Two spheres having equal mass 𝑚, charge 𝑞 and radius 𝑅, are moving towards each other. Both have speed 𝑢 at an instant when distance between their centers is 4𝑅. Minimum value of 𝑢 so that they touch each other is

Correct Answer:

$\sqrt{\frac{q^2}{16\pi {\epsilon }_{0}mR}}$

Explanation:

Consider two spherical conductors of mass m, charge q, and radius R as shown in the diagram below:

1. Initial State:
Initially, the distance between the centers of the two spheres is r₁ = 4R. Both spheres move directly towards each other with a speed u.
The total initial kinetic energy of the two-sphere system is:

$K_i=\frac{1}{2}m{u}^2+\frac{1}{2}m{u}^2=m{u}^2$

The initial electrostatic potential energy of interaction between the two spheres is:

$U_i=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{4R}$

2. Final State (Just Touching):
When the two spheres just touch each other, the distance between their centers becomes equal to the sum of their radii, i.e., r₂ = R + R = 2R.
For the minimum speed u required for them to touch, both spheres will momentarily come to rest relative to each other at the instant of contact. Thus, the final kinetic energy K_f = 0.
The final electrostatic potential energy at this instant is:

$U_f=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{2R}$

3. Applying Conservation of Mechanical Energy:
Since only conservative electrostatic forces are acting on the system, mechanical energy is conserved:

$K_i+U_i=K_f+U_f$

Substitute the values into the conservation equation:

$m{u}^2+\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{4R}=0+\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{2R}$

Rearranging terms to solve for u:

$m{u}^2=\frac{q^2}{4\pi {\epsilon }_{0}R}\left(\frac{1}{2}-\frac{1}{4}\right)$

$m{u}^2=\frac{q^2}{4\pi {\epsilon }_{0}R}\times \frac{1}{4}=\frac{q^2}{16\pi {\epsilon }_{0}R}$

Dividing by mass m:

$u^2=\frac{q^2}{16\pi {\epsilon }_{0}mR}$

Taking the square root on both sides gives the minimum speed u:

$u=\sqrt{\frac{q^2}{16\pi {\epsilon }_{0}mR}}$