Electric Charges: Forces on A and C Explained

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SUMMARY

The discussion centers on calculating the electric forces acting on two charged spheres, A and C, after C touches A. Initially, spheres A and B repel each other with a force of 2.0 x 10^5 Newtons. Upon touching, sphere C acquires half the charge of sphere A, resulting in both A and C having a charge of q/2. The force on A remains unchanged, while the force on C, when positioned between A and B, is calculated using the formula F1 = k(3/2)q*(1/2)q/r^2, leading to a new force expression based on their distances.

PREREQUISITES
  • Understanding of Coulomb's Law and electric forces
  • Knowledge of charge distribution in conductors
  • Familiarity with the concept of electric field strength
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study Coulomb's Law in detail, focusing on force calculations
  • Learn about charge distribution in conductive materials
  • Explore electric field concepts and their applications
  • Practice problems involving multiple charged objects and their interactions
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This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics and the behavior of charged objects in electric fields.

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Homework Statement


Two equally charged identical conducting spheres, A and B, repel each other with a force of 2.0 x 10^5 Newton. Another identical uncharged sphere, C, is touched to A and then moved over right next to B.
What is the electric force on A now?
What is the electric force on C after it has touched A and is halfway between A and B?


Homework Equations


F=K q^2/r^2



The Attempt at a Solution


I let q arbitrarily equal 1.0x 10^-6 Nm^2/C^2
I determined the distance to be 21.2 m
If A is negatively charged at 1.0 x 10^-6 does C touching it take half of that charge?
If so, does A and C now have a charge of 5.0 x 10^-7?
I'm not sure what happens to the charge when A and C touch.
If they are halved, then the electric force on A remains the same and
the force on C remains the same when it is at half the distance.
I would love some help with this, please.
 
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Don't assume arbitrary numerical values and make things messy. Let's suppose the charge on A and B is q each. If their dist is r, F = k*q^2/r^2.

After touching, the charge q will distribute itself equally on A and C, since they are conductors. So, A and C each has q/2.

If you now place C very close to B, then there is a net charge of 3q/2 there, and q/2 at A. So, force F1 = k(3/2)q*(1/2)q/r^2 = k*3q^2/(4r^2).

We can now find how many times is F1 of F.

Can you do the 2nd part now?
 
Yes, I've got it now. Thank you very much!
 

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