Question

Two square metal plates of side `1 m` are kept `0.01 m` apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicualr to an oil surface in the a tank filled with an insulating oil. The plates are connected to a battery of emf `55 V`. The plates are then lowered vertically into the oil at a speed of `0.001 ms^(-1)`. Calculate the current drawn from the battery during the process.
(Dielectric constant of oil `=11, epsilon_(0) = 8.85 xx 10^(-12) N^(-1) m^(-2))`.

Answer 1

Correct Answer - A::B::D
The adjacent figure is a case of parallel plate capacitor. The combined capacitance will be
`C=C_1+C_2`
`=(kepsilon_0(xxx1))/(d)+(epsilon_0(1-x)xx1)/(d)`
`C=(epsilon_0)/(d)[kx+1-x]` …(i)
Differentiating the above quation w.r.t time
`(dC)/(dt)=epsilon_0/d(k-1)(dx)/(dt)=epsilon_0/(d)(k-1)v`
where `v=(dx)/(dt)`
We know that `q=CV`, `(dq)/(dt)=V(dC)/(dt)`
From (iii) and (iv)
`I=V(epsilon_0)/(d)(k-1)v`
`I=(500xx8.85xx10^12)/(0.01)(11-1)xx0.001`
`=4.425xx10^-9 Amp.`