Question

Question 2.5: A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10 −12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-87

Answer 1

Use the formula, $\bf{C=\frac{\epsilon_0A}{d}}$
Here C is the capacitance of parallel plate capacitor
d is the separation between two parallel plates
A is the cross sectional area of plate.

given, $C_0=\frac{\epsilon_0A}{d}=8pF$
now, distance becomes half , e.g., d' = d/2
And space between them is filled with a substance of dielectric constant is 6
so, $C=\frac{K\epsilon_0A}{d/2}$
= 2KC₀
= 2 × 6 × 8 pF
= 96 pF

Answer 2

Answer:

C'=2×6C = 12C = 12×8=96pF