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VITEEE 2015 Paper

Section: Physics

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Question : 14 of 120

Marks: +1, -0

The capacitance of a parallel plate capacitor with air as medium is 3 µF. As a dielectric is introduced between the plates, the capacitance becomes 15 µF. The permittivity of the medium in

C^2 N^{-1} m^{-2}

is

8.15 x $10^{-11}$
0.44 x $10^{-10}$
15.2 x $10^{12}$
1.6 x $10^{-14}$

Solution:

Capacitance of air capacitor

C_0

=

\frac{\varepsilon_0 A}{d}

= 3 µF ... (i)

When a dielectric of permittivity

\varepsilon_r

and dielectric constant K is introduced between the plates, then

Capacitance, C =

\frac{K \varepsilon_0 A}{d}

= 15 µF ... (ii)

Dividing eq. (ii) by (i), we get

\frac{C}{C_0}

=

\frac{d}{\frac{\varepsilon_0 A}{d}}

=

\frac{15}{3}

⇒ K = - 5

∴ permittivity of the medium

\varepsilon_r

=

\varepsilon_0

K

= 8.85 ×

10^{-12}

× 5 = 0.44 ×

10^{-10}

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