Question

A parallel plate capacitor is having a plate area of $100c{{m}^{2}}$and separation between the plates is $1cm$. A glass plate ${{K}_{1}}=6$ of thickness $6mm$ and an ebonite plate $\left( {{K}_{2}}=4 \right)$ of thickness $4mm$ are inserted. Find the value of C.

& A.4.085pF \\\ & B.40.85pF \\\ & C.0.4085nF \\\ & D.40.85nF \\\ \end{aligned}$$

Answer 1

When two or more capacitors are kept in series, their reciprocal of equivalent or the effective capacitance is given as the sum of the reciprocal of their individual capacitances. This is the opposite in the case of resistors. Using this equation of effective capacitance, the answer is found. Hope this will help you to solve this question.

Complete step-by-step answer:
As the capacitors are in series the reciprocal of the equivalent capacitance can be found using the summation of the reciprocals of each and every capacitance included. Therefore in this case, we can write that,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$
We can cross multiply the terms,
$C=\dfrac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}$
As we all know, the capacitance of a material is given by the equation,
$C=\dfrac{K{{\varepsilon }_{0}}A}{d}$
Therefore now we can substitute this equation in effective capacitance.
$C=\dfrac{{{\varepsilon }_{0}}A{{K}_{1}}{{K}_{2}}}{{{K}_{1}}{{d}_{2}}+{{K}_{2}}{{d}_{1}}}$
Where ${{\varepsilon }_{0}}$ be the permittivity of the material.
Its value is given as,
${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}$
It is already mentioned in the question that the dielectric constants of the glass plate and ebonite plate is ${{K}_{1}}$ and ${{K}_{2}}$ respectively.
That is we can write that,
$\begin{aligned} & {{K}_{1}}=6 \\\ & {{K}_{2}}=4 \\\ \end{aligned}$
And the thickness of the glass plate and ebonite plate respectively are given as ${{d}_{1}}$ and ${{d}_{2}}$.
Therefore we can write that,

& {{d}_{1}}=6mm \\\ & {{d}_{2}}=4mm \\\ \end{aligned}$$ Area of the capacitor plate is written as, $$A=100c{{m}^{2}}$$ And the separation between the plates is given as, $$S=1c{{m}^{2}}$$ Substituting these values in the equation will give, $C=\dfrac{8.85\times {{10}^{-12}}\times {{10}^{-2}}\times 6\times 4}{\left( 6\times 6+4\times 4 \right)\times {{10}^{-3}}}$ Simplifying this will give, $C=4.085\times {{10}^{-11}}F$ $C=40.85pF$ Therefore the correct answer is option A.  **So, the correct answer is “Option A”.** **Note:** The capacitance of a circuit is meant by the property of the capacitor in order to store energy in it. The measure of capacity to store energy in a capacitor is the capacitance. The capacitance for a given substance or material is constant.