Question 93 of 180

Two metallic plates form a parallel plate capacitor. The distance between the plate is '$d$'. A metal sheet of thickness $\dfrac{d}{2}$ 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?

A$1:2$B$2:1$C$1:4$D$4:1$

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Answer: (B) $2:1$

Before the metal sheet is inserted, capacitance is given by $C=\dfrac{{\epsilon }_{0}A}{d}$.
After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance ${C}^{'}=\dfrac{{\epsilon }_{0}A}{\left(\dfrac{d}{4}\right)}=4C$.
The equivalent capacity is now $\dfrac{1}{C'}=\dfrac{1}{4C}+\dfrac{1}{4C}\Rightarrow C'=2C$.
The ratio is $\dfrac{C}{C'}=\dfrac{1}{2}=1:2$