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BITSAT 2018
Exam
BITSAT
Year
2018
Subject
Physics
Chapter
Capacitance

BITSAT 2018 · Physics · Capacitance

A parallel plate capacitor C with plates of unit area and...

A parallel plate capacitor CC with plates of unit area and separation dd is filled with a liquid of dielectric constant K=2K = 2, the level of liquid isd3,\dfrac{d}{3}, initially. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K
Suppose, the liquid level decreases at a constant speed vv, the time constant as a function of time is

Options

  1. A6ε0R5d+3vt\dfrac{6{ε}_{0}R}{5d + 3vt}
  2. B(15d+9vt)ε0R2d3−3dvt−gv2t2\dfrac{( 15d + 9vt ) {ε}_{0}R}{2d^{3} - 3dvt - gv^{2}t^{2}}
  3. C6ε0R5d−3vt\dfrac{6{ε}_{0}R}{5d - 3vt}
  4. D(15d−9vt)ε0R2d3+3dvt−9v2t2\dfrac{( 15d - 9vt ) {ε}_{0}R}{2d^{3} + 3dvt - 9v^{2}t^{2}}

Correct answer

A. 6ε0R5d+3vt\dfrac{6{ε}_{0}R}{5d + 3vt}

Step-by-step solution

We have, the time constant, τ=RC′…(i)τ = RC ' \ldots \left(i\right)

Now, C′=C1C2C1+C2C ' = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}

=(Aε0d−x)(KAε0x)Aε0d−x+KAε0x= \dfrac{\left(\dfrac{A{ε}_{0}}{d - x}\right)\left(\dfrac{KA{ε}_{0}}{x}\right)}{\dfrac{A{ε}_{0}}{d - x} + \dfrac{KA{ε}_{0}}{x}}

∴C′=KAε0x+K(d−x)∴ C ' = \dfrac{KA{ε}_{0}}{x + K ( d - x )}

Putting the value of C′C ' in Equation. (i), we get

τ=RKAε0d3−vt+k(d−d3+vt)[∵x=d3−vt]τ = \dfrac{RKA{ε}_{0}}{\dfrac{d}{3} - vt + k\left(d - \dfrac{d}{3} + vt\right)}\left[∵ x = \dfrac{d}{3} - vt\right]

Given, A=1A = 1 and K=2K = 2

∴τ=3×2ε0×Rd−3vt+6d−2d+6vt=6Rε05d+3vt∴ τ = \dfrac{3 \times 2{ε}_{0} \times R}{d - 3vt + 6d - 2d + 6vt} = \dfrac{6R{ε}_{0}}{5d + 3vt}

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