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 :
- 4 N
- 6 N
- 1 N
- 12 N
The Correct Option is B
Approach Solution - 1
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:
\[F_{PS} = \frac{3}{8} \times 16 = 6 \, \text{N}\]
Approach Solution -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 .
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