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(B) The initial Coulombian force between the two spheres is given by Coulomb's law:$F = \frac{k(q)(-q)}{d^2} = -\frac{kq^2}{d^2}$When $50\%$ of the ch

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$\frac{F}{4}$

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(B) The initial Coulombian force between the two spheres is given by Coulomb's law:$F = \frac{k(q)(-q)}{d^2} = -\frac{kq^2}{d^2}$When $50\%$ of the charge is transferred from sphere $B$ (charge $-q$) to sphere $A$ (charge $+q$),the amount of charge transferred is $\frac{q}{2}$.The new charges on the spheres are:$q_A = q - \frac{q}{2} = \frac{q}{2}$$q_B = -q + \frac{q}{2} = -\frac{q}{2}$The new Coulombian force $F^{\prime}$ is:$F^{\prime} = \frac{k(q_A)(q_B)}{d^2} = \frac{k(\frac{q}{2})(-\frac{q}{2})}{d^2}$$F^{\prime} = -\frac{kq^2}{4d^2}$Since $F = -\frac{kq^2}{d^2}$,we can substitute this into the expression for $F^{\prime}$:$F^{\prime} = \frac{F}{4}$

Two identical conducting spheres $A$ and $B$ having charges $+q$ and $-q$ respectively are kept at a distance $d$ apart and experience a Coulombian force $F$ between them. If $50\%$ of the charge is transferred from sphere $B$ to $A$,then the new Coulombian force between them is . . . . . . .

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