:

Capacitance is given as

  $C=\frac{q}{V}$

Here,

  $q$ is the charge

  $V$ is the voltage

Calculation:

Capacitance is given as

  $C=\frac{q}{V}$

Since the capacitor is connected to the battery therefore voltage across the capacitor will be constant.

$\begin{array}{l}4C=\frac{q}{V}\\ q\text{'}=4q\end{array}$

Conclusion:

Charge on the positive plate is changed by factor $4$.

To determine

(c)

Change in potential across the plate

Answer to Problem 102QAP

There will be no change in potential across the plates, as the capacitor is connected to the battery.

Explanation of Solution

Given:

Area is doubled, $A_2=2A$

Distance is halved, $d_2=\frac{d}{2}$

Formula used:

Capacitance is given as

  $C=\frac{q}{V}$

Here,

  $q$ is the charge

  $V$ is the voltage

Calculation:

Capacitance is given as

  $C=\frac{q}{V}$

Since the capacitor is connected to the battery therefore voltage across the capacitor will be constant.

Conclusion:

There will be no change in potential across the plates, as the capacitor is connected to the battery.

To determine

(d)

Change in potential energy of the capacitor

Answer to Problem 102QAP

The potential energy of the capacitor will be change by factor $4$.

Explanation of Solution

Given:

Area is doubled, $A_2=2A$

Distance is halved, $d_2=\frac{d}{2}$

Formula used:

The energy required to separate the capacitor is given as

  $U=\frac{1}{2}\frac{q^2}{C}$

Here,

  $C$ is the capacitance

  $V$ is the voltage

Calculation:

The energy required to triple the separation distance of the parallel plate capacitor is calculated as

  $U=\frac{1}{2}\frac{q^2}{C}$. .....(1)

The potential energy of the capacitor after increasing the distance

  $U_2=\frac{1}{2}\times \frac{{\left(4q\right)}^2}{4C}$. .....(2)

Dividing equation (2) by equation (1)

  $\begin{array}{c}\frac{U_2}{U}=\frac{\frac{1}{2}\times \frac{{\left(4q\right)}^2}{4C}}{\frac{1}{2}\times \frac{q^2}{C}}\\ U_2=4U\end{array}$

Conclusion:

The potential energy of the capacitor will be change by factor $4$.

To determine

(e)

• Change in capacitance across the capacitor
• Change in charge across the capacitor
• Change in potential across the capacitor
• Change in potential energy across the capacitor

Answer to Problem 102QAP

• Capacitance is changed by factor $4$
• Charge across the capacitor is constant
• Potential across the capacitor is changed by factor 1/4.
• Potential energy across the capacitor is changed by factor 1/4.

Explanation of Solution

Formula used:

The capacitance of the parallel plate capacitor is given as

  $C=\frac{{\epsilon }_{o}A}{d}$

Here,

  ${\epsilon }_o$ is the permittivity

  $A$ is the area

  $d$ is the distance between the plates

The energy required to separate the capacitor is given as

  $U=\frac{1}{2}C{V}^2$

Here,

  $C$ is the capacitance

  $V$ is the voltage

Calculation:

The capacitance of the parallel plate capacitor is given as

  $C=\frac{{\epsilon }_{o}A}{d}$

When the distance between the plates is halved and area is doubled, the capacitance become

  $\begin{array}{c}{C}_2=\frac{{\epsilon }_o\times 2A}{\frac{d}{2}}\\ =\frac{4{\epsilon }_{o}A}{d}\\ =4C\end{array}$

When the distance between the capacitor increases, the charge remain constant, therefor the charge across the capacitor remains constant.

To compensate for the change in capacitance, the voltage across the capacitance is reduced by factor 1/4.

The energy required to triple the separation distance of the parallel plate capacitor is calculated as

  $U=\frac{1}{2}C{V}^2$. .....(1)

The potential energy of the capacitor after increasing the distance

  $U_2=\frac{1}{2}\times 4C\times {\left(\frac{V}{4}\right)}^2$. .....(2)

Dividing equation (2) by equation (1)

  $\begin{array}{c}\frac{U_2}{U}=\frac{\frac{1}{2}\times 4C{\left(\frac{V}{4}\right)}^2}{\frac{1}{2}C{V}^2}\\ U_2=\frac{U}{4}\end{array}$

Conclusion:

• Capacitance is changed by factor $4$
• Charge across the capacitor is constant
• Potential across the capacitor is changed by factor 1/4.
• Potential energy across the capacitor is changed by factor 1/4.