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# A parallel plate capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$, the percentage increase in the capacitance will be:A) $400\%$B) $66.6\%$C) $33.3\%$D) $200\%$

Last updated date: 21st Apr 2024
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Hint: Dielectrics are materials that are insulating and have no free electrons. They do, however, exhibit a microscopic displacement of charges in the presence of an electric field. The insertion of a dielectric increases the capacitance of the capacitor due to this very reason. To find the increase, we subtract the initial capacitance without the dielectric from the final capacitance with the dielectric. Also this arrangement uses a series combination of capacitors.

Formulae used:
$C = \dfrac{A{\varepsilon_0}}{d}$
Where $C$ is the capacitance, $A$ is the area of the plates, ${\varepsilon _0 }$ is the permittivity of vacuum or free space and $d$ is the distance between the plates. When a dielectric of dielectric constant $k$ is added to the capacitor, the capacitance increases to
$C = \dfrac{kA{\varepsilon_0}}{d}$
Series combination of capacitors:
${C_s} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$

Complete step by step solution:
Without any dielectric, we know that the capacitance of a parallel plate capacitor is
$C = \dfrac{{A{\varepsilon _0 }}}{d}$
Where $C$ is the capacitance, $A$ is the area of the plates, ${\varepsilon _0 }$ is the permittivity of vacuum or free space and $d$ is the distance between the plates.
In this arrangement, dielectric has only been added to half the space and rest is still in air. This creates a combination of two capacitors with half the initial distance of the plates, same area of plates and different permittivity.
We will first find the permittivity for the half without the dielectric for ease.
Since the distance has now been halved, therefore
${C_1} = \dfrac{{A{\varepsilon _0 }}}{{\dfrac{d}{2}}}$
$\Rightarrow {C_1} = \dfrac{{2A{\varepsilon _0 }}}{d}$ $...(1)$
For the half with the dielectric, a new constant is multiplied and are is halved
${C_2} = \dfrac{{5A{\varepsilon _0 }}}{{\dfrac{d}{2}}}$
$\Rightarrow {C_2} = \dfrac{{10A{\varepsilon _0 }}}{d}$ $...(2)$
Therefore the final new capacitance is;
${C_s} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$
$\Rightarrow {C_s} = \dfrac{{\left( {\dfrac{{10A{\varepsilon _0 }}}{d}} \right)\dfrac{{2A{\varepsilon _0 }}}{d}}}{{\dfrac{{12A{\varepsilon _0 }}}{d}}}$
$\Rightarrow {C_s} = \dfrac{{20A{\varepsilon _0 }}}{{12d}} = \dfrac{{5A{\varepsilon _0 }}}{{3d}}$
The increase in capacitance is then
${C_s} - C = \dfrac{{5A{\varepsilon _0 }}}{{3d}} - \dfrac{{A{\varepsilon _0 }}}{d}$
$\Rightarrow {C_s} - C = \dfrac{{2A{\varepsilon _0 }}}{{3d}}$
Percentage increase is,
$\Rightarrow \dfrac{{{C_s} - C}}{C} \times 100 = \dfrac{{\dfrac{{2A{\varepsilon _0 }}}{{3d}}}}{{\dfrac{{A{\varepsilon _0 }}}{d}}} \times 100$
$\Rightarrow \dfrac{{{C_s} - C}}{C} \times 100 = \dfrac{2}{3} \times 100 = 66.6\%$

Therefore the percentage increase in the capacitance is (B), $66.6\%$.

Note: The insertion of a dielectric material between the capacitor plates breaks the original capacitor into a system of capacitors. The combination type (series/parallel) is decided by looking at the alignment of the material between the plates of the capacitor. The alignment given in the question makes the capacitor a series combination of two capacitors.