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Two identical small spheres carry charge of $Q_1$ and $Q_2$ with $Q_1>>Q_2.$ The charges are $d$ distance apart. The force they exert on one another is $F_1.$ The spheres are made to touch one another and then separated to distance $d$ apart. The force they exert on one another now is $F_2.$ Then $F_1/F_2$ is
$\frac{{4{Q_1}}}{{{Q_2}}}$
$\frac{{{Q_1}}}{{4{Q_2}}}$
$\frac{{4{Q_2}}}{{{Q_1}}}$
$\frac{{{Q_2}}}{{4{Q_1}}}$
Solution
In the 1st case we have:
$F_1=\frac{ kQ _1 Q_2}{d^2}\left(\right.$ By coulomb's law, $\left.k =\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 Nm ^2 / C ^2\right)$
After touching the spheres the charges distribute such that charge on sphere is: $q =\frac{ Q _1+ Q _2}{2} \approx \frac{ Q _1}{2}\left( Q _1> Q _2\right)$
So new force between them will be:
$F_2=\frac{ kq ^2}{ d ^2}=\frac{ kQ }{4 d ^2}$
Hence $\frac{F_1}{F_2}=\frac{4 Q_1 Q_2}{Q_1^2}=\frac{4 Q_2}{Q_1}$.
