posted by amber on .
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 7840 N/C. The mass of the water drop is 3.84 10-9 kg.
(a) Is the excess charge on the water drop positive or negative?
Directed upward? Then that is the direction a + would experience a force, so since the force is upward countering gravity, it must be positive? Correct?
The downward force from the weight of the drop is mg, where m is the mass of the drop and g is the acceleration due to gravity.
This must be balanced by the upward force from the field.
Let's say that the total charge on the drop is Q Coulombs,
then 11000Q is the force directed upwards. Since this must be equal to mg, we have
Q = mg/11000
= (4.79 x 10^-9)*(9.81)/11000
=4.27 x 10^-12 C
which is very small!
Note that an electron (or proton) has a charge of about 1.60 × 10^-19 C, so there are (4.27 x 10^-12)/(1.60 x 10^-19) = 26700000 or about 2.67 x 10 ^ 7 excess protons or electrons