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  4. Two identical small conducting sphere having charges Q1 and Q2 with (Q2 >>> Q1). The spheres are at ‘d ’ distance apart. The force they exert on each other is F1. The spheres are made to touch each other and then separated by distance d. The force they exert on each other is F2 . Then (F1/F2) is

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Q. Two identical small conducting sphere having charges $Q_1$ and $Q_2$ with ($Q_2 >>> Q_1$). The spheres are at ‘d ’ distance apart. The force they exert on each other is $F_1$. The spheres are made to touch each other and then separated by distance $d$. The force they exert on each other is $F_2$ . Then $\frac {F_1}{F_2} \,is$

Solution:

Before touching
$F_1 = \frac {1}{4\pi \epsilon_o} \frac{Q_1Q_2}{d^2}$
After touching charge will divide equally.
$Q{′}_1$ = $Q{′}_1$ = $\frac {Q_1+Q_2}{2}\simeq \frac{Q_2}{2}$
$F_2 = \frac {1}{4\pi \epsilon_o} \frac {Q^2_2}{4d^2}$
$\frac{F_{1}}{F_{2}}=\frac{\frac{1}{4\pi\varepsilon_{o}}\frac{Q_{1}Q_{2}}{d^{2}}}{\frac{1}{4\pi\varepsilon_{o}}\frac{Q^{2}_{2}}{4d^{2}}} = \frac{4Q_{1}}{Q_{2}}$