Chapter 25, Problem 49CP

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 κ is inserted a distance x into the capacitor as shown in Figure P25.49 (page 690). 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 = 2.0 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 P25.49

To determine

The equivalent capacitance of the device.

Answer to Problem 49CP

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

Explanation of Solution

Given Info: The sides of the metallic plate is $l$, the separation between the plates is $d$ the charge on the metallic plates are $-Q$ and $+Q$. The dielectric constant is $k$ of material.

The given Figure of the system is shown below.

Figure (1)

The system can be considered as two capacitors one with a dielectric and one without dielectric. The top plates of the two capacitors always have same potential

Formula to calculate the area of the with dielectric is,

$A_1=xl$

Here,

$A_1$ is the area of the with dielectric.

$x$ is the distance of plate in the dielectric medium.

Formula to calculate the capacitance with dielectric is,

$C_1=\frac{k{\epsilon }_0{A}_1}{d}$ (1)

Here,

${\epsilon }_0$ is the permittivity of the free space.

$C_1$ is the capacitance with dielectric.

Substitute $xl$ for $A_1$ in equation (1) to find the $C_1$.

$C_1=\frac{k{\epsilon }_{0}lx}{d}$

Formula to calculate the area of the without dielectric is,

$A_2=l\left(l-x\right)$

Here,

$A_2$ is the area of the without dielectric.

Formula to calculate the capacitance without dielectric is,

$C_2=\frac{{\epsilon }_0{A}_2}{d}$ (2)

Here,

${\epsilon }_0$ is the permittivity of the air.

$C_2$ is the capacitance without dielectric.

Substitute $l\left(l-x\right)$ for $A_2$ in equation (2) to find the $C_2$.

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

Since, the capacitors $C_1$ and $C_2$ are in parallel. Hence, the net equivalent capacitance is,

$C=C_1+C_2$ (3)

Substitute $\frac{k{\epsilon }_{0}lx}{d}$ for $C_1$ and $\frac{{\epsilon }_{0}l\left(l-x\right)}{d}$ for $C_2$ in equation (3) to find the $C$.

$\begin{array}{c}C=\frac{k{\epsilon }_{0}lx}{d}+\frac{{\epsilon }_{0}l\left(l-x\right)}{d}\\ =\frac{{\epsilon }_0\left[xl\left(k-1\right)+l^2\right]}{d}\\ =\frac{{\epsilon }_{0}l}{d}\left[x\left(k-1\right)+l\right]\end{array}$

Conclusion:

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

To determine

The energy stored in the capacitor.

Answer to Problem 49CP

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

Explanation of Solution

Given Info: The sides of the metallic plate is $l$, the separation between the plates is $d$ the charge on the metallic plates are $-Q$ and $+Q$. The dielectric constant is $k$ of material.

The equivalent capacitance of the device is,

$\frac{{\epsilon }_{0}l}{d}\left[x\left(k-1\right)+l\right]$.

Formula to calculate the energy stored in the capacitor is,

$U=\frac{Q^2}{2C}$

Substitute $\frac{{\epsilon }_{0}l}{d}\left[x\left(k-1\right)+l\right]$ for $C$ to find the $U$.

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

Conclusion:

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

To determine

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

Answer to Problem 49CP

The magnitude of the force exerted by the plates on the dielectric is $\frac{Q^{2}d}{2{\epsilon }_{0}l}\frac{d\left(k-1\right)}{{\left[x\left(k-1\right)+l\right]}^2}\stackrel{\wedge }{\text{i}}$ and the direction of the force is outward from the capacitor.

Explanation of Solution

Given Info: The sides of the metallic plate is $l$, the separation between the plates is $d$ the charge on the metallic plates are $-Q$ and $+Q$. The dielectric constant is $k$ of material.

As, the dielectric slab is inserted the potential energy of the system is affected. The $x$-component of the conservative force associated with this interaction.

Formula to calculate the force exerted by the plates on the dielectric is,

$F=-\frac{\partial U}{\partial x}$

Here,

$F$ is the force exerted by the plates on the dielectric.

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

$\begin{array}{c}F=-\frac{\partial \left(\frac{Q^{2}d}{2{\epsilon }_{0}l\left[x\left(k-1\right)+l\right]}\right)}{\partial x}\stackrel{\wedge }{\text{i}}\\ =\frac{Q^2}{2{\epsilon }_{0}l}\frac{d\left(k-1\right)}{{\left[x\left(k-1\right)+l\right]}^2}\stackrel{\wedge }{\text{i}}\end{array}$

The negative value of the force indicates the dielectric is pushed out from the capacitor.

Conclusion:

Therefore, the magnitude of the force exerted by the plates on the dielectric is $\frac{Q^{2}d}{2{\epsilon }_{0}l}\frac{\left(k-1\right)}{{\left[x\left(k-1\right)+l\right]}^2}\stackrel{\wedge }{\text{i}}$ and the direction of the force is outward from the capacitor.

To determine

The magnitude of the force exerted by the plates on the dielectric.

Answer to Problem 49CP

The magnitude of the force exerted by the plates on the dielectric is $205\stackrel{\wedge }{\text{i}}\text{μN}$.

Explanation of Solution

Given Info: The sides of the metallic plate is $5.00\text{cm}$, the separation between the plates is $2.00\text{mm}$, The dielectric constant is $4.50$ of material and the charge on capacitor is $2.00\times {10}^3\text{V}$

The distance of plate in the dielectric medium is,

$x=\frac{l}{2}$

Formula to calculate the charge is,

$Q=\frac{{\epsilon }_0{l}^2\Delta V}{d}$

Here,

$\Delta V$ is the potential of the capacitor.

The magnitude of the force exerted by the plates on the dielectric is,

$F=\frac{Q^{2}d}{2{\epsilon }_{0}l}\frac{\left(k-1\right)}{{\left[x\left(k-1\right)+l\right]}^2}\stackrel{\wedge }{\text{i}}$ (4)

Here,

$Q$ is the charge on the capacitor.

$k$ is the dielectric constant.

$d$ is the separation between the plates.

$l$ is the sides of the metallic plate.

Substitute $\frac{l}{2}$ for $x$ in equation (4).

$\begin{array}{c}F=\frac{Q^{2}d}{2{\epsilon }_{0}l}\frac{\left(k-1\right)}{{\left[\frac{l}{2}\left(k-1\right)+l\right]}^2}\stackrel{\wedge }{\text{i}}\\ =\frac{2Q^{2}d}{{\epsilon }_0{l}^3}\frac{\left(k-1\right)}{{\left[\left(k+1\right)\right]}^2}\stackrel{\wedge }{\text{i}}\end{array}$ (5)

Substitute $\frac{{\epsilon }_0{l}^2\Delta V}{d}$ for $Q$ in equation (5) to find the $F$

$\begin{array}{c}F=\frac{2{\left(\frac{{\epsilon }_0{l}^2\Delta V}{d}\right)}^{2}d}{{\epsilon }_0{l}^3}\frac{\left(k-1\right)}{{\left[\left(k+1\right)\right]}^2}\stackrel{\wedge }{\text{i}}\\ =\frac{2{\epsilon }_{0}l{\left(\Delta V\right)}^2\left(k-1\right)}{d{\left(k+1\right)}^2}\stackrel{\wedge }{\text{i}}\end{array}$ (6)

Substitute $5.00\text{cm}$ for $l$, $4.50$ for $k$, $2.00\times {10}^3\text{V}$ for $\Delta V$, $8.8542\times {10}^{-12}{{\text{C}}^2/\text{N}\text{.m}}^2$ for ${\epsilon }_0$ and $2.00\text{mm}$ for $d$ to find the $F$.

$\begin{array}{c}F=\frac{2{\epsilon }_{0}l{\left(\Delta V\right)}^2\left(k-1\right)}{d{\left(k+1\right)}^2}\stackrel{\wedge }{\text{i}}\\ =\frac{2\left(8.8542\times {10}^{-12}{{\text{C}}^2/\text{N}\text{.m}}^2\right){\left(2.00\times {10}^3\text{V}\right)}^2\left(5.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right){\left(\Delta V\right)}^2\left(4.50-1\right)}{2.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right){\left(4.50+1\right)}^2}\stackrel{\wedge }{\text{i}}\\ =205\times {10}^{-6}\stackrel{\wedge }{\text{i}}\text{N}\\ \text{=}205\stackrel{\wedge }{\text{i}}\text{μN}\end{array}$

Conclusion:

Therefore, the magnitude of the force exerted by the plates on the dielectric is $205\stackrel{\wedge }{\text{i}}\text{μN}$.