Question

Two identical metallic spheres A and B when placed at certain distance in air repel each other with a force of F. Another identical uncharged sphere C is first placed in contact with A and then in contact with B and finally placed at midpoint between spheres A and B. The force experienced by sphere C will be

• A. \(\frac{3}{2}F\)
• B. \(\frac{3}{4}F\)
• C. F
• D. 2F

Answer 1

The Correct Option is B

To solve the problem, we need to understand the impact of the interactions and the redistribution of charges among the spheres.

1. Initially, let's assume that the charge on spheres \(A\) and \(B\) is \(q\) each. Since they repel each other with force \(F\), we know that: \(F = \frac{kq^2}{r^2}\), where \(k\) is Coulomb's constant, and \(r\) is the distance between the spheres. 
2. Sphere \(C\) is initially uncharged.
3. When sphere \(C\) is brought into contact with sphere \(A\), the charge is shared between them. Since both spheres are identical, the charge will redistribute equally: \(q_{\text{after}} = \frac{q}{2}\) for both \(A\) and \(C\).
4. Next, sphere \(C\) is brought into contact with sphere \(B\). Again, the charge is shared equally between \(B\) and \(C\). So, the charge on each sphere becomes: \(q_{\text{B final}} = \frac{q + \frac{q}{2}}{2} = \frac{3q}{4}\) and \(q_{\text{C final}} = \frac{3q}{4}\).
5. Now, sphere \(C\) is placed at the midpoint between spheres \(A\) and \(B\).
6. The net force on sphere \(C\) due to spheres \(A\) and \(B\) must be considered. The force between sphere \(A\) and \(C\) is: \(F_{A \to C} = \frac{k \left(\frac{q}{2}\right) \left(\frac{3q}{4}\right)}{\left(\frac{r}{2}\right)^2} = \frac{3kq^2}{2r^2}\). The force between sphere \(B\) and \(C\) is the same due to symmetry: \(F_{B \to C} = \frac{k \left(\frac{3q}{4}\right)^2}{\left(\frac{r}{2}\right)^2} = \frac{9kq^2}{4r^2}\).
7. Adding these forces gives: \(F_{\text{net}} = \frac{3kq^2}{2r^2} - \frac{9kq^2}{4r^2} = \frac{3}{4}F\).

Thus, the force experienced by sphere \(C\) is \(\frac{3}{4}F\). Therefore, the correct answer is \(\frac{3}{4}F\).

Answer 2

When two identical sphere come in contact with each other, the total charge on them is equally distribute.

\(\frac{kQ^2}{d^2}\)=F

F=\(\frac{k9Q^2}{16×\frac{d^2}{4}}−\frac{k3Q^2}{8×\frac{d^2}{4}}\)
=\(\frac{9kQ^2}{4d^2}−\frac{3kQ^2}{2d^2}\)
=\(\frac{kQ^2}{d^2}[\frac{9}{4}−\frac{3}{2}]\)
=\(\frac{6}{8}F\)
=\(\frac{3}{4}F\)
So, the correct option is (B): \(\frac{3}{4}F\)