Question

Capacity of on air filled capacitor is 5 uf,
when a dielectric plate, (K = 4) is introduced
to cover one third area of each plate capacity
will become​

Answer 1

Solution :

Given capacitance of capacitor

C=∈Ad -----------(1)

Where A and d are area of each plate and seperation between two plates Now,

C′=K∈′0Ad′

=×∈0A(d/2)

=10(∈0Ad)

=10×8pF

c=80pF

Answer 2

Answer:

Capacity of the capacitor is 6.25μF.

Explanation:

When a dielectric slab is introduced in the partial part of a capacitor then the capacitor can be treated as multiple capacitors arranged in series or parallel according to the arrangement of dielectric slab.

Given one third area is covered with slab,

The capacitor can be treated as 2 capacitors in parallel whose capacitance are as follows,

$C_{1} =\frac{E_{0}A_{1} }{d}\\C_{2}= \frac{E_{0}A_{2} }{Kd}$

where A₁ and A₂ are A/3 and 2A/3 respectively.

Since C₁=5μF

so C₂=C₁/K=C₁/4=1.25μF

Since they are in parallel,

$C_{eq}=C_{1} +C_{2} \\ C_{eq}=(5+1.25)uF\\C_{eq}=6.25uF$

Hence the capacity of the capacitor is 6.25μF.