1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Two identical particles of mass m carry charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v. The closest distance of approach will be

A
14πε0Q2mv
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
14πε04Q2mv
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
14πε04Q2mv2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
14πε02Q2mv2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 14πε04Q2mv2Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be uSo using conservation of momentum we getmv=2mu or u=v2The initial energy of the system is given as 12mv2and the energy at the minimum distance is given as12m(v2)2+12m(v2)2+14πϵoQ2REquating the two energies we get12mv2=14mv2+14πϵoQ2Ror14mv2=14πϵoQ2RorR=4Q24πϵomv2

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Electric Potential Due to a Point Charge and System of Charges
PHYSICS
Watch in App
Join BYJU'S Learning Program