Question

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 $\frac{F_1}{F_2}is$

• A. $\frac{4Q_1}{Q_2}$
• B. $\frac{Q_1}{4Q_2}$
• C. $\frac{4Q_2}{Q_1}$
• D. $\frac{Q_2}{4Q_1}$

Answer 1

Answer: (A)

Before touching
$F_1=\frac{1}{4\pi {ϵ}_o}\frac{Q_1{Q}_2}{d^2}$
After touching charge will divide equally.
$Q{\prime }_1$ = $Q{\prime }_1$ = $\frac{Q_1+Q_2}{2}\simeq \frac{Q_2}{2}$
$F_2=\frac{1}{4\pi {ϵ}_o}\frac{Q_2^2}{4d^2}$
$\frac{F_1}{F_2}=\frac{\frac{1}{4\pi {\epsilon }_o}\frac{Q_1{Q}_2}{d^2}}{\frac{1}{4\pi {\epsilon }_o}\frac{Q_2^2}{4d^2}}=\frac{4Q_1}{Q_2}$