Question

Repeat Problem 70, assuming that a potential difference V=85.00 V, rather than the charge, is held constant.

Short answer

a) Capacitance after the slab is introduced is 0.708 pF.

b) Ratio of stored energy before to that after the slab is inserted is 0.600

c) Work is done on the slab is$1.2\times {10}^{-9}\mathrm{J}$

d The slab is sucked in.

Step-by-step solution

The given data

a) Thickness of slab$\mathrm{b}=2.00\times {10}^{-3}\mathrm{m}$

b) Plate area of capacitor$A=2.40\times {10}^{-4}{\mathrm{m}}^2$

c) Plate separation$d=5.00\times {10}^{-3}\mathrm{m}$

d) Charge$q=3.40\times {10}^{-6}\mathrm{c}$

e) Potential difference V=85 V

Understanding the concept of the work and energy

When the capacitor is charged, the energy is stored in it. This energy is written in terms of voltage 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 of the parallel plate capacitor to find the capacitance. An energy stored formula to calculate the ratio of energies before and after the slab is introduced. The work done on the slab can be calculated by taking the difference in energies.

Formulae:

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

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

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

Here,C is capacitance,${\epsilon }_0$ is permittivity of the free space,A is area of cross section, andd is 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.

(a) Calculation of the capacitance after the slab is introduced halfway between the plates

After the slab is introduced between the plates the lengthdis reduced tod-b.

Now, the new separation between the plates is given as:

$$\begin{gathered} d-b=5\mathrm{mm}-2\mathrm{mm} \\ =3\mathrm{mm} \\ =3\times {10}^{-3}\mathrm{m} \end{gathered}$$

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

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

Hence, the value of the capacitance is 0.78 pF.

(b) Calculation of the ratio of the stored energy before, to that after the slab is inserted

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

$U=\frac{1}{2}C{V}^2\dots \dots \dots \dots \dots \dots .\dots \dots \dots \dots \dots \dots \left(iv\right)$

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

$U^{\text{'}}=\frac{1}{2}{C}^{\text{'}}{V}^2\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(v\right)$

Taking the ratios of equation 1) and 2), we get the value of the stored energy ratio as:

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

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

$C=\frac{{\epsilon }_{0}A}{d}$

And the capacitance after slab is introduced can be given using equation (ii) as:

$C\text{'}=\frac{{\epsilon }_{0}A}{d-a}$

Now, substituting the values of capacitances in equation 3), we get the required ratio as:

$$\begin{gathered} \frac{U}{U\text{'}}=\frac{{\displaystyle \frac{{\epsilon }_{0}A}{d}}}{{\displaystyle \frac{{\epsilon }_{0}A}{d-a}}} \\ =\frac{d-b}{a} \\ =\frac{5.00\times {10}^{-3}\mathrm{m}-2.00\times {10}^{-3}\mathrm{m}}{5.00\times {10}^{-3}\mathrm{m}} \\ =0.600 \end{gathered}$$

Hence, the value of the ratio is 0.600.

(c) Calculation of the work done on the slab

Substituting the values of potential energies from equation (iii) and (iv) in equation (iii), we get the value of the work done on the slab as follows:

$$\begin{gathered} \mathrm{W}=\frac{1}{2}\mathrm{C}\text{'}{\mathrm{V}}^2-\frac{1}{2}{\mathrm{CV}}^2 \\ =\frac{1}{2}\left(\mathrm{C}\text{'}-\mathrm{C}\right){\mathrm{V}}^2 \\ =\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{2}\left(\frac{1}{\mathrm{d}-\mathrm{b}}-\frac{1}{\mathrm{d}}\right){\mathrm{V}}^2\left(\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{capacitances}\mathrm{from}\mathrm{part}\left(\mathrm{b}\right)\right) \\ =\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{2}\frac{\left(\mathrm{d}-\mathrm{b}+\mathrm{b}\right)}{\left(\mathrm{d}-\mathrm{b}\right)}{\mathrm{V}}^2 \\ =\frac{{\mathrm{\epsilon }}_0{\mathrm{AbV}}^2}{2\mathrm{d}\left(\mathrm{d}-\mathrm{b}\right)} \\ =\frac{8.85\times {10}^{-12}{\displaystyle \frac{\mathrm{C}}{{\mathrm{Nm}}^2}}\times 2.40\times {10}^{-4}{\mathrm{m}}^2\times 2.00\times {10}^{-3}\mathrm{m}\times {\left(85\mathrm{V}\right)}^2}{2\times 5.00\times {10}^{-3}\mathrm{m}\times 3.00\times {10}^{-3}\mathrm{m}} \\ =1.02\times {10}^{-9}\mathrm{J} \end{gathered}$$

Hence, the value of the work done is $1.02\times {10}^{-9}\mathrm{J}$.

(d) Calculation of the reason of the slab behaviour

Here, we are assuming that the potential difference is constant. It means slab is getting inserted when the battery is left attached. Itimplies that the force on the slab is not conservative. The charge distribution in the slab causes the slab to be sucked into the gap by the charge distribution on the plates.

Hence, the slab is get sucked in.