Question

Two metallic plates form a parallel plate capacitor. The distance between the plates is ‘ d ’. A metal sheet of thickness 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. 2 : 1
• B. 1 : 2
• C. 1 : 4
• D. 4 : 1

Answer 1

Answer: (A)

2 : 1

Answer 2

C$C_{eq}=\frac{e0A}{d-\frac{d}{2}+\frac{d}{2k}}=\frac{e0A}{\frac{d}{2}}=\frac{2e0A}{d}$

If

$C=\frac{e0A}{d}$

⇒$c_{eq}2c$ or $\frac{C_{new]}}{C_{c_dd}}=\frac{2}{1}$