Question

(a) A parallel plate capacitor of plate area 2 m
and plate separation 2 mm is charged to 1000 V
in vacuum. Calculate the capacitance of the
capacitor and charge on each plate.
(b) Now the battery is disconnected and a
dielectric slab of dielectric constant 5
introduced. What will be the capacitance of the
capacitor and charge on each plate?
batinnn the plates of a​

Answer 1

(a) area of parallel plate capacitor , A = 2m²

seperation between parallel plates, d = 2mm = 2 × 10^-3 m

voltage across capacitor, v = 1000 volts.

use formula, $C=\frac{\epsilon_0A}{d}$ to find capacitance of capacitor.

where , $\epsilon_0$ = 8.85 × 10^-12 C²/Nm²

i.e., C = 8.85 × 10^-12 × 2/2 × 10^-3

= 8.85 × 10^-9 F

and charge on capacitor , Q = CV

= 8.85 × 10^-9 × 1000

= 8.85 × 10^-6 C

(b) Now battery is disconnected and dielectric slab of dielectric constant 5 introduced.

so, charge must be conserved.

so, charge = 8.85 × 10^-6 C [ initial value ]

capacitance of capacitor, C' = kC where k is dielectric constant.

here k = 5 and C = 8.85 × 10^-9 F

so, capacitance = 5 × 8.85 × 10^-9 F

= 44.25 × 10^-9 F

= 4.425 × 10^-8 F