Question
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?
(a) Is the excess charge on the water drop positive or negative?
Answers
bobpursley
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?
DJ
positive
bill
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
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