Guest

A particle of mass 1 g and charge 1 microcoulomb is held at rest on a frictionless surface at a distance of 1 m from the fixed charge of 2 mC. if the particle is released, it will be repelled. the speed of the particle when it is at a distance of 10 m from the fixed charge is?

A particle of mass 1 g and charge 1 microcoulomb is held at rest on a frictionless surface at a distance of 1 m from the fixed charge of 2 mC. if the particle is released, it will be repelled. the speed of the particle when it is at a distance of 10 m from the fixed charge is?

Grade:12

2 Answers

ananthupotti
38 Points
7 years ago
 
Force eperienced by the particle=kq1q2/r^2
=k*1*(10^-6)*2*(10^-3)/1
=18 N.
acceleration of particle=f/m=18/1*(10^-3)
a=18000m/s^2.
since V^2=U^2+2*a*S,
Initial velocity of particle=0 m/s,
thus, V^2=2*18000*10
V^2=360000
V=600m/s
Hope everything is clear!!
 
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
We know, that electrostatic force is a conservative force.
Hence, the mechanical energy will remain conserved for the point charge
Hence, Ki + Ui = Kf + Uf
½ mu2 + kq1q2/r1 = ½ mv2 + kq1q2/r2
Hence, 0 + k x 10-6 x 10-3(1/1 – 1/10) = ½ x 10-3 x v2
Hence, v2 = 2 x 103 x 9 x 109 x 10-9 x 9/10
Hence, v2 = 16200
Hence, v = 127.28 m/s
 
Thanks and regards,
Kushagra

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free