Chapter 26, Problem 65AP

(a)

To determine

The stored energy in the system as function of $x$.

Answer to Problem 65AP

The energy stored in the system is $\frac{Q_0{}^{2}d\left(l-x\right)}{2{\epsilon }_0{l}^3}$.

Explanation of Solution

The following figure shows the metal block of length $l$ slide inside the two square plates of length $l$ for a distance $x$. The upper plate carries a positive charge of $+Q_0$ and the lower plate carries a negative charge of $-Q_0$.

Figure (1)

Any charge that appears on any of the plate of the capacitor will induce an equal magnitude and opposite sign of charge on the nearby surface of the slab. The slab has zero net charge and zero electric field inside it.

Write the expression for capacitance.

    $C=\frac{k{\epsilon }_{0}A}{d}$                                                                                                                   (I)

Here, $d$ is the distance between two plates, $A$ is the area of the plate, $C$ is the capacitance, ${\epsilon }_0$ is the free space permittivity and $k$ is the dielectric constant.

From figure (1), the area of the metal inside the capacitor is $lx$ and the area of the metal outside the capacitor is $l\left(l-x\right)$.

The value of dielectric constant for the partially filled capacitor with metal is $k=\infty$. Therefore, value of the capacitance for that portion of the capacitor is,

Substitute $lx$ for $A$ and $\infty$ for $k$ in Equation (I) to calculate $C_{\text{d}}$.

    $\begin{array}{c}{C}_{\text{d}}=\frac{\infty \left(lx\right){\epsilon }_0}{d}\\ =\infty \end{array}$

The value of capacitance for the portion where, metal is not inserted is,

Substitute $l\left(l-x\right)$ for $A$ and $1$ for $k$ in Equation (I) to calculate $C$.

    $C=\frac{{\epsilon }_0\left(l\right)\left(l-x\right)}{d}$

The value of the whole capacitance is,

Substitute $l^2$ for $A$ and $1$ for $k$ in Equation (I) to calculate $C_{\text{t}}$.

    $C_{\text{t}}=\frac{{\epsilon }_0\left(l^2\right)}{d}$

Write the expression for charge in the capacitor.

    $Q_{\text{t}}=C_{\text{t}}V$                                                                                                                  (II)

Here, $V$ is the voltage and $Q_{\text{t}}$ is the total charge in the capacitor.

For the whole conductor, the value of $Q_{\text{t}}$ is $Q_0$.

Substitute $\frac{{\epsilon }_0\left(l^2\right)}{d}$ for $C$ and $Q_0$ for $Q_{\text{t}}$ in Equation (II) to calculate $V$.

    $\begin{array}{c}{Q}_0=\left(\frac{{\epsilon }_0\left(l^2\right)}{d}\right)V\\ V=\frac{Q_{0}d}{{\epsilon }_0{\left(l\right)}^2}\end{array}$

The value of the charge for the portion where, metal is not inserted is,

Substitute $\frac{Q_{0}d}{{\epsilon }_0{\left(l\right)}^2}$ for $V$ and $\frac{{\epsilon }_0\left(l\right)\left(l-x\right)}{d}$ for $C_{\text{t}}$ in Equation (II) to calculate $Q$.

    $\begin{array}{c}Q=\left(\frac{Q_{0}d}{{\epsilon }_0{\left(l\right)}^2}\right)\left(\frac{{\epsilon }_0\left(l\right)\left(l-x\right)}{d}\right)\\ =\frac{Q_0\left(l-x\right)}{l}\end{array}$

Write the expression for the energy stored in the capacitor.

    $U=\frac{Q^2}{2C}$                                                                                                                   (III)

Here, $U$ is the energy stored.

Substitute $\frac{\left(l-x\right)Q_0}{l}$ for $Q$ and $\frac{{\epsilon }_0\left(l\right)\left(l-x\right)}{d}$ for $C$ in Equation (III) to calculate $U$.

    $\begin{array}{c}U=\frac{{\left(\frac{\left(l-x\right)Q_0}{l}\right)}^2}{2\left(\frac{{\epsilon }_0\left(l\right)\left(l-x\right)}{d}\right)}\\ =\left(\frac{{\left(l-x\right)}^2{Q}_0{}^2\left(d\right)}{2l^2{\epsilon }_0\left(l\right)\left(l-x\right)}\right)\\ =\frac{Q_0{}^{2}d\left(l-x\right)}{2{\epsilon }_0{l}^3}\end{array}$

Conclusion:

Therefore, energy stored in the system is $\frac{Q_0{}^{2}d\left(l-x\right)}{2{\epsilon }_0{l}^3}$.

(b)

To determine

The magnitude and direction of force that acts on the metallic block.

Answer to Problem 65AP

Magnitude of the force is $\frac{Q_0{}^2\left(d\right)}{2{\epsilon }_0{l}^3}$ towards the right side.

Explanation of Solution

Write the expression for force.

    $F=-\frac{dU}{dx}$                                                                                                                (IV)

Here, $F$ is the force.

Substitute $\frac{\left(l-x\right)Q_0{}^2\left(d\right)}{2{\epsilon }_0{l}^3}$ for $U$ in Equation (IV) to calculate $F$.

    $\begin{array}{c}F=\frac{-d}{dx}\left(\frac{\left(l-x\right)Q_0{}^2\left(d\right)}{2{\epsilon }_0{l}^3}\right)\\ =\frac{+Q_0{}^2\left(d\right)}{2{\epsilon }_0{l}^3}\end{array}$

Conclusion:

Therefore, the magnitude of the force is $\frac{Q_0{}^2\left(d\right)}{2{\epsilon }_0{l}^3}$ towards the right side.

(c)

To determine

The stress on the advancing front face of the block.

Answer to Problem 65AP

The stress on the advancing front face of the block is $\frac{Q_0{}^2}{2{\epsilon }_0{l}^4}$.

Explanation of Solution

The advancing front face area is,

    $A=ld$.

Write the expression for stress.

    $\sigma =\frac{F}{A}$                                                                                                                      (V)

Here, $\sigma$ is the stress, $F$ is the force and $A$ is the area.

Substitute $\frac{Q_0{}^2\left(d\right)}{2{\epsilon }_0{l}^3}$ for $F$ and $ld$ for $A$ in Equation (V) to calculate $\sigma$.

    $\begin{array}{c}\sigma =\frac{\left(\frac{Q_0{}^2\left(d\right)}{2{\epsilon }_0{l}^3}\right)}{\left(ld\right)}\\ =\frac{Q_0{}^2}{2{\epsilon }_0{l}^4}\end{array}$                                                                                                        (VI)

Therefore, the stress on the advancing front face of the block is $\frac{Q_0{}^2}{2{\epsilon }_0{l}^4}$.

(d)

To determine

The energy density in the electric field between the charged plates in terms of $Q_0$, $l$, $d$ and ${\epsilon }_0$.

Answer to Problem 65AP

The energy density in the electric field is $\frac{Q_0{}^2}{2{\epsilon }_0{l}^4}$.

Explanation of Solution

Write the expression for energy density.

    $u=\frac{1}{2}\left(\frac{{\sigma }^2}{{\epsilon }_0}\right)$                                                                                                            (VII)

Here, $u$ is the energy density.

Substitute $\frac{Q_0}{l^2}$ for $\sigma$ in Equation (VII) to calculate $u$.

    $\begin{array}{c}u=\frac{1}{2}\left(\frac{{\left(\frac{Q_0}{l^2}\right)}^2}{{\epsilon }_0}\right)\sqrt{a^2+b^2}\\ =\frac{Q_0{}^2}{2{\epsilon }_0{l}^4}\end{array}$                                                                                      (VIII)

Therefore, the energy density in the electric field between the charged plate is $\frac{Q_0{}^2}{2{\epsilon }_0{l}^4}$

(e)

To determine

The comparison of part (c) with part (d).

Answer to Problem 65AP

The stress is equal to the energy density in the electric field between the charged plates.

Explanation of Solution

Compare the energy density with the stress that is comparing the equation (VI) with equation (VIII).

    $\begin{array}{c}\sigma =u\\ \frac{Q_0{}^2}{2{\epsilon }_0{l}^4}=\frac{Q_0{}^2}{2{\epsilon }_0{l}^4}\end{array}$

Therefore, the stress is equal to the energy density in the electric field between the charged plates.