Chapter 20, Problem 66P

A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm². The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor.

To determine

The charges on plates before and after immersing in water.

Answer to Problem 66P

The charges on plates before and after immersing in water is $\underset{\_}{369\text{pC}}$.

Explanation of Solution

Write the equation for capacitance of the system.

  $C=\frac{\kappa {\epsilon }_{0}A}{d}$        (I)

Here, $k$ is the dielectric, $C$ is the capacitance of the capacitor, ${\epsilon }_0$ is the permittivity of free space, $A$ is the area and $d$ is distance.

Write the equation for charge stored in the capacitor by using (I).

  $Q=\left(\frac{\kappa {\epsilon }_{0}A}{d}\right)\left(\Delta V\right)$        (II)

Since the voltage supply is disconnected before immersing in water, charge on the plates is same before and after immersion.

Conclusion:

Substitute, $1.00$ for $\kappa$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$, $25.00{\text{cm}}^2$ for $A$, $250\text{V}$ for $\left(\Delta V\right)$, $1.50\text{cm}$ for $d$ in Equation (II) to find $Q$.

  $\begin{array}{c}Q=\frac{\left(1.00\right)\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)\left[\left(25.0{\text{cm}}^2\right){\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right]}{\left[\left(1.50\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\left(250\text{V}\right)\\ =369\times {10}^{-12}\text{C}\\ =369\text{ pC}\end{array}$

Thus, the charges on plates before and after immersing in water is $\underset{\_}{369\text{pC}}$.

To determine

The capacitance and potential difference after immersion.

Answer to Problem 66P

The capacitance and potential difference after immersion is $\underset{\_}{1.20\times {10}^{-10}\text{F}}$.and $\underset{\_}{3.10\text{V}}$ respectively.

Explanation of Solution

Write the equation for capacitance of the system after immersion.

  $C_f=\frac{{\kappa }_f{\epsilon }_{0}A}{d}$        (III)

Here, $k_f$ is the dielectric constant for distilled water, $C_f$ is the capacitance of the capacitor after immersion, ${\epsilon }_0$ is the permittivity of free space, $A$ is the area and $d$ is distance.

Write the expression for the potential difference after immersion.

$\left(\Delta V_f\right)=\frac{Q}{C_f}$        (IV)

Here, $\left(\Delta V_f\right)$ is the potential difference after immersion.

Conclusion:

Substitute, $80$ for $\kappa$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$, $25.00{\text{cm}}^2$ for $A$, $1.50\text{cm}$ for $d$ in Equation (III) to find $C_f$.

  $\begin{array}{c}{C}_f=\frac{\left(80\right)\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)\left[\left(25.00{\text{cm}}^2\right){\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right]}{\left[\left(1.50\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ =1.20\times {10}^{-10}\text{F}\end{array}$

Substitute, $369\times {10}^{-12}\text{C}$ for $Q$, $1.20\times {10}^{-10}\text{F}$ for $C_f$ in Equation (IV) to find $\left(\Delta V_f\right)$.

$\begin{array}{c}\left(\Delta V_f\right)=\frac{\left(369\times {10}^{-12}\text{C}\right)}{\left(1.20\times {10}^{-10}\text{F}\right)}\\ =3.10\text{V}\end{array}$

Thus, the capacitance and potential difference after immersion is $\underset{\_}{1.20\times {10}^{-10}\text{F}}$.and $\underset{\_}{3.10\text{V}}$ respectively.

To determine

The change in energy of the capacitor.

Answer to Problem 66P

The change in energy of the capacitor is $\underset{\_}{-45.5\text{nJ}}$.

Explanation of Solution

Write the expression for the initial capacitance.

$C=\frac{k{\epsilon }_{0}A}{d}$        (V)

Here, $C$ is the initial capacitance, $k$ is the dielectric constant for air.

Write the expression for the initial energy by using (V).

$\begin{array}{c}{U}_i=\frac{1}{2}C{\left(\Delta V\right)}^2\\ =\frac{1}{2}\left(\frac{k{\epsilon }_{0}A}{d}\right){\left(\Delta V\right)}^2\end{array}$        (VI)

Write the expression for the final energy by using (III) and (IV).

$\begin{array}{c}{U}_f=\frac{1}{2}{C}_f{\left(\Delta V_f\right)}^2\\ =\frac{1}{2}{C}_f{\left(\frac{Q}{C_f}\right)}^2\\ =\frac{1}{2}\frac{Q^2}{C_f}\end{array}$        (VII)

Write the expression for the change in energy by using (VI) and (VII).

$\begin{array}{c}\left(\Delta U\right)=U_f-U_i\\ =\frac{1}{2}\left[\left(\frac{Q^2}{C_f}\right)-\left(\left(\frac{k{\epsilon }_{0}A}{d}\right){\left(\Delta V\right)}^2\right)\right]\end{array}$        (VIII)

Conclusion:

Substitute, $369\times {10}^{-12}\text{C}$ for $Q$, $1.20\times {10}^{-10}\text{F}$ for $C_f$, $1.00$ for $\kappa$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$, $25.0{\text{cm}}^2$ for $A$, $1.50\text{cm}$ for $d$, $250\text{V}$ for $\left(\Delta V\right)$ in Equation (VIII) to find $\left(\Delta U\right)$.

$\begin{array}{c}\left(\Delta U\right)=\frac{1}{2}\left[\begin{array}{l}\left(\frac{{\left(369\times {10}^{-12}\text{C}\right)}^2}{\left(1.20\times {10}^{-10}\text{F}\right)}\right)-\\ \left(\frac{\left(1.00\right)\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)\left(25.0{\text{cm}}^2\right){\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}^2}{\left[\left(1.50\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right){\left(250\text{V}\right)}^2\end{array}\right]\\ =-4.55\times {10}^{-8}\text{J}\\ =-45.5\text{nJ}\end{array}$

Thus, the change in energy of the capacitor is $\underset{\_}{-45.5\text{nJ}}$.