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

4.43

Explanation

Let a be the side of the square plate
As shown in figure ,`C_1` and `C_2` are in parallel.Therefore total Capacity of capacitors in the position shown is
`C=C_1+C_2`
`C=(epsilon_0a(a-x))/d+(Kepsilon_0ax)/d therefore q=CV=(epsilon_0aV)/d(a-x+Kx)`
As plates are lowered in the oil,C increasese of charge stored will increase.
Therefore`i=(dq)/(dt)=(epsilon_0aV)/(d)(K-1)(dx)/(dt)`
Substaning the values `epsilon_0=8.85xx10^(-13)C^(2)//Nm^(2)`
`a=1m, V=500 "volt",d=0.01m,K=11` and `(dx)/(dt)=` speed of plate `=0.001 m//s` We get current `i=((8.85xx10^(-128))(1)(500)(11-1)(0.0001))/((0.001)), i=4.43xx10^(-9)A`