Chapter 27, Problem 88PQ

A parallel-plate capacitor has square plates of side s = 2.50 cm and plate separation d = 2.50 mm. The capacitor is charged by a battery to a charge Q = 4.00 μC, after which the battery is disconnected. A porcelain dielectric (κ = 6.5) is then inserted a distance y = 1.00 cm into the capacitor (Fig. P27.88). Hint: Consider the system as two capacitors connected in parallel. a. What is the effective capacitance of this capacitor? b. How much energy is stored in the capacitor? c. What are the magnitude and direction of the force exerted on the dielectric by the plates of the capacitor?

Figure P27.88

To determine

Effective capacitance of the capacitor.

Answer to Problem 88PQ

Effective capacitance of the capacitor is $7.08\text{pF}$.

Explanation of Solution

Write the expression to find the capacitance of the parallel plate capacitor with dielectric.

  $C_1=\frac{\kappa {\epsilon }_{0}sy}{d}$                                                                                                  (I)

Here, $\kappa$ is the dielectric constant, ${\epsilon }_0$ is the permittivity in free space, $C_1$ is the capacitance of the parallel plate capacitor with dielectric, $s$ is the length of the plate of capacitor, $y$ is the height of dielectric in capacitor, $d$ is the separation between the plates of the capacitor.

Substitute $6.5$ for $\kappa$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$, $2.50\text{cm}$ for $s$, $1.0\text{0cm}$ for $y$ and $2.50\text{mm}$ for $d$ to find the capacitance of the parallel plate capacitor with dielectric.

  $\begin{array}{c}{C}_1=\frac{\left(6.5\right)\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)\left(1.0\text{0cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}{\left(2.50\text{mm}\frac{{10}^{-3}\text{m}}{1.0\text{mm}}\right)}\\ =5.75\text{pF}\end{array}$

Write the expression to find the capacitance of the parallel plate which is unfilled.

  $C_2=\frac{{\epsilon }_{0}s\left(s-y\right)}{d}$                                                                                                  (II)

Here, ${\epsilon }_0$ is the permittivity in free space, $C_2$ is the capacitance of the parallel plate capacitor which is unfilled, $s$ is the length of the plate of capacitor, $y$ is the height of dielectric in capacitor, $d$ is the separation between the plates of the capacitor.

Substitute  $8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$, $2.50\text{cm}$ for $s$, $1.0\text{0cm}$ for $y$ and $2.50\text{mm}$ for $d$ to find the capacitance of the parallel plate capacitor with dielectric.

  $\begin{array}{c}{C}_2=\frac{\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}-1.0\text{0cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}{\left(2.50\text{mm}\frac{{10}^{-3}\text{m}}{1.0\text{mm}}\right)}\\ =\frac{\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)\left(1.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}{\left(2.50\text{mm}\frac{{10}^{-3}\text{m}}{1.0\text{mm}}\right)}\\ =1.33\text{pF}\end{array}$

The system can be considered as two capacitors are connected in parallel.

Write the expression to find the equivalent capacitance of the two capacitors connected in parallel.

  $C=C_1+C_2$

Here, $C$ is the equivalent capacitance.

Conclusion:

Substitute $5.75\text{pF}$ for $C_1$ and $1.33\text{pF}$ for $C_2$ to find the effective capacitance of the capacitor.

  $\begin{array}{c}C=5.75\text{pF}+1.33\text{pF}\\ =7.08\text{pF}\end{array}$

Therefore, effective capacitance of the capacitor is $7.08\text{pF}$.

To determine

The energy stored in the capacitor.

Answer to Problem 88PQ

The energy stored in the capacitor is $1.13\text{J}$.

Explanation of Solution

Write the expression to find the equivalent capacitance of the two capacitors connected in parallel.

  $C=C_1+C_2$

Here, $C$ is the equivalent capacitance.

Substitute equations (I) and (II) in above expression to find the equivalent capacitance

  $\begin{array}{c}C=\frac{\kappa {\epsilon }_{0}sy}{d}+\frac{{\epsilon }_{0}s\left(s-y\right)}{d}\\ =\frac{{\epsilon }_{0}s\left(s+y\left(\kappa -1\right)\right)}{d}\end{array}$                                                                                       (III)

Write the equation for energy stored in the capacitor.

  $U=\frac{Q^2}{2C}$

Here, $U$ is the energy stored in the capacitor, $Q$ is the charge on the capacitor, $C$ is the equivalent capacitance of the capacitor.

Substitute equation (III) in above expression to find the energy stored in the capacitor.

  $\begin{array}{c}U=\frac{Q^2}{2\left(\frac{{\epsilon }_{0}s\left(s+y\left(\kappa -1\right)\right)}{d}\right)}\\ =\frac{Q^{2}d}{2{\epsilon }_{0}s\left(s+y\left(\kappa -1\right)\right)}\end{array}$

Conclusion:

Substitute $4.00\mu \text{C}$ for $Q$,  $8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$, $2.50\text{cm}$ for $s$, $1.0\text{0cm}$ for $y$ and $2.50\text{mm}$ for $d$, $6.5$ for $\kappa$ to find the energy stored in the capacitor.

  $\begin{array}{c}U=\frac{{\left(4.00\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}^2\left(2.50\text{mm}\frac{{10}^{-3}\text{m}}{1.0\text{mm}}\right)}{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}+1.0\text{0cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\left(6.5-1\right)\right)}\\ =\frac{{\left(4.00\times {10}^{-6}\text{C}\right)}^2}{2\left(7.08\times {10}^{-12}\text{F}\right)}\\ =1.13\text{J}\end{array}$

Therefore, the energy stored in the capacitor is $1.13\text{J}$.

To determine

The magnitude and direction of the force exerted on dielectric by the plates of capacitor.

Answer to Problem 88PQ

The magnitude and direction of the force exerted on dielectric by the plates of capacitor is $77.7\text{N}\widehat{i}$.

Explanation of Solution

Write the equation for energy stored in the capacitor.

  $U=\frac{Q^2}{2C}$

Here, $U$ is the energy stored in the capacitor, $Q$ is the charge on the capacitor, $C$ is the equivalent capacitance of the capacitor.

Substitute equation (III) in above expression to find the energy stored in the capacitor.

  $\begin{array}{c}U=\frac{Q^2}{2\left(\frac{{\epsilon }_{0}s\left(s+y\left(\kappa -1\right)\right)}{d}\right)}\\ =\frac{Q^{2}d}{2{\epsilon }_{0}s\left(s+y\left(\kappa -1\right)\right)}\end{array}$                                                                                  (IV)

Write the expression to find the force exerted on dielectric by the plates of capacitor.

  $\overrightarrow{F}=-\left(\frac{dU}{dy}\right)\widehat{i}$

Substitute equation (IV) in above expression to find the force exerted on dielectric by the plates of capacitor.

  $\begin{array}{c}\overrightarrow{F}=-\left(\frac{d\left(\frac{Q^{2}d}{2{\epsilon }_{0}s\left(s+y\left(\kappa -1\right)\right)}\right)}{dy}\right)\widehat{i}\\ =\frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_{0}s{\left(s+y\left(\kappa -1\right)\right)}^2}\widehat{i}\end{array}$

Conclusion:

Substitute $4.00\mu \text{C}$ for $Q$,  $8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$, $2.50\text{cm}$ for $s$, $1.0\text{0cm}$ for $y$, $2.50\text{mm}$ for $d$, $6.5$ for $\kappa$ to find the magnitude and direction of the force exerted on dielectric by the plates of capacitor.

  $\begin{array}{c}\overrightarrow{F}=\frac{{\left(4.00\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}^2\left(2.50\text{mm}\frac{{10}^{-3}\text{m}}{1.0\text{mm}}\right)\left(6.50-1\right)}{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right){\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}+1.0\text{0cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\left(6.5-1\right)\right)}^2}\widehat{i}\\ =77.7\text{N}\widehat{i}\end{array}$

Therefore, the magnitude and direction of the force exerted on dielectric by the plates of capacitor is $77.7\text{N}\widehat{i}$.