Question

A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each plate is $A{m^2}$ and the separation is $d$ meter. The dielectric constants are ${K_1}$ and ${K_2}$ respectively. Its capacitance (in farad) will be:
(A) $\dfrac{{2{\varepsilon _0}A}}{d}\left( {\dfrac{{{K_1} + {K_2}}}{{{K_1}{K_2}}}} \right)$
(B) $\dfrac{{2{\varepsilon _0}A}}{d}\left( {\dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right)$
(C) $\dfrac{{{\varepsilon _0}A}}{d}\left( {\dfrac{{{K_1} + {K_2}}}{{2{K_1}{K_2}}}} \right)$
(D) $\dfrac{{{\varepsilon _0}A{K_1}{K_2}}}{{2\left( {{d_2}{K_1} + {d_1}{K_2}} \right)}}$

Answer 1

Hint : Here, use the given figure and also the information given in the question such that the space between the parallel plates is divided into two equal parts hence the distance between two plates is divides as $\dfrac{d}{2}$ and $\dfrac{d}{2}$ . Use formula for effective capacitance of capacitors in series combination.

Complete Step By Step Answer:
Here, in the above figure and the information given says that the parallel plate capacitor is half filled with dielectric ${K_1}$ and half with ${K_2}$ and the total area is $A{m^2}$ and the distance is divided between two dielectrics is $d$ . Let ${C_1}$ be the capacitance of capacitor with ${K_1}$ and ${C_2}$ be the capacitance of capacitor with dielectric ${K_2}$
Thus, we have formula for calculating the capacitance as
$C = \dfrac{{KA{\varepsilon _0}}}{d}$ ….(general formula for capacitance)
Thus we have,
${C_1} = \dfrac{{{K_1}A{\varepsilon _0}}}{{\dfrac{d}{2}}}$
$\Rightarrow {C_1} = \dfrac{{2{K_1}A{\varepsilon _0}}}{d}$ …. $(1)$
Similarly,
${C_2} = \dfrac{{2{K_2}A{\varepsilon _0}}}{d}$ …. $(2)$
They are in series. Thus, we have to use the formula for effective capacitance in series combination.
So,
${C_{eff}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
$\Rightarrow {C_{eff}} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$
From equations $(1)$ and $(2)$ , we can write above equation as:
$\Rightarrow {C_{eff}} = \dfrac{{\left( {\dfrac{{2{K_1}A{\varepsilon _0}}}{d}} \right)\left( {\dfrac{{2{K_2}A{\varepsilon _0}}}{d}} \right)}}{{\left( {\dfrac{{2{K_1}A{\varepsilon _0}}}{d} + \dfrac{{2{K_2}A{\varepsilon _0}}}{d}} \right)}}$
$\Rightarrow {C_{eff}} = \dfrac{{4\left( {{K_1}{K_2}\dfrac{{{A^2}{\varepsilon _0}^2}}{{{d^2}}}} \right)}}{{\left( {2{K_1} + 2{K_2}} \right)\dfrac{{A{\varepsilon _0}}}{d}}}$
$\Rightarrow {C_{eff}} = \dfrac{4}{2}\left( {\dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right)\dfrac{{A{\varepsilon _0}}}{d}$
$\Rightarrow {C_{eff}} = 2\dfrac{{A{\varepsilon _0}}}{d}\left( {\dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right)$
Thus, the required answer is given by $\dfrac{{2A{\varepsilon _0}}}{d}\left( {\dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right)$
The correct answer is option B.

Note :
Here, we can observe that the capacitor is filled with two different dielectrics which divides the capacitor from between in two equal parts. Here we see that the gap between two parallel plates is divided between two in two equal parts. And also these two capacitors are connected in series that is why we use the series combination formula for calculating capacitance in the capacitor.