91Ó°ÊÓ

• Thickness of slab$b=2.00×{10}^{-3}\mathrm{m}$
• Plate area of the capacitor$A=2.40×{10}^{-4}{\mathrm{m}}^2$
• Plate separation$d=5.00×{10}^{-3}\mathrm{m}$
• Charge $q=3.40×{10}^{-6}C$

When the capacitor is charged, the energy is stored in it. This energy is written in terms of charge and capacitance of the capacitor. Some work is done when the slab is inserted between the plates. The work done is equal to the change in the energy when the slab is inserted between the plates.

We need to use the capacitance formula and energy stored formula of parallel plate capacitor to find the capacitance and the energy stored respectively. Using the work-energy principle, we can find the work done on the slab.

Formulae:

The energy stored between the capacitor plates,$U=\frac{q^2}{2C}$ ...(i)

Here, q is charge, U is energy stored in capacitor, and C is capacitance of the capacitor.

The capacitance of the capacitor plates,$C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$ ...(ii)

Here, C is capacitance, ${\mathrm{Î\mu }}_0$ is the permittivity of the free space, A is the area of cross-section, and d is the separation between the plates.

The work done due to the energy change, $W=∆U$ ...(iii)

Here, W is work done, and $∆U$is the change in energy.

After the slab is introduced between the plates the length d is reduced to $d-b$.

Now, the new separation value can be given as:

$$\begin{gathered} d-b=5.00\mathrm{mm}-2.00\mathrm{mm} \\ =3.00\mathrm{mm} \\ =3.00×{10}^{-3}\mathrm{m} \end{gathered}$$

The capacitance for the new length is then given using equation (ii) by,

$$\begin{gathered} C\text{'}=\frac{{\mathrm{Î\mu }}_{0}A}{d-b} \\ =\frac{\left(8.85×{10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{{\mathrm{Nm}}^2}}\right)\left(2.40×{10}^{-4}{\mathrm{m}}^2\right)}{3.00×{10}^{-3}\mathrm{m}} \\ =0.708\mathrm{pE} \end{gathered}$$

Hence, the value of the capacitance is $0.708\mathrm{pE}$.

The energy stored in the capacitor before the slab is introduced can be given using equation (i) as follows:

$U=\frac{q^2}{2C}........................................................\left(\mathrm{iv}\right)$

And the energy stored in the capacitor after the slab is introduced that is given using equation (i) as:

$$\begin{gathered} U\text{'}=\frac{1}{2}C\text{'}{V}^2 \\ =\frac{q^2}{2C\text{'}}....................................................\left(\mathrm{v}\right) \end{gathered}$$

Taking the ratios of equation (iv) and (v), we get the value of the required energy stored ratio as follows:

role="math" localid="1661836854264" $$\begin{gathered} \frac{U}{U\text{'}}=\frac{{\displaystyle \frac{q^2}{2C}}}{{\displaystyle \frac{q^2}{C\text{'}}}} \\ =\frac{C\text{'}}{C}........................................\left(\mathrm{vi}\right) \end{gathered}$$

The capacitance of capacitor before the slab introduced is given using equation (ii) as follows:

$C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$

And the capacitance after slab is introduced that can be given using calculations from part (a) as follows:

$C\text{'}=\frac{{\mathrm{Î\mu }}_{0}A}{d-b}$

Now, substituting the values of capacitances in equation (vi), we get the value of the ratio of the stored energy values as:

$$\begin{gathered} \frac{U}{U\text{'}}=\frac{\left({\displaystyle \frac{{\mathrm{Î\mu }}_{0}A}{d-b}}\right)}{{\displaystyle \frac{{\mathrm{Î\mu }}_{0}A}{d}}} \\ =\frac{d}{d-b} \\ =\frac{\left(5.00×{10}^{-3}\mathrm{m}\right)}{\left(5.00×{10}^{-3}\mathrm{m}-2.00×{10}^{-3}\mathrm{m}\right)} \\ =1.67 \end{gathered}$$

Hence, the value of the ratio is $1.67$.

Substituting the value of equation (i) in equation (iii), we get the value of the work done on the slab as follows:

$$\begin{gathered} W=\frac{q^2}{2C\text{'}}-\frac{q^2}{2C} \\ =\frac{q^2}{2}\left(\frac{1}{C\text{'}}-\frac{1}{C}\right) \\ =\frac{q^2}{2{\mathrm{Î\mu }}_{0}A}\left(\left(d-b\right)-d\right)\left(\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{capacitances}\mathrm{from}\mathrm{part}\left(\mathrm{b}\right)\right) \\ =-\frac{q^{2}b}{2{\mathrm{Î\mu }}_{0}A} \\ =-\frac{{\left(3.40×{10}^{-6}\mathrm{C}\right)}^2\left(2×{10}^{-3}\mathrm{m}\right)}{2×8.85×{10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{{\mathrm{Nm}}^2}}×2.40×{10}^{-4}{\mathrm{m}}^2} \\ =-5.44\mathrm{J} \end{gathered}$$

Hence, the value of the work done is $-5.44\mathrm{J}$.

Since the work done is negative. The slab is sucked in.