Two identical small spheres carry charge of Q | vedclass

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Question

In the 1st case we have:

$F_1=\frac{ kQ _1 Q_2}{d^2}\left(ight.$ By coulomb's law, $\left.k =\frac{1}{4 \pi \epsilon_0}=9 	imes 10^9 Nm ^2 /

Vedclass · Accepted Answer

$\frac{{4{Q_2}}}{{{Q_1}}}$

Vedclass · Answer

In the 1st case we have:

$F_1=\frac{ kQ _1 Q_2}{d^2}\left(ight.$ By coulomb's law, $\left.k =\frac{1}{4 \pi \epsilon_0}=9 	imes 10^9 Nm ^2 / C ^2ight)$

After touching the spheres the charges distribute such that charge on sphere is: $q =\frac{ Q _1+ Q _2}{2} \approx \frac{ Q _1}{2}\left( Q _1> Q _2ight)$

So new force between them will be:

$F_2=\frac{ kq ^2}{ d ^2}=\frac{ kQ }{4 d ^2}$

Hence $\frac{F_1}{F_2}=\frac{4 Q_1 Q_2}{Q_1^2}=\frac{4 Q_2}{Q_1}$.

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