(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…

at the surface &

at the centre &

When a dielectric slab is inserted between the plates of capacitance there is induced charge density \(\displaystyle \sigma_{\mathrm{P}}\) which opposes the original charge density \(\displaystyle (\sigma)\) on the plate of capacitance.

Electric field with dielectric medium is

\[\begin{aligned} & \mathrm{E}=\frac{\left(\sigma-\sigma_{\mathrm{P}}\right)}{\varepsilon_{0}} \\ & \mathrm{~V}=\mathrm{E} \times \mathrm{d}=\frac{\left(\sigma-\sigma_{\mathrm{P}}\right)}{\varepsilon_{0}} \mathrm{d} \end{aligned} \]

\[\left(\sigma-\sigma_{\mathrm{P}}\right)=\frac{\sigma}{\mathrm{K}} \]

\[\mathrm{V}=\frac{\sigma \mathrm{d}}{\varepsilon_{0} \mathrm{K}}=\frac{\mathrm{Qd}}{\mathrm{~A} \varepsilon_{0} \mathrm{K}} \]

\[\mathrm{C}=\frac{Q}{\mathrm{V}}=\frac{\mathrm{K} \varepsilon_{0} A}{\mathrm{d}} \]