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Two electric field problems - pick one or both!

  • #1
Here are two problems, both of which I'm not really sure where to start. The simpler (?) one first:

A 1.10-mm-diameter glass sphere has a charge of +1.10 nC. What speed does an electron need to orbit the sphere 1.80mm above the surface?

I thought of using v = (2*pi*r)/T because it is circular motion, but I can only derive "r" (I got r = 0.00235m, though I'm not sure if this is right. I know the sphere acts as a point charge and measured the 1.8mm and the radius of the sphere.). I don't feel like the electric field equations would get me anywhere.

Question 2 (and more complex):

The ink drops have a mass m = 1.00×10−11kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 22.0m/s . The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length D0 = 1.65cm, where there is a uniform vertical electric field with magnitude E = 8.10×104N/C .

If a drop is to be deflected a distance d = 0.250mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is 1000kg/m3 , and ignore the effects of gravity.

153611.jpg

This one ... not a clue where to begin. I've got my electric field formulas ready to go. I'd appreciate any nudges you guys can give on either problem.
 

Answers and Replies

  • #2
832
30
For the first, you can think of the Coulomb force as the equivalent of the Newton gravitational force for charges (this is not quite right, accelerated charges would radiate, but I think that's all what the problem is about, just neglect radiation).

In the second, you must use some kinematics (same thing here about radiation, just neglect it), with the acceleration given by the Coulomb force for the charge. You see that the positive charge will be accelerated when it enters the region with the uniform electric field, right? then just use ##x_f=x_0+v_it+\frac{1}{2}at^2##, where a is the acceleration given by the Coulomb force.
 
  • #3
For the first, I've got nothing. Generically, I developed:

Q = (16*pi*epsilon-naught*m*r^3)/(eT^2)

I don't think this is it. I solved for T then and got 7.35x10^-8, but I don't think I'm on the right track.



For the second, I found all sorts of kinematic values - can't connect them to q, though. I found values for a, t, and theta, but can't connect them to find charge. Don't think my theta is right though either, which was 2.53x10^-6.
 
  • #4
832
30
Well, the charge q appears in the Coulomb force, which gives you the acceleration, right?
 

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