A capacitor is made of two circular plates of radius R each, separated by a dJstance d `lt lt` R. The capacitor is connected to a constant voltage. A th.in conducting disc of radius r `lt lt` R and thickness t `lt lt` R is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m.

Explanation

Initially the thin conducting ruse is placed at the centre of the bottom plate. Whole plate is equipotential surface. If electric field applied to disc is E, then the electric field between the plates of capacitor ` E =(V)/(d)`
Let charge q. is transferred to the disc during the process. Hence, by Gaussu.s law,
`ointEds =(q)/(in_0)`
`:. q = in_(0)(V)/(d)pir^(2)" " [becauseE= (V)/(d) ds=pir^(2)]`
The repulsive force acting on the disc is in upward direction, so
F= qE
=` E_(0). (V)/(d).pi r^(2) . (V)/(d) [ because E= (V)/(d)]`
`= (V^(2))/(d^(2)).pir^(2) in_(0)`
This repulsive force will be balanced by weight (mg) of disc,
`:. (V^(2))/(d^(2))pi r^(2) in_(0)` = mg
If the disc is to be lifted then minimum voltage,
`V= sqrt((mgd^(2))/(piin_(0)r^(2)))`
is required equation .