A charged oil drop remains stationary when situated between two parallel horizontal metal plates 25 mm apart and a p.d. of 1000 V is applied to the plates. Find the charge on the drop if it has a mass of 5 × 10-15kg.
Draw a sketch of the electric field between the plates and state if the field is everywhere uniform.
Given:
Separation of plates (d) = 25mm
Mass of drop (m) = 5*10-15kg
Potential difference (U) = 1000V
Charge on the drop (q) =?
if the oil drop remains stationary, then
The force on oil drop due to electric field = weight of the drop
i.e. qE=mg
or, "q = \\frac{mg}{E} = \\frac{mg}{\\frac{u}{d}} = \\frac{mgd}{u}" "= \\frac{5*10^{-15}*10*25*10^{-3}}{1000} = 1.25*10^{-18}C"
The charge on the oil drop q =1.25×10-18C
Hence the required charge on the oil drop is 1.25×10-18C
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