What Happens to the Force Between Two Touching Charged Spheres When Separated?

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When two touching charged spheres are separated, the charges redistribute evenly, resulting in each sphere having the same magnitude of charge, but retaining their original signs. The positively charged sphere, having a larger initial magnitude, will still be positive after contact, while the negatively charged sphere will remain negative. This charge distribution leads to a repulsive force between the spheres when they are separated. The force can be calculated using the formula F = (8.99E9 N*m^2/C^2)*(q1*q2)/r^2, where q1 and q2 are the charges on each sphere. Ultimately, the net force between the spheres will be repulsive due to their equal magnitudes of charge.
TheMadCapBeta
Two conducting spheres (of the same size), one negatively charged, one positively charged are at a finite distance apart. Magnitude of neg charged sphere < pos charged sphere.


When the spheres are brought together and touched, and brought back to initial distance, what is the net force between the two spheres?


The first question [unwritten here] was computed by finding the force charge between q1 and q1


F = (8.99E9 N*m^2/C^2)*(q1*q2)/r^2

-I know when the two spheres are physically juxtaposed, the charged is then evenly distributed:

Qt = q1 + q2

So, after the distribution, each charge would be Qt/2.

My question is, would the charges then become neutral, therefore there will result in no repulsion or attraction? Or, would the according spheres retain their sign charge but the magnitudes will become equal: Qt/2? Or, would the charge with the larger magnitude be the pervasive charge and both will become positive when touched? Therefore, the two spheres would repulse instead of attract. Since the positive charge was the dominant charge in terms of magnitude, they would become positive.

In any case, the resultant electric force between the would be:

F = (8.99x10^9 N*m^2/C^2)*(Qt/2)*(Qt/2)/r^2

Can anyone verify any of my assumptions?

Thanks
 
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The magnitude of the positive ball is greater, so both charges are left with a slightly positive charge. I think they would then repel each other.
 


Based on the given information, it can be assumed that the two spheres have opposite charges, with the positively charged sphere having a larger magnitude of charge. When the spheres are brought together and touched, the charges will redistribute and become evenly distributed on each sphere. This means that the net charge on each sphere will be half of the original charge, resulting in both spheres having the same magnitude of charge. This does not mean that the charges will become neutral, as they will still retain their signs (one positive and one negative).

Therefore, the resulting force between the two spheres will depend on the distance between them and the magnitude of the charges. If the distance between the spheres remains the same, the force will be repulsive since both spheres now have the same magnitude of charge. The equation for the force between two charged spheres is F = (8.99x10^9 N*m^2/C^2)*(q1*q2)/r^2, where q1 and q2 are the charges on each sphere and r is the distance between them.

In summary, when the two spheres are brought together and touched, the net force between them will be repulsive due to the redistribution of charges. The charges on each sphere will become equal in magnitude, but they will still retain their original signs.
 
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