Force between conducting spheres

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When two identical conducting spheres with charges of +6µC and -2µC are touched together, they share their charges equally, resulting in both spheres having a charge of +2µC. After being separated again, the new force between them is calculated using the formula F=(kq1q2)/r². Since both spheres now have the same positive charge, the force becomes repulsive. The resulting force is -F/3, indicating a change from attractive to repulsive due to the equal charges. The initial assumption about the net charge remaining constant was incorrect, as the redistribution of charge alters the force significantly.
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X and Y are two identical conducting spheres separated by a distance d. X has a charge of +6µC and Y has a charge of -2µC. The electric force between them is +F (i.e. attractive). The spheres are touched together and are then returned to their original separation d. What is the new value of the force between them?

Any help is greatly appreciated. I assumed that since the net charge should remain the same after, there would be no change in the value of the force, but the answer scheme gives the answer as -F/3.
 
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What happens with the charge if the conducting spheres touch each other?
 
thanks for the reply man, i think i got it. The charges become +2µC on both sphere X and Y, at which point it's just subbing the before and after values for the charge into F=(kq1q2)/r2, and since, they're now like charged, the force becomes a repulsive one, hence -F/3
 
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