Question 1:
difficultA parallel plate condenser is filled with two dielectrics as shown in figure. Area of each plate is A metre² and the separation is d metre. The dielectric constants are K1 and K2 respectively. Its capacitance in farad will be :
Both the parts can be taken as separate capacitors connected in parallel.
So, C=C1+ C2=\(\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} +K_{2}\right)}{2}\)
Question 2:
difficultThe capacity of a parallel plate condenser is \(C_0\). If a dielectric of relative permittivity \(\varepsilon_r\) and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes \(C\). The value of \(\frac{C}{C_0}\) will be :
The initial capacitance is \(C_0 = \frac{\varepsilon_0 A}{d}\). With a dielectric slab of thickness \(t = d/4\), the new capacitance is \(C = frac{\varepsilon_0 A}{d - t + \frac{t}{\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4\varepsilon_r}}\). Simplifying this gives \(C = C_0 \frac{4\varepsilon_r}{3\varepsilon_r + 1}\).