JEE Challenger
More from Electrostatic Potential and Capacitance
PhyMedAdv 2018 • Paper 1

Dielectric Permittivity in Switched Multi-Capacitor Circuit

Consider three identical capacitors C1,C2C_1, C_2, and C3C_3, each possessing a capacitance of 1.0 μF1.0\text{ }\mu\text{F}, which are initially uncharged. As illustrated in the circuit diagram, they are connected to a cell of electromotive force (emf) V0=8 VV_0 = 8\text{ V}, and capacitor C1C_1 is completely filled with a dielectric substance having relative permittivity ϵr\epsilon_r. Initially, switch S1S_1 is closed with switch S2S_2 remaining open. Once capacitor C3C_3 achieves full charge, switch S1S_1 is opened and switch S2S_2 is concurrently closed. After the circuit settles into electrostatic equilibrium, the charge residing on C3C_3 is measured to be 5 μC5\text{ }\mu\text{C}. Determine the value of ϵr\epsilon_r.

Question Diagram 1
Official Numerical Answer1.5

Topics & Concepts

Electrostatic Potential and CapacitanceCapacitorsDielectrics

Step-by-Step Solution

To determine the relative permittivity ϵr\epsilon_r, we analyze the circuit in two stages:

Stage 1: Switch S1S_1 is closed and S2S_2 is open

  • Since switch S2S_2 is open, the middle branch containing capacitors C1C_1 and C2C_2 remains uncharged.
  • Capacitor C3C_3 is connected directly across the cell of emf V0=8 VV_0 = 8\text{ V} through switch S1S_1.
  • Given the capacitance of C3C_3 is C0=1.0 μFC_0 = 1.0\text{ }\mu\text{F}, the initial charge stored on C3C_3 when fully charged is: Q0=C3V0=(1.0 μF)×(8 V)=8 μCQ_0 = C_3 V_0 = (1.0\text{ }\mu\text{F}) \times (8\text{ V}) = 8\text{ }\mu\text{C}

Stage 2: Switch S1S_1 is opened and S2S_2 is closed

  • Opening S1S_1 disconnects the cell from the circuit, isolating the charges.

  • Closing S2S_2 connects the series combination of capacitors C1C_1 and C2C_2 in parallel with capacitor C3C_3.

  • Let CsC_s be the equivalent capacitance of the series combination of C1C_1 and C2C_2: C1=ϵrC0=ϵr(1.0 μF)C_1 = \epsilon_r C_0 = \epsilon_r (1.0\text{ }\mu\text{F}) C2=C0=1.0 μFC_2 = C_0 = 1.0\text{ }\mu\text{F} Cs=C1C2C1+C2=ϵrϵr+1 μFC_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{\epsilon_r}{\epsilon_r + 1}\text{ }\mu\text{F}

  • By charge conservation, the initial charge Q0Q_0 on the top plate of C3C_3 redistributes between C3C_3 and the series combination of C1C_1 and C2C_2: Q0=Q3′+QsQ_0 = Q_3' + Q_s where:

    • Q3′=5 μCQ_3' = 5\text{ }\mu\text{C} is the final charge on C3C_3.
    • QsQ_s is the charge on the series branch of C1C_1 and C2C_2.

  • Thus, the charge on the series combination is: Qs=Q0−Q3′=8 μC−5 μC=3 μCQ_s = Q_0 - Q_3' = 8\text{ }\mu\text{C} - 5\text{ }\mu\text{C} = 3\text{ }\mu\text{C}

  • Since C3C_3 and the branch with CsC_s are in parallel, they share the same final potential difference VfV_f: Vf=Q3′C3=5 μC1.0 μF=5 VV_f = \frac{Q_3'}{C_3} = \frac{5\text{ }\mu\text{C}}{1.0\text{ }\mu\text{F}} = 5\text{ V}

  • The equivalent capacitance CsC_s can also be expressed in terms of QsQ_s and VfV_f: Cs=QsVf=3 μC5 V=0.6 μFC_s = \frac{Q_s}{V_f} = \frac{3\text{ }\mu\text{C}}{5\text{ V}} = 0.6\text{ }\mu\text{F}

  • Equating the two expressions for CsC_s: ϵrϵr+1=0.6=35\frac{\epsilon_r}{\epsilon_r + 1} = 0.6 = \frac{3}{5}

  • Solving for ϵr\epsilon_r: 5ϵr=3(ϵr+1)5\epsilon_r = 3(\epsilon_r + 1) 5ϵr=3ϵr+35\epsilon_r = 3\epsilon_r + 3 2ϵr=3  ⟹  ϵr=1.52\epsilon_r = 3 \implies \epsilon_r = 1.5

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