91Ó°ÊÓ

a) Area of the plates,$A=0.12m^2$

b) Separation of the plates,$d=1.2cm$

c) Potential difference,$V=120V$

d) Thickness of the slab,$b=4mm$

e) Dielectric constant of the material,$\mathrm{κ}=4.8$

If the space between theplates of a capacitor is completely filled with a dielectric material, the capacitance Cis increased by a factor k , called the dielectric constant, which is characteristic of the material. In a region thatis completely filled by adielectric, all electrostatic equations containing must be modified by replacing${\mathit{Î\mu }}_{\mathbf{0}}$ with$\mathit{κ}{\mathit{Î\mu }}_{\mathbf{0}}$ .

Using the formula for the capacitance of a parallel plate capacitor, we can find the capacitance before and after the slab is inserted. Using the equation, we can find the charge before and after the slab is inserted and also the magnitude of the electric field. By using the relation for electric field without dielectrics and with dialectics, we can find the magnitude of the electric field. We can find the external work done from the equation.

Formulae:

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

The charge stored between the capacitor plates,$q=CV$ ...(ii)

The electric field between the plates without dielectric,$E=\frac{q}{{\mathrm{Î\mu }}_{0}A}$ ...(iii)

The electric field if the plates filled with the dielectric material,$E\text{'}=\frac{E}{\mathrm{κ}}$ ...(iv)

The potential difference between the plates,$V=E.d$ ...(v)

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

The work done by a body due to the energy change, $W_{eq}=∆U$ ...(vii)

By substituting the values in equation (i) with no dielectric, we can get the value of capacitance before the slab is inserted as follows:

$$\begin{gathered} C_0=\frac{\left(8.85×{10}^{-12}{C}^2/N.m^2\right)\left(0.12m^2\right)}{1.2×{10}^{-2}m} \\ =8.85×{10}^{-11}F \\ ≈8.9×{10}^{-11}F \\ =89pF \end{gathered}$$

Hence, the value of the capacitance is 89pF .

The electric field in the free space is given using equation (iii) as follows:
$E_1=\frac{q}{{\mathrm{Î\mu }}_{0}A}$ …(±¹¾±¾±¾±)

But inside the slab, the electric field in the presence of a dielectric can be given using the above value in equation (iv) as follows:

$E_2=\frac{q}{\mathrm{κ}{\mathrm{Î\mu }}_{0}A}$

Since the total potential is the sum of the potential inside and the outside the slab, thus, the value of the potential is given using the above value in equation (v) as follows:

$$\begin{gathered} V_0=E_1\left(d-b\right)+E_{2}b \\ =\left(\frac{q}{{\mathrm{Î\mu }}_{0}A}\right)\left(d-b+\frac{b}{k}\right) \end{gathered}$$ …(¾±³æ)

Substituting the above value in equation (ii), we can get the total capacitance when the slab is in place as follows:

$$\begin{gathered} C=\frac{q}{\left({\displaystyle \frac{q}{{\mathrm{Î\mu }}_{0}A}}\right)\left(d-b+{\displaystyle \frac{b}{k}}\right)} \\ =\frac{{\mathrm{Î\mu }}_{0}AK}{\left(d-b\right)k+b} \\ =\frac{\left(8.85×{10}^{-12}{C}^2/N.m^2\right)\left(0,12m^2\right)\left(4.8\right)}{\left(1.2×{10}^{-2}m-0.4×{10}^{-2}m\right)4.8+\left(4×{10}^{-3}m\right)} \\ =\frac{5.1×{10}^{-12}}{8×{10}^{-3}×4.8+4×{10}^{-3}} \\ =1.2×{10}^{-10}F \\ =1.2×{10}^{2}F \end{gathered}$$

Hence, the value of the capacitance is$1.2×{10}^{2}F$ .

The freechargebefore the slab is insertedis given using the given data in equation (ii) as follows:

$$\begin{gathered} q_f=\left(8.9×{10}^{-11}F\right)\left(120V\right) \\ =11×{10}^{-9}C \\ =11nC \end{gathered}$$

Hence, the value of the free charge is 11nC .

Now, the dielectric slab is inserted in the plates but since the battery is disconnected, the total charge will remain constant. Thus, the value of the charge is 11nC .

By substituting the values in equation (viii), we can get the magnitude of the electric field as follows:

$$\begin{gathered} E=\frac{11×{10}^{-9}C}{\left(8.85×{10}^{-12}{C}^2/N.m\right)\left(0.12m^2\right)} \\ =10000V/m \\ =10kV/m \end{gathered}$$

Hence, the value of the field is 10kV/m .

Now, substituting the above value in equation (iv), we can get the value of the electric field in the dielectric space as follows:

$$\begin{gathered} E_2=\frac{10000V}{4.8m} \\ =2100V/m \\ =2.1kV/m \end{gathered}$$

Hence, the value of the field is 2.1 kV/m .

The potential difference across the plates when the slab is in place is given using the calculated value in equation (ix) (derived in part (ix) calculations)as follows:

By substituting the values, we can get

$$\begin{gathered} V=\left(2100V/m\right)×\left(4×{10}^{-3}m\right)+10000V/m\left(12×{10}^{-3}m-4×{10}^{-3}m\right) \\ =88V \end{gathered}$$

Hence, the value of the potential difference is 88V .

The external work done when the slab is inserted within the capacitor plates can be given using equation (vi) in equation (vii) as follows:

($U_0$ is the potential energy without dielectrics and U is the potential energy with dielectrics.)

$$\begin{gathered} W_{ext}=\frac{q^2}{2}\left(\frac{1}{C}-\frac{1}{C_0}\right) \\ =\frac{{\left(11×{10}^{-9}C\right)}^2}{2}\left(\frac{1}{1.2×{10}^{-10}F}-\frac{1}{8.85×{10}^{-11}F}\right) \\ =6×{10}^{-17}×\left(-2.9×{10}^9\right)C^2/F \\ =-1.7×{10}^{-7}J \end{gathered}$$

Hence, the value of the work done is$-1.7×{10}^{-7}J$ .