An air cylindrical capacitor eith a `dc` voltage `V = 200 V` applied across it is being submerged vertivally into a vessel filled with water at a velocity `v = 5.0 mm//s`. The electrodes of the capacitors are separated by a distance `d = 2.0 mm`, the mean curvature radius of the electrodes is equal to `r = 50 mm`. Find the current flowing in this case along lead wires, if `d lt lt r`.

Explanation

Since `d lt lt r`, the capacitance of the given capacitor can be calculated using the formula for a parallel plate capacitor. Therefore if the water (permittivity `epsilon`) is introduced up to a height `x` and the capacitor is of length `l`, we have,
`C = (epsilon epsilon_(0) pi r x)/(d) + (epsilon_(0) (l-x) 2pi r)/(d) = (epsilon_(0) 2pi r)/(d) (epsilon x + 1 - x)`
Hence charge on the plate at that instant, `q = CV`
Again we know that the electric current intensity,
`I = (dq)/(dt) = (d (CV))/(dt)`
`= (V epsilon_(0) 2pi r)/(d) (d(epsilon x + 1- x))/(dt) (epsilon - 1) (dx)/(dt)`
But, `(dx)/(dt) = v`,
So, `I = (2pi r epsilon_(0) (epsilon - 1) V)/(d)v = 0.11 mu A`