# Magnitude of charge on oil drop (electric field)

1. Mar 21, 2007

### kbyws37

A positively charged oil drop of mass 1.00 × 10−15 kg is placed in the region of a uniform electric field between two oppositely charged, horizontal plates. The drop is found to remain stationary under the influence of the Earth's gravitational field and the uniform electric field of 0.680 × 104 N/C. What is the magnitude of the charge on the drop?

I am not sure where to start. I am given a mass so would I start with

U = -(G(m1)(m2))/r

2. Mar 21, 2007

### robb_

If the drop is stationary, what can you say about the net force on the drop?

3. Mar 21, 2007

### kbyws37

would force = 0?

4. Mar 21, 2007

### robb_

do you mean net force?

5. Mar 21, 2007

### kbyws37

Sorry, I'm still confused.
Wouldn't gravity be the force that allows it to drop?
Which equation would be good to use?

6. Mar 21, 2007

### robb_

The force of gravity acts down on the drop, but the drop is charged and it sits (stationary) between two charged plates. Since it is stationary, what can you say about the acceleration and net force on it?

7. Mar 21, 2007

### kbyws37

There is no acceleration and thus net force = 0

8. Mar 21, 2007

### robb_

Good, now what two forces act on the drop?

9. Mar 21, 2007

### kbyws37

gravity and drag force?

10. Mar 21, 2007

### robb_

Drag force??
A positive charge is in an electric field.

11. Mar 21, 2007

### kbyws37

gravity and the weight of the drop

12. Mar 21, 2007

### robb_

those are the same character!(force)
I gave you a big hint.