# Electric fields of oil

1. Feb 17, 2010

### rafterman

1. The problem statement, all variables and given/known data

I am looking to calculate the horizontal and vertical components of an electric field at the point of the oil drop can someone tell me if I am on the correct tracks by using the following equation.

2. Relevant equations

E=kQ/r2

3. The attempt at a solution

2. Feb 17, 2010

### collinsmark

Could you provide more problem details?

Does this problem involve Robert Millikan's and Harvey Fletcher's famous oil drop experiment used to measure the elementary charge of an electron? In that case, an electric field was created using a pair of parallel plates. And in the case of parallel plates, the electric field can be approximated as constant vector (i.e. not a function of r).

3. Feb 18, 2010

### rafterman

Hi
It does involve the oil drop experiment, Plates are 8mm apart with the top plate is maintained at a potential of +500V relative to the lower one. The mass of the oil drop is 4.08x10-15 kg with a neg charge = -4e.
There is also a fixed charge Q=+200 x10-11 C 2.00mm horizontally from the oil drop which does not effect the charge distribution on the plates. g=9.81

4. Feb 18, 2010

### collinsmark

Fortunately, the electric field caused by the fixed charge is perpendicular to the electric field field caused by the parallel plates (from the point of view of the oil drop, in this particular configuration). So you can treat both separately.

The only thing that affects the horizontal component of the electric field is fixed charge (from the point of view of the oil drop). So yes, you were on the right track with the E=kQ/r2 equation for that part.

The only thing that effects the vertical component (in this problem) is the parallel plates. Approximate the electric field between the parallel plates as a constant (but still having a specific direction). You can calculate the electric field due to this finding the negative slope of V, where V is expressed as a function of height.

You can calculate the slope of V using simple algebra. Remember, the slope of a straight line is the "rise over the run". Express how V changes (goes from 0 to 500 V -- the rise) with respect to how the height changes (goes from 0 to 8 mm -- the run). The electric field is the negative of this slope.

E = - (slope of V)

(There's enough information given here for you E in a parallel plate situation, but you might find it easier to look in your textbook for an equation that gives you E for parallel plates, as a function of V [the voltage] and d [the distance between the parallel plates]. Deriving the equations yourself is always fun, but I suggest as a first step, simply find the equation in your textbook)

5. Feb 19, 2010

### rafterman

Thats fantastic, thanks for the help. I have found the other equation for parallel plates.

One last question tho, to calculate the net force on the oil drop would I be correct in using the equation:-

Fnet = V/d - mg

6. Feb 19, 2010

### collinsmark

Close, at least for the net force concerning the parallel plates and gravitational force. But still not quite. The electric field represents the force per unit charge. So you're going to need a q in there somewhere, where the q represents the charge of the oil drop, not the fixed charge, or the charge on the plates (i.e., $$\vec F = q \vec E$$). Also be careful with your signs. Take note of the sign on the equation that relates E, V, and d (I don't think it's right in your above equation). Also, if you assume that the oil drop is negatively charged, that sign can make a difference in interpreting the final results.

And lastly, remember, there is another perpendicular component to the force, the comes from the electric field caused by the fixed point charge. That's not addressed in your above equation. Of course, this force has no effect on the net vertical force. (So if your above equation only applies to the net vertical force only, you can ignore this last comment.)

7. Feb 21, 2010

### rafterman

Hi

Thanks for all the help I could not have asked for any better.

You get five stars from me!

Cheers

8. Feb 23, 2010

### coz

Can I just check something. Once you have worked out the horizontal and vertical components you have a value for Ex and Ey. To then work out the net force on the oil drop could you use:

Fnet = (Ex + Ey) x q

Thanks

9. Feb 23, 2010

### collinsmark

Just be careful with your terminology. Remember, you are working with vectors. It may be more appropriate to write,

$$F_{\mbox{net, electric}} = q(E_x \hat x + E_y \hat y)$$

And although rafterman didn't specify this, I'm guessing that the gravitational force balances out the y component of the electric force (the gravitational force is not shown in the above equation, but probably needs to be considered as part of the project).

10. Feb 23, 2010

### coz

I'm sorry to ask another question, but I think I'm missing something here.

For the horizontal component I get
Ex = 0
Ey = -4.5x104 NC-1

For the vertical component I get
Ey = 0
Ex = 6.25x104 NC-1

To calculate the net force on the oil drop would it be better to use

Felectric = qE(r) and Fgravitational mg(r)

and combine these two equations? Thanks again

11. Feb 23, 2010

### collinsmark

Yes, that's fine, but just make sure to work with the components separately. I assume the gravitation force is completely in the y direction. If so, you can add that to the y component of the electric force (it has no impact on the x component of the electric force), to get the total net force.