Two identical metallic spheres,having unequal opposite charges are placed at a distance of 0.50 m apart in air. After bringing them in contact with each other,they are placed at the same distance apart.Now the force of repulsion between them is 0.108 N. Calculate the final charge oneach of them.
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After bringing them in contact with each other, the spheres acquire equal charges, let us say, Q due to charge redistribution. We have, q1=q2=Q, r=0.50m as the spheres are put back to their initial positions and given that F=0.108N. Therefore, the final charge on both the spheres is either +√3μC or −√3μC.
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