Calculating Electric Field of a Charged Particle at 1m Distance

In summary, the charged soot particle has a charge of 7.25 x 10^-16 C and has a radius of 5 x 10^-7 m. The electric field at a distance of 1.00 meter from the particle is 26065200 N/C.
  • #1
Soaring Crane
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0
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.
 
Last edited:
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  • #2
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
But why did the problem include 1 micrometer as the diameter for the charged soot particle? Is this just extra info?
 
  • #4
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
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??
 
Last edited:
  • #6
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
Soaring Crane said:
Would one use

E = (1/[4pi*epsilon_0])[(q)/(r^2)]
Yes.

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
As StatusX explained, r is the distance from the particle (its center) to the point where you want to find the field, not the radius of the particle. So r = 1m.

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

1. What is the formula for calculating the electric field of a charged particle at 1m distance?

The formula for calculating the electric field of a charged particle at 1m distance is E = kQ/r^2, where E is the electric field in newtons per coulomb (N/C), k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the charge of the particle in coulombs (C), and r is the distance from the particle in meters (m).

2. How do you determine the direction of the electric field at 1m distance?

The direction of the electric field at 1m distance can be determined by using the principle of superposition. The direction of the electric field will be in the direction of the net force on a positive test charge placed at that point.

3. Can the electric field at 1m distance be negative?

Yes, the electric field at 1m distance can be negative. This indicates that the force on a positive test charge would be in the opposite direction of the electric field. A negative electric field can occur when the charged particle is negatively charged.

4. How does the distance from the charged particle affect the electric field?

The electric field is inversely proportional to the square of the distance from the charged particle. This means that as the distance increases, the electric field decreases. So, at a distance of 2m, the electric field would be 1/4 of the value at 1m distance.

5. What if there are multiple charged particles present? How do you calculate the electric field at 1m distance?

If there are multiple charged particles present, the electric field at 1m distance can be calculated by using the principle of superposition. This means that the total electric field at that point is the vector sum of the individual electric fields from each charged particle present. The direction of the electric field will be determined by the net force on a positive test charge at that point.

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