A parallel- plate capacitor has plate area `100cm^2` and plate separation `1.0cm`. A glass plate (dielectric constant 4.0)are inserted one over the other the fill the space between the plate of the capacitor.Find the new capacitance.

Options

A. 44pF, 211.2pF
B. 4.4pF, 211.2pF
C. 4.4pF, 21.12pF
D. 21.12pF, 44pF

Answer

C

Explanation

`A= 50 cm^2 = 50 xx10^(-4) m^2 , d = 1 xx10^(-2) m`
V = 200 Volt
When battery is disconnected -
`k = 4.8`
Before, `C_(@)=(Aepsilon_0)/ d =(50 xx10^(-4)xx8.85 xx10^(-12))/(1xx10^(-2))=4.42` pF
After, `C=kC_(@) = 4.8 xx4.42 pF = 21.21 ` pF