In free space, a particle A of charge 1 μC is held fixed at a point P. Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is [Take1/(4πε₀)=9×10⁹ N m² C⁻²]

Question

Accepted Answer

(4) 2.0×10³ m s⁻¹. Both charges are q=10⁻⁶ C, so kq²=9×10⁹×10⁻¹²=9×10⁻³. Initial energy at rᵢ=10⁻³ m: Uᵢ=(9×10⁻³)/(10⁻³)=9 J. At r_f=9×10⁻³ m: U_f=(9×10⁻³)/(9×10⁻³)=1 J. Energy released: 1/2mv²=Uᵢ-U_f=8 J. Taking the mass as 4 mg=4×10⁻⁶ kg: v²=(16)/(4×10⁻⁶)=4×10⁶, so v=2.0×10³ m s⁻¹. Note the printed data: with the mass read literally as 4 μg=4×10⁻⁹ kg the same 8 J gives v=6.3×10⁴ m s⁻¹, which is not among the options, so the released key corresponds to a mass of 4 mg.

In free space, a particle A of charge 1 μC is held fixed at a point P. Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is [Take1/(4πε₀)=9×10⁹ N m² C⁻²]

Step-by-step solution

Why the other options are wrong