91Ó°ÊÓ

From sample problem 25.06

• Area of the plates, $A=115c{m}^2$
• Separation between the plates, d = 1.24 cm
• The potential difference,$V_0=85.5V$
• Separation of the Gaussian surfaces,$b=0.780cm$
• Dielectric constant of the material,$\mathrm{κ}=2.61$

If the space between the plates of a capacitor is completely filled with a dielectric material, the capacitance C is increased by a factor k, called the dielectric constant, which is characteristic of the material. In a region that is completely filled by a dielectric, all electrostatic equations containing${\mathit{Î\mu }}_{\mathbf{0}}$ must be modified by replacing${\mathrm{Î\mu }}_0$ with$\mathrm{κ}{\mathrm{Î\mu }}_0$.

By using the concept of the electric field, we can find the total potential. From that value of potential, we can find the capacitance of the capacitor. Also, according to the equation, we can calculate the total charge on the capacitor plate. Then by using the relation between charge and electric field we can calculate the electric field in the gap. And finally, by using the equation, we can find the electric field when the slab is in place.

Formulae:

The electric field between the capacitor plates due to dielectric constant, $E=\frac{q}{\mathrm{κ}{\mathrm{Î\mu }}_{0}A}$ …(¾±)

Here, E is the electric field, $\mathrm{κ}$is the dielectric constant of the material,${\mathrm{Î\mu }}_0$ is the permittivity of the free space, q is the electric charge, A is the cross-section of the plates.

The capacitance value between the plates,$C=Q/V$ …(¾±¾±)

The potential difference between the plates, $V=E.d$ …(¾±¾±¾±)

The electric field in the free space (that is air, $\mathrm{κ}=1$) is given using equation (i) as:

$E_1=\frac{q}{{\mathrm{Î\mu }}_{0}A}$ …(¾±±¹)

But inside the slab, the electric field due to the dielectric material is given using equation (i) as:

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

Thus, we get the relation between both fields as:

$E_2=\frac{E_1}{k}$ …(±¹)

The total potential is the sum of the potential inside and outside the slab can be given using the given data and above value in equation (iii) as follows:

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

By substituting the above value in equation (ii), we can get the value of capacitance as follows:

$$\begin{gathered} C=\frac{q}{\left({\displaystyle \frac{q}{{\mathrm{Î\mu }}_{0}A}}\right)\left(d-b+{\displaystyle \frac{b}{k}}\right)} \\ =\frac{{\mathrm{Î\mu }}_{0}AK}{\left(d-b\right)k+b} \\ =\frac{\left(8.85×{10}^{-12}{C}^2/N.m^2\right)\left(115×{10}^{-4}{m}^2\right)\left(2.61\right)}{\left(0.0124m-0.00780m\right)×2.61+\left(0.00780m\right)} \\ =13.4pF \end{gathered}$$

Hence, the value of the capacitance is 13.4 pF .

Now, using the given data and the above capacitance value in equation (ii), we can get the value of the charge as follows:

$$\begin{gathered} q=\left(13.4×{10}^{-12}F\right)\left(85.5V\right) \\ =1.15nC \end{gathered}$$

Hence, the value of the charge is 1.15 nC .

The magnitude of the electric field in the gap with air filled can be given using equation (iv) and the given data as follows:

$$\begin{gathered} E_1=\frac{1.15×{10}^{-9}C}{\left(8.85×{10}^{-12}{C}^2/N.m^2\right)\left(115×{10}^{-4}{m}^2\right)} \\ =1.13×{10}^{4}N/C \end{gathered}$$

Hence, the value of the field is$1.13×{10}^{4}N/C$ .

By using the above field value in equation (iv), we can find the electric field in the slab after the slab in place as follows:

$$\begin{gathered} E_2=\frac{113×{10}^{4}N/C}{2.61} \\ =4.33×{10}^{3}N/C \end{gathered}$$

Hence, the value of the electric field is$4.33×{10}^{3}N/C$ .