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Soot Particles 
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#1
Feb1105, 08:29 PM

P: 484

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1 x 10^5 newtons per coulomb while acted upon by a drag force of 7.24 x 10^11 newtons. What is the charge q1 on the particle? Ignore the effects of gravity.
Express your answer in coulombs to three significant figures. q1 = 7.25 x 10^16 C by F = E*q If this charged soot particle is now isolated (that is, removed from the electric field described in the previous part), what will be the magnitude of the electric field due to the particle at distance 1.00 meter from the particle? Now I know this must be incorporated: E = F/q, where F = (1/[4pi*epsilon_0])*[k]*[(q1q2)/(r^2)] But where is the diameter or radius of the particle used? And what is r (1 m?) and q2 (is it the same as q1)? Thanks again. 


#2
Feb1105, 08:55 PM

HW Helper
P: 2,567

First of all, you're looking for the field, not the force, so there is no q2 factor (E=F/q, remember). The particle is probably supposed to be considered as a point particle. That means it has no size, it is just a tiny point of charge. All you need is coulomb's law to get the field. (notice there is no place in the coulomb equation to put the diameter. if the particle did have a diameter, you would have to to an integral to get the force)



#3
Feb1205, 09:20 AM

P: 484

But why did the problem include 1 micrometer as the diameter for the charged soot particle? Is this just extra info?



#4
Feb1205, 09:35 AM

HW Helper
P: 2,567

Soot Particles
Oh, I didn't see that. Well since they give no more information, you can probably assume it's spherical, which means you can still use coulomb's law on it. I don't know why they gave it. Did you have to calculate the drag force yourself using it? Maybe it was just to throw you off.



#5
Feb1205, 10:03 AM

P: 484

No, the drag force is given.
Would one use E = (1/[4pi*epsilon_0])[(q)/(r^2)] q = charge (7.25 x 10^16 C) and r is radius (5 x 10^7 m) 1/[4pi*epsilon_0] = 8.988 x 10^9 Plugging everything in formula, I got 26065200 N/C, but this wrong. What did I do incorrectly? Is this even correct?? 


#6
Feb1205, 11:48 AM

P: 484

F_drag = 7.24 x 10^11 N, so F_E is = to this?
What is E_q? In the prob, I have to find mag. of electric field E. So I have to solve for E (N/C) = F/q. How does finding q help? 


#7
Feb1205, 03:39 PM

Mentor
P: 41,453

(Note: Forget my last postI obviously didn't read the entire question! In fact, I should delete it!) 


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