A Parallel Plate Capacitor Having Plates Of Area S And Plate Separation D Has Capacitance C1 In Air. When Two Dielectrics Of Different Relative Permittivities (Ε1=1 And Ε2=4) Are Introduced Between The Two Plates. The Capacitance Becomes C2. What Will Be The Ratio Of C2/C1?

Sol:

$$C = 2\varepsilon _{0} \frac{s}{d}\\{C}’ = 4\varepsilon _{0} \frac{s}{d}$$

They are in series, their equivalent =

$$\frac{4\varepsilon _{0} \frac{s}{d} * 2\varepsilon _{0} \frac{s}{d}}{4\varepsilon _{0} \frac{s}{d} + 2\varepsilon _{0} \frac{s}{d}}\\C_{3} = \frac{2\varepsilon _{0}s}{2d} = \frac{\varepsilon _{0}s}{d}\\C_{2} = \frac{4\varepsilon _{0}s}{3d} + \frac{\varepsilon _{0}s}{d} = \frac{7}{3}\varepsilon _{0}\frac{s}{d}\\C_{1} = \varepsilon _{0}\frac{s}{d}\\\frac{C_{2}}{C_{1}} = \frac{7}{3}$$

Therefore, the ratio of$$\frac{C_{2}}{C_{1}} = \frac{7}{3}$$

Explore more such questions and answers at BYJU’S.