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 `d//3` initially. Suppose the liquid level decreases at a constant speed v, the time constant as a function of time t is-

Options

A. `(6 epsi_(0) R)/(5d + 3//t)`
B. `((15d+9vt) epsi_(0)R)/(2d^(3) -3dvt- gv^(2) t^(2))`
C. `(6 epsi_(0)R)/(5d- 3vt)`
D. `((15d- 9vt) epsi_(0)R)/(2d^(3) + 3dvt- 9v^(2) t^(2))`

Answer

A

Explanation

We have the time constant, `tau = RC.`…(i)
Now, `C.= (C_(1)C_(2))/(C_(1) + C_(2)) = (((A epsi_(0))/(d-x))((KA epsi_(0))/(x)))/((A epsi_(0))/(d-x) + (KA epsi_(0))/(x))`
`therefore C.= (KA epsi_(0))/(x+ K(d-x))`
Putting the value of C. in Eq (i), we get
`tau= (RK A epsi_(0))/((d)/(3)- vt + k (d- (d)/(3) + vt)) [ because x= (d)/(3)- vt]`
Given, A= 1 and K=2
`therefore tau= (3 xx 2 epsi_(0) xx R)/(d-3vt+ 6d - 2d+ 6vt)=(6R epsi_(0))/(5d+ 3vt)`