Question

A parallel plate capacitor has a plate area of $100{m^2}$ and plate separation of $10m.$ The space between the plates is filled up to a thickness of $5m$ with a material of dielectric constant $10$. The resultant capacitance of the system is ‘$x$’ pF. The value of ${\varepsilon _0} = 8.85 \times {10^{ - 12}}F{m^{ - 1}}$. The value of ‘$x$’ to the nearest integer is ___

Answer 1

Hint: In order to solve this question, we will use the general formula for the capacitance of a parallel plate capacitor and by using given parameters values, we will solve for the resultant capacitance of the parallel plate capacitor.

Formula used:
For a parallel plate capacitor, if area of plates is A, distance between the plates is d, thickness of the dielectric constant filled is t, dielectric constant is k, relative permittivity of free space is ${\varepsilon _0}$ then, resultant capacitance C is given by,
$C = \dfrac{{{\varepsilon _0}A}}{{d - t + \dfrac{t}{k}}}$

Complete step by step solution:
We have given the following parameters value for a parallel plate capacitor as
$A = 100{m^2} \\
\Rightarrow d = 10m \\
\Rightarrow t = 5m \\
\Rightarrow k = 10 \\
{\varepsilon _0} = 8.85 \times {10^{ - 12}}F{m^{ - 1}} \\ $
We need to calculate the resultant capacitance of the parallel plate capacitor $x$ in units of pF. Using Formula $C = \dfrac{{{\varepsilon _0}A}}{{d - t + \dfrac{t}{k}}}$ and putting the values we get,
$C = \dfrac{{8.85 \times {{10}^{ - 12}} \times 100}}{{10 - 5 + \dfrac{5}{{10}}}}$
here C denotes for x so,
$x = \dfrac{{885 \times {{10}^{ - 12}}}}{{5.5}} \\
\therefore x = 161pF \\ $
Hence, the value of $x$ is $161$ in the units of pF.

Note: It should be remembered that the unit of capacitance is Farad and Pico-farad is a smaller unit of capacitance which is related to Farad as $1pF = {10^{ - 12}}F$, always convert all the physical quantities units into the required form to get the correct value in such numerical.