Two parallel plates of area $\mathbf{100}{\mathbf{cm}}^{\mathbf{2}}$are given charges of equal magnitudes $\mathbf{8}\mathbf{.}\mathbf{9}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{C}$, but opposite signs. The electric field within the dielectric material filling the space between the plates is$\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{x}{\mathbf{10}}^{\mathbf{6}}\frac{\mathbf{V}}{\mathbf{m}}$(a)Calculate the dielectric constant of the material(b)Determine the magnitude of charge induced on each dielectric surface.

1. Area of the plates,$\mathrm{A}=100{\mathrm{cm}}^2$
2. Magnitude of the equal charges,$\mathrm{q}=8.9\times {10}^{-7}\mathrm{C}$
3. The charges have equal magnitudes and opposite signs.
4. Electric field within the dielectric material,$\mathrm{E}=1.4\times {10}^6\mathrm{V}/\mathrm{m}$

When a dielectric is present,Gauss’ law may be generalized to,

${\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\oint }\mathit{k}\overrightarrow{\mathbf{E}}\mathbf{.}\mathit{d}\overrightarrow{\mathbf{A}}\mathbf{=}\mathit{q}$

Here, is a free charge; any induced surface charge is accounted for by including the dielectric constant k inside the integral.

By using the concept of Gauss’s law with dielectric, we can find the dielectric constant of the material and the charge induced on the dielectric surface.

Formulae:

The electric field between the capacitor plates with dielectric,$E=\frac{q}{k{\epsilon }_{0}A}$ …(i)

Here, is the electric field,is the dielectric constant of the material,${\epsilon }_0$is the permittivity of the free space, q is electric charge and A is the area of cross-section.

The net charge dielectric induced on each plate, $q\text{'}=q\left(1-\frac{1}{K}\right)$ …(ii)

Here, q is electric charge and q' is induced electric charge, k is the dielectric constant of the material.

By substituting the given values in equation (i), we can calculate the value of the dielectric constant of the material as follows:

$$\begin{gathered} \mathrm{k}=\frac{8.9\times {10}^{-7}\mathrm{C}}{\left(1.4\times {10}^6\mathrm{V}/\mathrm{m}\right)\times \left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.\mathrm{m}\right)\times \left(100\times {10}^{-4}{\mathrm{m}}^2\right)} \\ =7.2 \end{gathered}$$

Hence, the value of the dielectric constant is 7.2 .

The magnitude of the induced charge on each dielectric constant is given using equation (ii) as follows:

$$\begin{gathered} \mathrm{q}\text{'}=\mathrm{q}-\frac{\mathrm{q}}{\mathrm{k}} \\ =\mathrm{q}\left(1-\frac{1}{\mathrm{k}}\right) \\ =8.9\times {10}^{-7}\mathrm{C}\left(1-\frac{1}{7.2}\right) \\ =7.7\times {10}^{-7}\mathrm{C} \end{gathered}$$

Hence, the value of the induced charge is$7.7\times {10}^{-7}\mathrm{C}$s .