Question

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 :

• 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: (C)

$\frac{4Q_2}{Q_1}$

Answer 2

Here's how to solve this problem:

1. Initial Force ($F_1$): Use Coulomb's Law:

   $F_1 = k \frac{Q_1 Q_2}{d^2}$

2. Charge Redistribution: When the spheres touch, the total charge ($Q_1 + Q_2$) is distributed equally between them. So, each sphere has a new charge $q = \frac{Q_1 + Q_2}{2}$.

3. New Force ($F_2$):

   $F_2 = k \frac{q^2}{d^2} = k \frac{(\frac{Q_1 + Q_2}{2})^2}{d^2} = k \frac{(Q_1 + Q_2)^2}{4d^2}$

4. Ratio $F_1/F_2$:

   $\frac{F_1}{F_2} = \frac{k \frac{Q_1 Q_2}{d^2}}{k \frac{(Q_1 + Q_2)^2}{4d^2}} = \frac{4Q_1 Q_2}{(Q_1 + Q_2)^2}$

5. Approximation: Since $Q_1 >> Q_2$, we can approximate $Q_1 + Q_2 \approx Q_1$.

   $\frac{F_1}{F_2} \approx \frac{4Q_1 Q_2}{Q_1^2} = \frac{4Q_2}{Q_1}$

Therefore, the ratio $F_1/F_2$ is approximately $\frac{4Q_2}{Q_1}$.