# Thread:

1. ## electromagnetic

Hello, i was doing my work and am unsure about some of these questions. i am hoping someone here can help me out.

1. Because Earth has an electric field of 150N/C, water droplets can remain stationary in the atmosphere if they are charges. if the avg water droplet is 0.030 mm in diameter, how many electron does it need to remain stationary?

2. it asks to draw the magnetic field lines and trajectories of the negatively charged particles. i know how the field lines go but wont the electrons go in the opposite direction?

2. Originally Posted by MarkC
Hello, i was doing my work and am unsure about some of these questions. i am hoping someone here can help me out.

1. Because Earth has an electric field of 150N/C, water droplets can remain stationary in the atmosphere if they are charges. if the avg water droplet is 0.030 mm in diameter, how many electron does it need to remain stationary?

2. it asks to draw the magnetic field lines and trajectories of the negatively charged particles. i know how the field lines go but wont the electrons go in the opposite direction?

The electic force is balancing the gravitational force. So find the weight of the droplet and then the size of the electric force. From that you can calculate the number of charges.

For the second problem, yes, negative charges move in the opposite direction that positively charged particles move.

-Dan

3. Im still having a bit of difficulty with the rain drop question. i found the mass and then set Eq=Fg
Fq/q = Gm1m2/r^2
and then solved for q after simplifying. is that right? because i got 3.9E12 electrons which seems like a bit too much

4. Originally Posted by MarkC
Im still having a bit of difficulty with the rain drop question. i found the mass and then set Eq=Fg
Fq/q = Gm1m2/r^2
and then solved for q after simplifying. is that right? because i got 3.9E12 electrons which seems like a bit too much
It is a bit too much.

The radius of the water drop is 0.00150 cm. Given a density of water as 1 g/cm^3 I get a weight of $1.4137 \times 10^{-10}~N$ for the water droplet.

Thus I get a charge on the water droplet of
$q = \frac{mg}{E} = 9.2363 \times 10^{-13}~C$
to balance the water droplet in the E field

and finally
$n = \frac{q}{e} = 5772700$ electrons.

-Dan
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