Question 712: Electrostatics & Capacitance

Question

A parallel plate capacitor having plates of area S and plate separation d, has capacitance C
1
in air. When two dielectrics of different relative permittivities ($\varepsilon $
1
= 2 and $\varepsilon $
2
= 4) are introduced between the two plates as shown in the figure, the capacitance becomes C
2
. The ratio ${{{C_2}} \over {{C_1}}}$ is
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l0wn7s3z/a2616b50-b1b4-4fe0-a2fa-f1eb091b0abb/7d16c100-a6d9-11ec-bd73-d1f19edd5c44/file-1l0wn7s40.png?format=png)

Accepted Answer

We can think of this configuration to be made up of three parts:
(a) Capacitor of plate area s/2, plate separation d, and filled with a dielectric of relative permittivity $\in$
1
= 2 (lower half). The capacitance of this part is ${C_a} = { \in _1}{ \in _0}{{s/2} \over d} = { \in _0}{s \over d} = {C_1}$.
(b) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$
1
= 2 (left upper half). The capacitance of this part is ${C_b} = { \in _1}{ \in _0}{{s/2} \over {d/2}} = 2{ \in _0}{s \over d} = 2{C_1}$.
(c) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$
2
= 4 (right upper half). The capacitance of this part is ${C_c} = { \in _2}{ \in _0}{{s/2} \over {d/2}} = 4{ \in _0}{s \over d} = 4{C_1}$.
The capacitors C
b
and C
c
are connected in series with equivalent capacitance
${C_{bc}} = {{{C_b}{C_c}} \over {{C_b} + {C_c}}} = {{(2{C_1})(4{C_1})} \over {(2{C_1}) + (4{C_1})}} = {4 \over 3}{C_1}$.
The capacitor C
bc
is connected in parallel with C
a
. The equivalent capacitance is
$\left. {{C_2} = {C_{abc}} = {C_a}} \right\|{C_{bc}} = {C_1} + {4 \over 3}{C_1} = {7 \over 3}{C_1}$.

Answer

${6 \over 5}$

Answer

${5 \over 3}$

Answer

${7 \over 5}$

Answer

${7 \over 3}$

Electrostatics & Capacitance

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