A charge water drop of radius `r` is in equilibrium in an electric field.If charge on it is equal to charge of an electron,then intensity of electric field will be : Density of water =`rho` ,assume gravity......??

A charged water drop whose radius is 0.1mum is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be (g=10ms^(-1))

What is the intensity of the electric field at the centre of a charged ring?

Calculate the radius of a charged water drop which remains just suspended in equilibrium in the earth's electric field. Charge in the water drop is equal to that of an electron. Magnitude of the earth's electric field is 10^(-2)"stat"V.cm^(-1) . [e=4.805xx10^(-10) esu of charge, g=980cm.s^(-2) ]

Electric field due to point charge | Electric field lines|Force on a charge in uniform electric field

The electric field intensity on the surface of a charged conductor is

The electric field intensity at the surface of a charged conductor is

What is the electric field intensity inside a charged conductor?

Electric Field Due to Variable Charge Density