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← JEE Advanced 2018 · Paper 1
PhysicsNumerical+3Question 11 of 54

JEE Advanced 2018 · Paper 1, Physics question 11

Three identical capacitors C1C_1, C2C_2 and C3C_3 have a capacitance of 1.0 μ1.0\ \muF each and they are uncharged initially. They are connected in a circuit as shown in the figure and C1C_1 is then filled completely with a dielectric material of relative permittivity ϵr\epsilon_r. The cell electromotive force (emf) V0=8V_0 = 8 V. First the switch S1S_1 is closed while the switch S2S_2 is kept open. When the capacitor C3C_3 is fully charged, S1S_1 is opened and S2S_2 is closed simultaneously. When all the capacitors reach equilibrium, the charge on C3C_3 is found to be 5 μ5\ \muC. The value of ϵr\epsilon_r = ______.

Three parallel branches between a top and a bottom wire: left, the cell V₀ in series with switch S₁; middle, switch S₂ in series with C₁ (dielectric-filled, hatched) and C₂; right, C₃.
Answer and worked solution

Official answer: 1.5

First stage. With S1S_1 closed, C3C_3 charges to V0V_0: Q0=1×8=8 μQ_0=1\times8=8\ \muC.

Second stage. S1S_1 opens and S2S_2 closes, so C3C_3 shares its charge with C1C_1 and C2C_2 in series:
Cs=ϵr⋅1ϵr+1 μF.C_s=\frac{\epsilon_r\cdot1}{\epsilon_r+1}\ \mu\text{F}.

Charge conservation. Finally C3C_3 holds 5 μC, so the common voltage is 5 V. The series pair took the other 3 μC:
Cs×5=3  ⟹  Cs=0.6  ⟹  ϵrϵr+1=0.6  ⟹  ϵr=1.5.C_s\times5=3\implies C_s=0.6\implies\frac{\epsilon_r}{\epsilon_r+1}=0.6\implies\epsilon_r=1.5.

Answer: 1.5