Calculating Charges of Identical Tiny Conducting Spheres

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The problem involves two identical conducting spheres, one positively charged and the other negatively charged, which attract each other with a force of 0.748 N when 0.404 m apart. After touching, they share their charges and then repel each other with a force of 0.59 N at the same distance. The equations of electrostatics, specifically Coulomb's law, are applied to determine the charges. Charge conservation is crucial in solving for the original charges of the spheres. The solution requires calculating the final charge after they touch and using the forces to backtrack to the original charges.
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Homework Statement


You have two identical tiny conducting spheres: sphere A carries positive charge qA, and sphere B carries negative charge qB. First the spheres are placed distance d = 0.404 m apart, and they attract each other with a force F1 = 0.748 N. Then the spheres are brought together, touch each other, and are brought back to distance 0.404 m apart. Now the spheres are both positive, and they repel each other with force F2 = 0.59 N. Find the original charge on each sphere.



Homework Equations


F=k q1q2/r^2



The Attempt at a Solution


after the spheres touch i get qf=(q1+q2)/2
and i solve for the charge on the spheres after they touch and get .1035uC as my answer but I am not sure how to find original charge after that my proffesor never did an example like this any help is welcome and appreciated
 
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Charge is conserved. Make use of that along with F1.
Besides, this belongs in the introductory physics section.
 
Last edited:
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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