Two metallic plates form a parallel plate capacitor. The distance between the plate is ' $d$ '. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

Question

TestHub · Accepted Answer

Before the metal sheet is inserted, capacitance is given by $C = \frac{ε_{0} A}{d}$ .

After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance $C^{'} = \frac{ε_{0} A}{(\frac{d}{4})} = 4 C$ .
The equivalent capacity is now $\frac{1}{C'} = \frac{1}{4 C} + \frac{1}{4 C} \Rightarrow C' = 2 C$ .

The ratio is $\frac{C}{C'} = \frac{1}{2} = 1 : 2$

Physics - Capacitor Question with Solution | TestHub

Options:

Doubts & Discussion