(a) (i) Obtain the expression for the capacitance of a parallel plate capacitor with a dielectric medium between its plates. (ii) A charge of 6 μ C…

When a dielectric slab is inserted between the plates of capacitor, there is induced charge density $\displaystyle \sigma_P$ which opposes the original charge density ($\displaystyle \sigma$) on the plate of capacitance.

Electric field with dielectric medium is

\[E = \frac{(\sigma - \sigma_P)}{\varepsilon_0}\]

\[V = E \times d = \frac{(\sigma - \sigma_P)}{\varepsilon_0} d\]

\[(\sigma - \sigma_P) = \frac{\sigma}{K}\]

\[V = \frac{\sigma\, d}{\varepsilon_0 K} = \frac{Qd}{A\varepsilon_0 K}\]

\[C = \frac{Q}{V} = \frac{K\varepsilon_0 A}{d}\]

Electric potential due to a point charge

\[V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}\]

At the surface

\[V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} = \frac{9\times 10^{9}\times 6\times 10^{-6}}{0.2}\]

\[V = 2.7\times 10^{5}\ \text{V}\]

Since electric field inside the hollow sphere is zero, hence V is same as that of the surface and remains constant throughout the volume.

\[V = 2.7\times 10^{5}\ \text{V}\]

The Flux through the Gaussian surface is

\[= E \times 4\pi R^{2}\]

In this case Gaussian surface encloses no charge.

Hence $\displaystyle E \times 4\pi R^{2} = 0$

\[E = 0\](ii) Field outside the shell-

Electric flux through Gaussian surface

\[E \times 4\pi r^{2} = \frac{(\sigma\, 4\pi R^{2})}{\varepsilon_0}\]

Charge enclosed by the Gaussian surface

\[E \times 4\pi r^{2} = \frac{(\sigma\, 4\pi R^{2})}{\varepsilon_0}\]

Using Gauss's law:

\[\int \vec{E}\cdot \vec{ds} = \frac{Q}{\varepsilon_0}\]

\[E \times 4\pi r^{2} = \frac{(\sigma\, 4\pi R^{2})}{\varepsilon_0}\]

\[E = \frac{\sigma}{\varepsilon_0}\frac{R^{2}}{r^{2}} = \frac{q}{4\pi\varepsilon_0 r^{2}}\]

For conducting sheet,

Electric field due to a conducting sheet

\[E_c = \frac{\sigma}{\varepsilon_0}\]

For non-conducting sheet

\[E_{nc} = \frac{\sigma}{2\varepsilon_0}\]

Since surface charge density is same.

\[2E_{nc} = E_c\]