What Is The Sum of Foures On Two Equally Charged Pith Balls

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Homework Statement



I have 2 equally charged pith balls suspended by a string repelling eachother at a 28 degree angle. I know that Tension, Gravity, and the electostatic force are all acting. I found tension by creating a force diagram. T= cos14mg. I also found g=mg. I'm having trouble finding out what the electrostatic force would be based on tension and gravity.

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The Attempt at a Solution



I believed that Fe= Fg-T, is this correct?
 

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  • #2
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Homework Statement



I have 2 equally charged pith balls suspended by a string repelling eachother at a 28 degree angle. I know that Tension, Gravity, and the electostatic force are all acting. I found tension by creating a force diagram.
Can I assume that by "28o" you mean the angle from the vertical to one of the strings? (As opposed the the total angle between strings?)
T= cos14mg.
You'll have to redo that equation. The gravitational force corresponds to a component of the tension; not the other way around. You've got the cos14o in the wrong place.
I also found g=mg.
I think you mean Fg = mg. Yes, that's fine. :approve:
I'm having trouble finding out what the electrostatic force would be based on tension and gravity.
Fg = mg is one of the two components of the tension. The other component of the tension is the electrical force.
I believed that Fe= Fg-T, is this correct?
Not so correct. But you can relate T and Fe using trigonometry. [Hint: Take a look at your free body diagram. If it is drawn correctly, Fg and Fe are perpendicular, forming a right triangle. T is the hypotenuse. :wink:]

[Edit: Another hint: You will also need to use Coulomb's law, of course.]
 
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Thanks alot! And by 28 i mean the angle between the 2 pith balls so it would actually be 14 degrees between the Fg and T. I think i could then use the angle to solve for the hypotenuse/ tension. I would then use Pythagorean's Theorem to solve for the other side of the triangle which is Fe. I understand how to get the charge after this point i'm just confused on how to acquire Fe.
 
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Thanks alot! And by 28 i mean the angle between the 2 pith balls so it would actually be 14 degrees between the Fg and T.
Okay, that makes sense. :approve:
I think i could then use the angle to solve for the hypotenuse/ tension.
I still think you're missing something. The tension is the hypotenuse. :wink:
I would then use Pythagorean's Theorem to solve for the other side of the triangle which is Fe.
You know that the tension is the hypotenuse. You can find the two components of the tension by using the relationships,

[tex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/tex]

[tex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/tex]
I understand how to get the charge after this point i'm just confused on how to acquire Fe.
Moving on the horizontal direction only (we can come back to the vertical direction later)... There are two horizontal forces. There is the associated component of the tension discussed above. There is also the force described by Coulomb's law.

Since nothing is accelerating, you know that the sum of the corresponding tension component + Column force must be equal to zero (because if it wasn't something would be accelerating). In other words, the forces are equal in magnitude and opposite in direction.
 
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Ah ok i found the horizontal component of tension. This would be half of the electrostatic force correct?
 
  • #6
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Ah ok i found the horizontal component of tension. This would be half of the electrostatic force correct?
No, it would be the full force.

But there is one thing to be careful with here. Coulomb's law states,
Fe = k(q1q2/r2)​
But in our situation, r is not the distance from the center line to a given pith ball, rather it is the distance all the way from one pith ball to the other (i.e. it's twice the distance from the center-line). That's particularly important because r gets squared. :wink:
 
  • #7
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Right i almost forgot! But would u mind explaining why it is the full electrostatic force? My reasoning for it beeing half is your only working with half of the triangle. Therefore wouldn't the force only be half because your only using half of the triangle?
 
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Right i almost forgot! But would u mind explaining why it is the full electrostatic force? My reasoning for it beeing half is your only working with half of the triangle. Therefore wouldn't the force only be half because your only using half of the triangle?
Each pith ball exerts the same force on the other, just in the opposite direction. It's Newton's third law of motion. For every action, there is an equal and opposite reaction.

But we can (and we are) taking one of the pith balls in isolation. We're treating the electric force Fe as an external force. But the first pith ball still has the same force acting on it whether we consider the second pith ball or not. So Fe = k(q1q2/r2) is the electrical force, regardless.

The only "tricky" part for this problem is to remember that r is twice the distance to the center-line.
 
  • #9
Hi, I have the exact same lab excercise, except we also have to find the electric field of the pith ball.

I did exactly as you had mentioned above, I found a the force, used Coloumb's law to find the charge.

And to find the electric field, I simply used the formula,

E = Kq/(r^2)

where

K = 8991804694
q = (what I found using Coloumb's law)
r = the distance between the two pith balls?

I am unsure because I had found my charge to be a really small number,

3.21 * 10 ^ -10

and for my electric field, I am getting about 8000, which to me is a lot higher than I expected. Is an electric field of 8000 reasonable?

All calculations are in SI units.
 
  • #10
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Hello laventine5678,

Welcome to Physics Forums!

Usually when starting a new problem, even if it is sort of similar to an existing one, you should start a new thread. I'll respond to this one anyway, but in the future start a new thread for a new problem.

You haven't given very much information, such as the exact angle, distance between the pith balls, mass of the pith balls, lengths of the strings, etc. It would be nice if you showed your work.

Also, are you supposed to calculate the electric field due to a single pith ball across all space, or just at the particular point in space containing the second pith ball?

and for my electric field, I am getting about 8000, which to me is a lot higher than I expected. Is an electric field of 8000 reasonable?

All calculations are in SI units.
Assuming that you are supposed to calculate the electric field due to one of the pith balls, at the point in space containing the second pith ball, and assuming that your calculation for charge is correct, then a value of ~8000 [N/C] (and you should really show your units by the way too) is reasonable if the distance between the pith balls is roughly a couple of centimeters.
 

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