Question

Two identical conducting spheres P and S with charge Q on each, repel each other with a force 16N. A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :

• A. 4 N
• B. 6 N
• C. 1 N
• D. 12 N

Answer 1

The Correct Option is B

The initial force of repulsion between \( P \) and \( S \) is:
\[F_{PS} \propto Q^2\]
\[F_{PS} = 16 \, \text{N}\]
1. When the uncharged sphere \( R \) is brought in contact with \( P \), the charge on \( P \) and \( R \) redistributes equally because they are identical spheres. Thus, after contact with \( P \):
\[\text{Charge on each of } P \text{ and } R = \frac{Q}{2}\]
2. Next, \( R \) is brought in contact with \( S \), and the charge will again redistribute equally between \( R \) and \( S \). After this contact:
\[\text{Charge on each of } S \text{ and } R = \frac{3Q}{4}\]
Now, \( P \) has a charge of \( \frac{Q}{2} \) and \( S \) has a charge of \( \frac{3Q}{4} \). The new force of repulsion between \( P \) and \( S \) is:
\[F_{PS} \propto \frac{Q}{2} \times \frac{3Q}{4} = \frac{3Q^2}{8}\]
Since the initial force \( F_{PS} \) was 16 N, we have:
\[F_{PS} = \frac{3}{8} \times 16 = 6 \, \text{N}\]

Answer 2

To determine the new force of repulsion between the spheres after the third sphere (R) is brought into contact with P and S, let's break down the scenario step-by-step:

Step 1: Initial Condition

Initially, the two identical conducting spheres P and S each have a charge \(Q\) and repulse each other with a force of 16 N.

According to Coulomb's Law, the force between two charges is given by:

\(F = \frac{k \cdot Q_1 \cdot Q_2}{r^2}\)

Given that \(Q_1 = Q_2 = Q\) and the force \(F = 16 \text{ N}\), we can write:

\(16 = \frac{k \cdot Q^2}{r^2} \quad \text{(Equation 1)}\)

Step 2: Bringing Sphere R in Contact with P

Sphere R is initially uncharged. When R is brought in contact with P, the total charge is shared equally between them because they are identical. Therefore, the charge on each becomes:

\(Q_{\text{new}} = \frac{Q + 0}{2} = \frac{Q}{2}\)

Step 3: Bringing Sphere R in Contact with S

S is still charged with \(Q\), and R now has a charge of \(\frac{Q}{2}\). When R is brought in contact with S, they share the charge equally. Therefore, the charge on each becomes:

\(Q_{\text{new}} = \frac{Q + \frac{Q}{2}}{2} = \frac{3Q}{4}\)

Step 4: Final Charges on P and S

Now, P has a charge of \(\frac{Q}{2}\) and S has a charge of \(\frac{3Q}{4}\).

Step 5: Calculating the New Force of Repulsion

Using Coulomb's Law again to find the new force:

\(F_{\text{new}} = \frac{k \cdot \left(\frac{Q}{2}\right) \cdot \left(\frac{3Q}{4}\right)}{r^2}\)

Simplifying, we find:

\(F_{\text{new}} = \frac{k \cdot \frac{3Q^2}{8}}{r^2}\)

Using Equation 1, we substitute \(k \cdot Q^2/r^2 = 16\):

\(F_{\text{new}} = \frac{3}{8} \cdot 16 = 6 \text{ N}\)

Thus, the new force of repulsion between P and S is 6 N.

Therefore, the correct answer is 6 N .