1

Question

A particle of mass 1 kg and charge −2μC(initially at a very large distance) is projected with initial velocity 200 m/s towards a positive charged particle of charge 2μC which is fixed. The distance of 1 kg particle from heavy particle when speed of 1 kg particle is equal to twice of initial value is x107 m. Find the value of ‘x’.

Open in App

Solution

Initial separation is large so, initial electrostatic potential energy is 0

12mu2+0=12m.(2u)2−(14πϵ0)q2r

r=2q24πϵ0.3mu2=9×109×2×(2×10−6)23×1×(200)2

r=6107 m

12mu2+0=12m.(2u)2−(14πϵ0)q2r

r=2q24πϵ0.3mu2=9×109×2×(2×10−6)23×1×(200)2

r=6107 m

1

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program