Question

b) The capacitance of a parallel-plate capacitor with vacuum
is 5 uF. If a dielectric slab with K = 1.5 is inserted between
the plates, the capacitance will be equal to​

Answer 1

Answer:

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Explanation:

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C' = K C

Ans is 7.5 ūF

Answer 2

Answer:

The capacitance will be equal to​ 7.5μF.

Explanation:

The capacitance of the capacitor with vacuum is given as,

$C=\frac{\epsilon_oA}{d}$       (1)

Where,

C=capacitance of the capacitor

ε₀=permitivity of free space

A=area of the plates of the capacitor

d=distance between the plates of the capacitor

From the question we have,

C=5μF

K=1.5

But substituting the value of C in equation (1) we get;

$5\mu F=\frac{\epsilon_oA}{d}$       (2)

When the dielectric slab is inserted between the plates then capacitance becomes,

$C'=\frac{K\epsilon_oA}{d}$         (3)

K=dielectric constant

By using equation (2) and placing the value of K in equation (1) we get;

$C'=1.5\times 5\mu F=7.5\mu F$

Hence, the capacitance will be equal to​ 7.5μF.