Chapter 20, Problem 85P

A capacitor is constructed from two square, metallic plates of sides ℓ and separation d. Charges +Q and −Q are placed on the plates, and the power supply is then removed. A material of dielectric constant K is inserted a distance x into the capacitor as shown in Figure P20.85. Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (d) Obtain a numerical value for the force when x = ℓ/2, assuming ℓ = 5.00 cm, d = 2.00 mm, the dielectric is glass (κ = 4.50), and the capacitor was charged to 2.00 × 10³ V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel.

Figure P20.85

To determine

The equivalent capacitance of the device.

Answer to Problem 85P

The equivalent capacitance of the device is $\frac{{\epsilon }_{0}l}{d}\left[l+x\left(\kappa -1\right)\right]$.

Explanation of Solution

Write the expression for the equivalent capacitance.

$C_{eq}=C_1+C_2$        (I)

Here, $C_{eq}$ is the equivalent capacitance of the circuit and $C_1,C_2$ are capacitance connected in the circuit.

Write the expression for the area with dielectrics.

$A=lx$        (II)

Here, $A$ is the area, $l$ is the length and $x$ is the width.

Write the equation for capacitance by using equation (II).

  $C_1=\frac{\kappa {\epsilon }_{0}lx}{d}$        (III)

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

Write the expression for the area without dielectrics.

$A=l\left(l-x\right)$        (IV)

Write the equation for capacitance by using equation (IV).

  $C_2=\frac{{\epsilon }_{0}l\left(l-x\right)}{d}$        (V)

Conclusion:

Substitute, $\frac{\kappa {\epsilon }_{0}lx}{d}$ for $C_1$, $\frac{{\epsilon }_{0}l\left(l-x\right)}{d}$ for $C_2$ in Equation (I) to find $C_{eq}$.

  $\begin{array}{c}{C}_{eq}=\left[\frac{\kappa {\epsilon }_{0}lx}{d}\right]+\left[\frac{{\epsilon }_{0}l\left(l-x\right)}{d}\right]\\ =\frac{{\epsilon }_{0}l}{d}\left[l+x\left(\kappa -1\right)\right]\end{array}$

Thus, the equivalent capacitance of the device is $\frac{{\epsilon }_{0}l}{d}\left[l+x\left(\kappa -1\right)\right]$.

To determine

The energy stored in the capacitor.

Answer to Problem 85P

The energy stored in the capacitor is $\underset{\_}{\frac{Q^{2}d}{2{\epsilon }_{0}l\left[l+x\left(\kappa -1\right)\right]}}$.

Explanation of Solution

Write the expression for the stored energy.

$U=\frac{Q^2}{2C}$        (VI)

Here, $Q$ is the charge and $U$ is the energy.

Conclusion:

Substitute, $\frac{{\epsilon }_{0}l}{d}\left[l+x\left(\kappa -1\right)\right]$ for $C$ in Equation (VI) to find $U$.

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

Thus, the energy stored in the capacitor is $\underset{\_}{\frac{Q^{2}d}{2{\epsilon }_{0}l\left[l+x\left(\kappa -1\right)\right]}}$.

To determine

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

Answer to Problem 85P

The direction and the magnitude of the force exerted by the plates on the dielectrics is $\underset{\_}{\frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_{0}l\left[l+x{\left(\kappa -1\right)}^2\right]}}$ and directed along positive x-axis.

Explanation of Solution

Write the expression exerted force.

$F=\frac{dU}{dx}$        (V)

Conclusion:

Substitute, $\frac{Q^{2}d}{2{\epsilon }_{0}l\left[l+x\left(\kappa -1\right)\right]}$ for $U$ in Equation (V) to find $F$.

  $\begin{array}{c}F=\frac{d\left[\frac{Q^{2}d}{2{\epsilon }_{0}l\left[l+x\left(\kappa -1\right)\right]}\right]}{dx}\\ =\frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_{0}l\left[l+x{\left(\kappa -1\right)}^2\right]}\end{array}$        (VI)

For $x=0$, the value of force is,

$\begin{array}{c}F=\frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_{0}l\left[l+\left(0\right){\left(\kappa -1\right)}^2\right]}\\ =\frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_0{l}^3}\end{array}$

For $x=l$, the value of force is,

$\begin{array}{c}F=\frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_{0}l\left[l+l{\left(\kappa -1\right)}^2\right]}\\ \approx \frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_0{l}^3{\kappa }^2}\end{array}$

The force is directed along positive x-direction.

Thus, the direction and the magnitude of the force exerted by the plates on the dielectrics is $\underset{\_}{\frac{Q^{2}d\left(\kappa -1\right)}{2{\epsilon }_{0}l\left[l+x{\left(\kappa -1\right)}^2\right]}}$ and directed along positive x-axis.

To determine

The numerical value of force.

Answer to Problem 85P

The numerical value of force is $\underset{\_}{205\text{μN}}$.

Explanation of Solution

Write the expression for the area.

$A=l^2$        (VII)

Write the expression for the initial capacitance.

$C_0=\frac{{\epsilon }_0{l}^2}{d}$        (VIII)

Here, $C_0$ is the initial capacitance.

Write the expression for the charge.

$\begin{array}{c}Q=C_0\left(\Delta V\right)\\ =\frac{{\epsilon }_0{l}^2\left(\Delta V\right)}{d}\end{array}$        (IX)

Write the expression for the given condition.

$x=\frac{l}{2}$        (X)

Write the expression for the force by using (VI), (VII), (VIII), (IX) and (X).

$F=\frac{2{\epsilon }_{0}l{\left(\Delta V\right)}^2\left(\kappa -1\right)}{d{\left(\kappa +1\right)}^2}$        (XI)

Conclusion:

Substitute, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$, $5.00\text{cm}$ for $l$, $2.00\text{mm}$ for $d$, $4.50$ for $\kappa$, $2.00\times {10}^3\text{V}$ for $\left(\Delta V\right)$ in Equation (XI) to find $F$.

  $\begin{array}{c}F=\frac{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)\left[\left(5.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]{\left(2.00\times {10}^3\text{V}\right)}^2\left(4.50-1\right)}{\left[\left(2.00\text{mm}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right]{\left(4.50+1\right)}^2}\\ =205\times {10}^{-6}\text{N}\\ =205\text{μN}\end{array}$

Thus, the numerical value of force is $\underset{\_}{205\text{μN}}$.