Electrostatics: 3 balls on a string, calculate net forces.

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SUMMARY

The discussion centers on calculating the net electrostatic forces acting on three charged balls arranged in a line, specifically focusing on Ball B and Ball C. Using Coulomb's Law, represented by the equation F=kq1q2/d^2, participants calculated the net force on Ball B as -4.34x10^24 N. The conversation clarified that the problem does not involve a fourth ball, and the net force on Ball C should be determined by summing the forces exerted by Balls A and B. Participants emphasized the importance of vector direction in force calculations.

PREREQUISITES
  • Coulomb's Law for electrostatic force calculations
  • Understanding of vector addition in physics
  • Basic knowledge of electric charge units (Coulombs)
  • Familiarity with the concept of net force
NEXT STEPS
  • Study Coulomb's Law in detail, including its applications and limitations
  • Learn about vector addition and how to resolve forces in different directions
  • Explore the concept of electric fields and their relation to electrostatic forces
  • Investigate the implications of charge distribution on net forces in multiple charge systems
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Students studying electrostatics, physics educators, and anyone interested in understanding the principles of electrostatic forces and their calculations.

pugfug90
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Homework Statement


3 balls are on a ?string?/line?. Each are equally displaced, 72cm. Ball A has a charge of -50e5c, Ball B a charge of +25e2, Ball C -10e10.
1. What is the net force applied on the middle ball (Ball B)?
2. Assuming a force to the right(ungiven name of Ball D?) (of Ball C) is positive, find the net force applied on Ball C..


Homework Equations


F=kq1q2/d^2
k=9x10^9 n*m^2/c^2

The Attempt at a Solution



These are all theoretical values.. I know the basic structure of the problem, but totally guesstimating the values. With that said.. To try and get net force of Ball B..
I did
F(ab)=kq(a)q(b)/d^2
F(bc)=kq(b)q(c0/d^2
...
F(ab)+F(bc)=net force applied on Ball B.. I think that part was easy..
net force=-4.34x10^24, work on screenshot
physstanks.gif

===
Kind of lost for Part 2.. How do I find out the force for D? Or the charge of D? Or the distance (baby blue)?

On the test, what I think I did was add charges A+B, set equal to C+D, or A+B-C=D.. Then use that value of charge for D, assume 72cm (I want to know why I should assume), calculate force.. I don't think that's how it works though..
===
I thought up an interesting theory just now.. All charges (balls) have same net forces, but vary +/-..?
Here's another pic
physstanks2.gif

I forgot to draw it, but there's supposed to be a "blue vector equal to magenta vector".
 
Last edited:
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Did the problem actually state that there was a fourth ball? Or did you just assume that's what was meant? I'm guessing that part 2 was just to find the net force on ball C. (The stuff about to the right being positive--that's just so you use the correct sign convention.)
 
Problem did not say 4th ball (I assumed so because there was a positive "force" to right of Ball C)

PS, my 1st question, answered, love this forums.
Loving to see that more people are willing to spend their time to help others. I may be inspired to be a good physics teacher :D

So Doc Al, I added a 3rd edit.. thoughts?
 
Last edited by a moderator:
My thoughts? Get rid of ball D!

The force on ball C is just F(A on C) + F(B on C). Figure them out and add them up! (Since forces are vectors, directions--signs--matter.)
 
Umm..
Would that mean for net force of A, add Force(charge of A+C)+Force(charge a+b)?

With Ball B, I supposed that I used the one on the left and the one on the right..

Or am I not looking at it from the right perspective?

How would this problem be different if there was no force on the right?
 
pugfug90 said:
Umm..
Would that mean for net force of A, add Force(charge of A+C)+Force(charge a+b)?

With Ball B, I supposed that I used the one on the left and the one on the right..

Or am I not looking at it from the right perspective?
There are three balls. They exert electrostatic forces on each other. To find the net force on anyone (say ball A) add up the individual forces on that ball (A) from the other two (B & C). That means calculating the force that B exerts on A and the force that C exerts on A. (To find the individual forces between any two balls, use Coulomb's law.) Follow the same logic to find the net force on any of the balls.

How would this problem be different if there was no force on the right?
If I understand the problem correctly, there is no extra "force on the right"--that's just your misinterpretation of the instructions to use positive numbers to represent a force to the right (and negative numbers for forces to the left).
 

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