Chapter 17, Problem 83P

(a)

To determine

The charge on the plates, electric field between the plates and the energy stored in the parallel plate capacitor with air in between.

Answer to Problem 83P

The charge on the capacitor is $556\text{ pC}$, the electric field between the plates is $2.00{\text{ kVm}}^{-1}$ and the energy stored on the capacitor with the air inside is $5.56\text{ nJ}$.

Explanation of Solution

The area of the plate is $314{\text{cm}}^2$, the applied potential difference is $20.0\text{V}$, and the separation between the plates is $1.00\text{cm}$.

Write the expression for capacitance

  $C={\in }_{0}A\left(\frac{\kappa }{d}\right)$                                                                     (I)

Here, $C$ is the capacitance of the capacitor, ${\in }_0$ is the permittivity of free space, $A$ is the area of the electrodes, $\kappa$ is the dielectric constant and $d$ is the distance between the electrodes.

Substitute $8.85\times {10}^{-12}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-2}$ for ${\in }_0$, $314{\text{cm}}^2$ for $A$, $1.00\text{cm}$ for $d$ and $1.00054$ for $\kappa$ in (I) to find $C$

    $\begin{array}{c}C=\frac{\left(1.00054\right)\left(8.85\times {10}^{-12}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-2}\right)\left(314{\text{cm}}^2\right)}{\left(1.00\text{cm}\right)}\\ =\frac{\left(1.00054\right)\left(8.85\times {10}^{-12}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-2}\right)\left(314\times {10}^{-4}{\text{m}}^2\right)}{\left(1.00\times {10}^{-2}\text{m}\right)}\cdot \frac{\text{1}\text{F}}{\text{1}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-1}}\\ =2.78\times {10}^{-11}\text{}\text{F}\end{array}$

Thus, the capacitance of the capacitor is $2.78\times {10}^{-11}\text{}\text{F}$.

Write the expression for charge stored on each plate of the capacitor

  $Q=C\Delta V$                                                              (II)

Here, $Q$ is the magnitude of charge on each plate of the capacitor, $C$ is the capacitance of the capacitor and $\Delta V$ is the applied potential difference.

Substitute  $20.0\text{V}$ for $\Delta V$ and $2.78\times {10}^{-11}\text{}\text{F}$ for $C$ in (II) to find $Q$

    $\begin{array}{c}Q=\left(2.78\times {10}^{-11}\text{}\text{F}\right)\left(20.0\text{V}\right)\\ =556\times {10}^{-12}\text{}\text{VF}\cdot \frac{\text{1C}}{\text{1VF}}\\ =556\text{ pC}\end{array}$

Thus, the charge on the capacitor plates with the air in-between is $556\text{ pC}$.

Write the expression for electric field

  $E=\frac{\Delta V}{d}$                                                                    (III)

Here, $E$ is the electric field.

Substitute  $20.0\text{V}$ for $\Delta V$ and $1.00\text{cm}$ for $d$ in (III) to find $E$

    $\begin{array}{c}E=\frac{20.0\text{V}}{1.00\text{cm}}\\ =\frac{20.0\text{V}}{1.00\times {10}^{-2}\text{m}}\\ =2.00\times {10}^{-3}{\text{Vm}}^{-1}\\ =2.00{\text{ kVm}}^{-1}\end{array}$

Thus, the electric field between the parallel plates is $2.00{\text{ kVm}}^{-1}$.

Write the expression for energy stored in the capacitor

    $U=\frac{1}{2}C{\left(\Delta V\right)}^2$                                                                      (IV)

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

Substitute  $20.0\text{V}$ for $\Delta V$ and $2.78\times {10}^{-11}\text{}\text{F}$ for $C$ in (IV) to find $U$

    $\begin{array}{c}U=\frac{1}{2}\left(2.78\times {10}^{-11}\text{}\text{F}\right){\left(20.0\text{V}\right)}^2\\ =5.56\times {10}^{-9}\text{}{\text{FV}}^2\cdot \frac{\text{1J}}{{\text{1V}}^2\text{F}}\\ =5.56\text{ nJ}\end{array}$

Thus, the energy stored on the capacitor with the air inside is $5.56\text{ nJ}$.

(b)

To determine

The charge on the plates, the electric field, the potential difference between them and the energy stored in the capacitor after introducing a slab of strontium titanate.

Answer to Problem 83P

After introducing strontium titanate between the plates, the charge on the capacitor is $172\text{ nC}$, the electric field between the plates is $2.00{\text{ kVm}}^{-1}$, the potential difference across the plates is $20.0\text{V}$ and the energy stored on the capacitor is $1.72\text{ μJ}$.

Explanation of Solution

The dielectric constant of the strontium titanate is $310$.

Since the battery is still connected to the plate, the potential difference doesn’t change and hence it is $20.0\text{V}$.

Substitute $8.85\times {10}^{-12}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-2}$ for ${\in }_0$, $314{\text{cm}}^2$ for $A$, $1.00\text{cm}$ for $d$ and $310$ for $\kappa$ in (I) to find $C$

    $\begin{array}{c}C=\frac{\left(310\right)\left(8.85\times {10}^{-12}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-2}\right)\left(314{\text{cm}}^2\right)}{\left(1.00\text{cm}\right)}\\ =\frac{\left(310\right)\left(8.85\times {10}^{-12}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-2}\right)\left(314\times {10}^{-4}{\text{m}}^2\right)}{\left(1.00\times {10}^{-2}\text{m}\right)}\cdot \frac{\text{1}\text{F}}{\text{1}{\text{N}}^{-1}{\text{C}}^2{\text{m}}^{-1}}\\ =8.62\times {10}^{-9}\text{}\text{F}\end{array}$

Thus, the capacitance of the capacitor is $8.62\times {10}^{-11}\text{}\text{F}$.

Substitute  $20.0\text{V}$ for $\Delta V$ and $8.62\times {10}^{-11}\text{}\text{F}$ for $C$ in (II) to find $Q$

    $\begin{array}{c}Q=\left(8.62\times {10}^{-9}\text{}\text{F}\right)\left(20.0\text{V}\right)\\ =172\times {10}^{-9}\text{}\text{VF}\cdot \frac{\text{1C}}{\text{1VF}}\\ =172\text{ nC}\end{array}$

Thus, the charge on the capacitor plates is $172\text{ nC}$.

Substitute  $20.0\text{V}$ for $\Delta V$ and $1.00\text{cm}$ for $d$ in (III) to find $E$

    $\begin{array}{c}E=\frac{20.0\text{V}}{1.00\text{cm}}\\ =\frac{20.0\text{V}}{1.00\times {10}^{-2}\text{m}}\\ =2.00\times {10}^{-3}{\text{Vm}}^{-1}\\ =2.00{\text{ kVm}}^{-1}\end{array}$

Thus, the electric field between the parallel plates is $2.00{\text{ kVm}}^{-1}$.

Substitute  $20.0\text{V}$ for $\Delta V$ and $8.62\times {10}^{-9}\text{}\text{F}$ for $C$ in (IV) to find $U$

    $\begin{array}{c}U=\frac{1}{2}\left(8.62\times {10}^{-9}\text{}\text{F}\right){\left(20.0\text{V}\right)}^2\\ =1.72\times {10}^{-6}\text{}{\text{FV}}^2\cdot \frac{\text{1J}}{{\text{1V}}^2\text{F}}\\ =1.72\text{ μJ}\end{array}$

Thus, the energy stored on the capacitor is $1.72\text{ μJ}$.