JEE Advanced 2015 · Paper 2 · Physics Q14

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PhysicsMultiple correct+4 / -2Question 14 of 60

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 $\dfrac{C_2}{C_1}$ is

A parallel-plate capacitor (plates on the left (+) and right (−), separation d). The lower half of the plate area (S/2) is filled across the whole gap with dielectric ε₁. The upper half (S/2) has ε₁ filling the half of the gap next to the + plate and ε₂ filling the half (d/2) next to the − plate.

6/5

5/3

7/5

7/3

Answer and worked solution

Official answer: D

Let $C_1=\varepsilon_0S/d$ .

Lower half (area $S/2$ , filled with $\varepsilon_1=2$ ):
$C_a=\frac{2\varepsilon_0(S/2)}d=C_1.$

Upper half ( $\varepsilon_1$ and $\varepsilon_2$ , each $d/2$ thick, in series):

the $\varepsilon_1$ layer: $\dfrac{2\varepsilon_0(S/2)}{d/2}=2C_1$ ;
the $\varepsilon_2$ layer: $4C_1$ ;
in series: $\dfrac{2\cdot4}{6}C_1=\dfrac43C_1$ .

Total (in parallel). $C_2=\dfrac73C_1$ .

(D)

Topics:Dielectrics Series and parallel combinations

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JEE Advanced 2015 · Paper 2, Physics question 14