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?

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

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?

positive

To determine the excess charge on the water drop, we need to consider the forces acting on it.

In this scenario, the water drop is suspended motionless in air, which means that the gravitational force pulling it downward is balanced by an upward force. The electric field provides this upward force and counteracts the gravitational force.

We know that the electric field has a magnitude of 7840 N/C and is directed upward. Since the electric field is uniform, the force on the water drop can be calculated using the formula:

F = q * E

Where F is the force, q is the charge, and E is the electric field. The force in this case is equal to the gravitational force pulling the water drop downward.

Let's calculate the gravitational force first:

F_gravity = m * g

Where F_gravity is the gravitational force, m is the mass of the water drop (3.84x10^-9 kg), and g is the acceleration due to gravity (approximated as 9.8 m/s^2).

F_gravity = (3.84x10^-9 kg) * (9.8 m/s^2) = 3.7632x10^-8 N

Now, equating the gravitational force to the electric force:

3.7632x10^-8 N = q * (7840 N/C)

Solving for q:

q = (3.7632x10^-8 N) / (7840 N/C) = 4.8x10^-12 C

Since the excess charge on the water drop is positive, we can conclude that the excess charge on the water drop is positive.