(I) The capacitance of a parallel plate capacitor with dielectric slab (t < d)
+q, –q = the charges on the capacitor plates
\(+q_i, - q_i\) = Induced charges on the faces of the dielectric slab
\(E_0\) → electric field intensity in air between the plates
E → the reduced value of electric field intensityinside the dielectric slab.
When a dielectric slab of thickness thickness t<d d is introduced between the two plates of the capacitor the electric field reduces to E due to the polarisation of the dielectric. The potential difference between the two plates is given by
\(V = V_i+ + V_t + V_2\)
\(V = E_0 d_1 + Et + E_0d_2\ ..(1)\)
Here E is the reduced value of electric field intensity
\(\vec E = \vec E_0 + \vec E _i .\ Here\ \vec E_i\) is the electric field due to the induced charges \([+q_i\ and\ -q_i]\)
\(E = \sqrt{E_o^2 + E_i^2 + 2E_0E_1\cos180^\circ}\)
\( = \sqrt{(E_0 - E_i)^2}\)
\(E =E_0 - E_i\)
Also the dielectric constant K is given by
\(K = \frac{E_0}{E}\ ..(2)\)
\(E_0 = \frac{\sigma}{\varepsilon_0} = \frac{q}{A \varepsilon_0}\ ..(3)\)
From equations (1), (2) and (3)
\(V = E_0[d_1+ d_2] + \frac{E_0}{K}t\)
\(V = \frac{q}{A\varepsilon_0}[d -t + \frac{t}{K}]\ ..(4)\)
The capacitance of the capacitor on the introduction of the dielectric slab is
\(C = \frac{q}{V}\ ..(5)\)
From (4) and (5)
\(C = \frac{\varepsilon _0A}{d -t + \frac{t}{k}}\)
If t = d, then \(C = K\frac{\varepsilon_0A}{d}\Rightarrow C = KC_0\) \(Here\ C_0 = \frac{\varepsilon_0A}{d}\)
Since K >1 therefore \(C>C_0\)
(II) For a metallic slab K is infinitely large, therefore \(C = \frac{\varepsilon_0a}{d-t}\)
