A parallel plate capacitor with air between the plate has a capacitance of $15$ pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant $3.5$ . Then the capacitance becomes $x /4$ pF.
The value of $x$ is

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TestHub · Accepted Answer

The initial capacitance of the parallel plate capacitor with air is given by:

$C_{1} = \frac{ϵ _{0} A}{d} = 15 pF$

When the separation between the plates is doubled ( $d^{'} = 2 d$ ) and the space is filled with a dielectric of constant $K = 3.5$ , the new capacitance becomes:

$C_{2} = \frac{K ϵ _{0} A}{d ^{'}} = \frac{3.5 ϵ _{0} A}{2 d}$

Substituting the value of $\frac{ϵ _{0} A}{d}$ from the initial condition:

$C_{2} = \frac{3.5}{2} \times 15 = \frac{7}{4} \times 15 = \frac{105}{4} pF$

Given that the new capacitance is $\frac{x}{4} pF$ , we can compare the two expressions:

$\frac{x}{4} = \frac{105}{4}$

$x = 105$

Answer: $105$

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