NEET 2025 Physics Coulomb's Law MCQ Question

To solve the problem, we will analyze the situation step by step, focusing on how charge redistribution occurs between the spheres and how it affects the force of repulsion.

Step 1: Initial Setup

Initially, we have two identical conducting spheres A and B, each carrying a charge $q$. The force of repulsion $F$ between them can be calculated using Coulomb's Law:

$F = k \frac{q^2}{r^2}$

where $k$ is Coulomb's constant and $r$ is the distance between the centers of the spheres.

Step 2: Bringing in the Uncharged Sphere

When a third identical uncharged conducting sphere (let's call it sphere C) is brought in contact with sphere A, the charge redistributes. Since spheres A and C are identical and in contact, the total charge $q$ on sphere A will be evenly distributed between them.

After contact:

• Charge on sphere A: $\frac{q}{2}$
• Charge on sphere C: $\frac{q}{2}$

Now, when sphere C is brought in contact with sphere B, the charge will redistribute again. The total charge on sphere B is still $q$, and now it has to share this charge with sphere C.

After contact with sphere B:

• Total charge on spheres B and C: $q + \frac{q}{2} = \frac{3q}{2}$
• Charge on sphere B after contact: $\frac{3q}{4}$
• Charge on sphere C after contact: $\frac{3q}{4}$ (as it gains $\frac{q}{4}$)

Step 3: Final Charges on Spheres A and B

After all the contacts, we have:

• Charge on sphere A: $\frac{q}{2}$
• Charge on sphere B: $\frac{3q}{4}$

Step 4: Calculating New Force of Repulsion

Now, we can use Coulomb's Law again to find the new force of repulsion $R$ between spheres A and B:

$R = k \frac{(q_A)(q_B)}{r^2}$

Substituting the charges we found: $R = k \frac{\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{r^2} = k \frac{\frac{3q^2}{8}}{r^2}$

Step 5: Relating New Force to Initial Force

Recall that the initial force $F$ was given by: $F = k \frac{q^2}{r^2}$

Now, we can express the new force $R$ in terms of the original force $F$:

$R = \frac{3}{8} k \frac{q^2}{r^2} = \frac{3}{8} F$

Thus, the new force of repulsion between the spheres A and B after the charge redistribution is:

$R = \frac{3F}{8}$

Conclusion

Therefore, the correct answer is option D: $\frac{3F}{8}$.

Why Other Options Are Incorrect

• Option A ($\frac{3F}{5}$): This value does not correspond to any charge distribution scenario derived from the problem.
• Option B ($\frac{2F}{3}$): Similar to option A, this value does not reflect the calculated redistribution of charges.
• Option C ($\frac{F}{2}$): Again, this does not match our derived force based on the final charges on the spheres.

In conclusion, the analysis and calculations show that the new force of repulsion is indeed $\frac{3F}{8}$, confirming the correctness of option D.