A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2 The level of liquid is dl 3 initially. Suppose the liquid level decreases at a constant speed v , the time constant as a function of time t is

Question

A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2 The level of liquid is dl 3 initially. Suppose the liquid level decreases at a constant speed v , the time constant as a function of time t is (A)  (6 ε0R/5d+3vt)  (B)  ((15d+9vt)ε0R/2d2-3dvt-9v2t2)  (C)  (6 ε0R/5d-3vt)  (D)  ((15d-9vt)ε0R/2d2+3dvt-9v2t2)

Tardigrade Admin · Accepted Answer

After time t, thickeness of liquid will remain ( (d/3)-vt ). Now, time constant as function of time τc=CR= (ε0(1).R/(d- (d)3+vt)+ (d/3-vt)2 (Applying C= (ε0A/d-t+ (t)k)) = (6 ε0R/5d+3vt)

Q. A parallel plate capacitor $C$ with plates of unit area and separation d is filled with a liquid of dielectric constant $K = 2$ The level of liquid is dl $3$ initially. Suppose the liquid level decreases at a constant speed $v$ , the time constant as a function of time $t$ is

$\frac{6 ε _{0} R}{5 d + 3 v t}$

$\frac{( 15 d + 9 v t ) ε _{0} R}{2 d ^{2} - 3 d v t - 9 v ^{2} t ^{2}}$

$\frac{6 ε _{0} R}{5 d - 3 v t}$

$\frac{( 15 d - 9 v t ) ε _{0} R}{2 d ^{2} + 3 d v t - 9 v ^{2} t ^{2}}$

After time $t$ , thickeness of liquid will remain $(\frac{d}{3} - v t) .$
Now, time constant as function of time
$τ_{c} = CR = \frac{ε _{0} ( 1 ) . R}{( d - \frac{d}{3} + v t ) + \frac{d /3 - v t}{2}}$
(Applying $C = \frac{ε _{0} A}{d - t + \frac{t}{k}}$ )
$= \frac{6 ε _{0} R}{5 d + 3 v t}$

Q. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K=2 The level of liquid is dl 3 initially. Suppose the liquid level decreases at a constant speed v , the time constant as a function of time t is

5d+3vt6ε0​R​

2d2−3dvt−9v2t2(15d+9vt)ε0​R​

5d−3vt6ε0​R​

2d2+3dvt−9v2t2(15d−9vt)ε0​R​