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(C) When two identical conducting spheres are brought into contact,the total charge is redistributed equally between them.Total charge $Q = 2.1 \, nC

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) When two identical conducting spheres are brought into contact,the total charge is redistributed equally between them.Total charge $Q = 2.1 \, nC + (-0.1 \, nC) = 2.0 \, nC$.Charge on each sphere after contact $q = \frac{Q}{2} = \frac{2.0 \, nC}{2} = 1.0 \, nC = 1.0 	imes 10^{-9} \, C$.The distance between the spheres is $r = 0.5 \, m$.The electrostatic force is given by Coulomb's Law: $F = k \frac{q_{1}q_{2}}{r^{2}}$.Substituting the values: $F = (9 	imes 10^{9}) 	imes \frac{(1.0 	imes 10^{-9}) 	imes (1.0 	imes 10^{-9})}{(0.5)^{2}}$.$F = \frac{9 	imes 10^{9} 	imes 10^{-18}}{0.25} = \frac{9 	imes 10^{-9}}{0.25} = 36 	imes 10^{-9} \, N$.Thus,the force is $36 	imes 10^{-9} \, N$.

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