Two spheres of electric charges +2 nC and -8 | vedclass

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(B) Initial charges are $q_1 = +2 \ nC$ and $q_2 = -8 \ nC$. The magnitude of the initial attractive force is $|F| = \frac{k |q_1 q_2|}{d^2} = \frac{k

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$\frac{3d}{4}$

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(B) Initial charges are $q_1 = +2 \ nC$ and $q_2 = -8 \ nC$. The magnitude of the initial attractive force is $|F| = \frac{k |q_1 q_2|}{d^2} = \frac{k (2 	imes 8) 	imes 10^{-18}}{d^2} = \frac{16k 	imes 10^{-18}}{d^2}$.When the spheres touch,the total charge is shared equally: $q_{new} = \frac{q_1 + q_2}{2} = \frac{2 - 8}{2} = -3 \ nC$. Both spheres now have a charge of $-3 \ nC$.Let the new distance be $d'$. The new repulsive force is $|F'| = \frac{k |q_{new} q_{new}|}{(d')^2} = \frac{k (3 	imes 3) 	imes 10^{-18}}{(d')^2} = \frac{9k 	imes 10^{-18}}{(d')^2}$.Given $|F| = |F'|$,we have $\frac{16k 	imes 10^{-18}}{d^2} = \frac{9k 	imes 10^{-18}}{(d')^2}$.$\frac{16}{d^2} = \frac{9}{(d')^2} \Rightarrow (d')^2 = \frac{9}{16} d^2$.Taking the square root,$d' = \frac{3}{4} d$.

List-$I$	List-$II$
$P. Q_1, Q_2, Q_3, Q_4$ all positive	$1. +x$
$Q. Q_1, Q_2$ positive; $Q_3, Q_4$ negative	$2. -x$
$R. Q_1, Q_4$ positive; $Q_2, Q_3$ negative	$3. +y$
$S. Q_1, Q_3$ positive; $Q_2, Q_4$ negative	$4. -y$

Two spheres of electric charges $+2 \ nC$ and $-8 \ nC$ are placed at a distance '$d$' apart. If they are allowed to touch each other,what is the new distance between them to get a repulsive force of the same magnitude as before?

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