Two point charges A and B with magnitudes +4q | vedclass

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(C) The initial force acting between charges $+4q$ and $-4q$ is given by Coulomb's Law:$F = \frac{k(4q)(-4q)}{r^2} = \frac{-16kq^2}{r^2} \quad \dots(i

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$\frac{9}{16} F$

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(C) The initial force acting between charges $+4q$ and $-4q$ is given by Coulomb's Law:$F = \frac{k(4q)(-4q)}{r^2} = \frac{-16kq^2}{r^2} \quad \dots(i)$When $25\%$ of the charge from point $A$ is transferred to point $B$:Charge transferred $= 0.25 	imes 4q = 1q$.New charge on $A$ $(q_1)$ $= 4q - 1q = 3q$.New charge on $B$ $(q_2)$ $= -4q + 1q = -3q$.The new force $F'$ between the charges is:$F' = \frac{k(3q)(-3q)}{r^2} = \frac{-9kq^2}{r^2}$.Comparing $F'$ with $F$:$F' = \frac{9}{16} 	imes \left( \frac{-16kq^2}{r^2} ight) = \frac{9}{16} F$.

Two point charges $A$ and $B$ with magnitudes $+4q$ and $-4q$ are placed along a line separated by a distance $r$. The force acting between them is $F$. If $25\%$ of the charge from point $A$ is transferred to point $B$,the force between the charges now becomes:

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