An oil drop with a radius of
1µm is held stationary in a constant electric field of
3.65×104N∕C due to excess electrons on it. The oil drop's density is
1.26g∕cm3. To find the number of excess electrons on the oil drop, consider the following details:
Convert the given measurements:
Radius of the drop,
r=1µm=10−6m.
Density of the oil,
ρoil , is converted to:
ρoil =1.26g∕cm3=1.26×103kg∕m3Since the droplet is stationary, the gravitational force acting downward is balanced by the electric force acting upward:
Weight of the droplet
= Force due to the electric field
Mathematically, this can be expressed as:
πr3ρoil ⋅g=qEwhere
q is the charge due to excess electrons. If
n is the number of excess electrons, then:
q=neSubstitute this into the previous equation:
πr3ρoil⋅g=neESolving for
n :
n=Substitute the known values into the equation to calculate
n :
n=| 4×3.14×(10−6)3×1.26×103×10 |
| 3×1.6×10−19×3.65×104 |
=9.03≈9Thus, there are approximately 9 excess electrons on the oil drop.