Chapter 16, Problem 59E

(a)

To determine

The charge on the plates.

Answer to Problem 59E

The charge on the plates is $8.85\times {10}^{-8}\text{C}$.

Explanation of Solution

Write the expression for the capacitance.

  $C=\frac{A{\epsilon }_0}{l}$        (I)

Here, ${\epsilon }_0$ is the permittivity of free space, $A$ is the area, and $l$ is the distance between the capacitors.

Write the expression for the capacitance.

  $Q=CV$        (II)

Here, $Q$ is the charge and $V$ is the potential difference.

Solve (I) and (II) for $Q$,

$Q=\left(\frac{A{\epsilon }_0}{l}\right)V$        (III)

Conclusion:

Substitute $1000\text{V}$ for $V$, and $0.01\text{m}$ for $l$, $0.1{\text{m}}^2$ for $A$, and $8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (III) to find $Q$.

  $\begin{array}{c}Q=\frac{\left(1000\text{V}\right)\left(0.1{\text{m}}^2\right)\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}{\left(0.01\text{m}\right)}\\ =8.85\times {10}^{-8}\text{C}\end{array}$

Therefore, charge on the plates is $8.85\times {10}^{-8}\text{C}$.

(b)

To determine

The energy stored in the capacitor.

Answer to Problem 59E

The energy stored in the capacitor is $4.42\times {10}^{-5}$.

Explanation of Solution

Write the expression for the energy stored in the capacitor.

  $U=\frac{1}{2}C{V}^2$        (IV)

Here, $U$ is the energy stored in the capacitor.

Solve for $U$ using (IV) and (I).

  $U=\frac{1}{2}\left(\frac{A{\epsilon }_0}{l}\right)V^2$        (V)

Conclusion:

Substitute $1000\text{V}$ for $V$, and $0.01\text{m}$ for $l$, $0.1{\text{m}}^2$ for $A$, and $8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (V) to find $U$.

  $\begin{array}{c}U=\frac{1}{2}\left(\frac{\left(0.1{\text{m}}^2\right)\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}{0.01\text{m}}\right){\left(1000\text{V}\right)}^2\\ =4.42\times {10}^{-5}\text{J}\end{array}$

Therefore, the energy stored in the capacitor is $4.42\times {10}^{-5}$

(c)

To determine

The energy and potential difference of the capacitor under the insertion of dielectric slab.

Answer to Problem 59E

The energy and potential difference of the capacitor under the insertion of dielectric slab are $\underset{\_}{4.42\times {10}^{-6}\text{J}}$ and $\underset{\_}{100\text{V}}$.

Explanation of Solution

Write the expression for the capacitance under the insertion of dielectric slab is,

  $C_K=\frac{K{\epsilon }_{0}A}{l}$        (VI)

Here, $C_K$ is the capacitance under the insertion of dielectric slab and $K$ is the dielectric constant.

Write the potential difference for the capacitance under the insertion of dielectric slab.

  $V_K=\frac{Q}{C_K}$        (VII)

Here, $V_K$ is the potential difference for the capacitance under the insertion of dielectric slab.

Write the expression for the energy stored in the capacitor under the insertion of dielectric slab.

  $U_K=\frac{1}{2}{C}_K{V}_K^2$        (VIII)

Here, $U_K$ is the energy stored in the capacitor under the insertion of dielectric slab.

Conclusion:

Substitute $10$ for $K$, and $0.01\text{m}$ for $l$, $0.1{\text{m}}^2$ for $A$, and $8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (VI) to find $C_K$.

  $\begin{array}{c}{C}_K=\frac{\left(10\right)\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(0.1{\text{m}}^2\right)}{0.01\text{m}}\\ =8.85\times {10}^{-10}\text{F}\end{array}$

Substitute $8.85\times {10}^{-8}\text{C}$ for $Q$ and $8.85\times {10}^{-10}\text{F}$ for $C_K$ in equation (VII) to find $V_K$.

  $\begin{array}{c}{V}_K=\frac{8.85\times {10}^{-8}\text{C}}{8.85\times {10}^{-10}\text{F}}\\ =100\text{V}\end{array}$

Substitute $100\text{V}$ for $V_K$ and $8.85\times {10}^{-10}\text{F}$ for $C_K$ in equation (VIII) to find $U_K$.

  $\begin{array}{c}U=\frac{1}{2}\left(8.85\times {10}^{-10}\text{F}\right){\left(100\text{V}\right)}^2\\ =4.42\times {10}^{-6}\text{J}\end{array}$

Therefore, the energy and potential difference of the capacitor under the insertion of dielectric slab are $\underset{\_}{4.42\times {10}^{-6}\text{J}}$ and $\underset{\_}{100\text{V}}$.

(d)

To determine

The explanation for the change.

Answer to Problem 59E

The energy stored reduces by the factor $K$ due to the insertion of dielectric constant.

Explanation of Solution

Rearrange the equation (II) to get $V$.

  $V=\frac{Q}{C}$        (IX)

Write the energy stored in the capacitor of medium dielectric constant.

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

  $U=\frac{Q^2}{2C}$        (X)

Write the energy stored in the capacitor of medium dielectric constant using (X).

  $U_K=\frac{Q^2}{2K{C}_0}$        (XI)

Here, $C_0$ is the dielectric constant of air.

Conclusion:

Solve the equation (X) and (XI).

$U=\frac{U_K}{K}$

Therefore, the energy stored reduces by the factor $K$ due to the insertion of dielectric constant.