A point charge Q is placed on a table and another particle of charge q and mass m is placed in air in such a way that it remains suspended at the same location in a state of equilibrium. What should be the distance of particle from point charge fixed on table
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Answer:
A charged particle of charge Q is held fixed and another charged particle of mass m and charge q (of the same sign) is released from a distance r. The impulse of the force exerted by the external agent on the fixed charge by the time distance between Q and q becomes 2r is:
December 20, 2019
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Parveen Rekhaa
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ANSWER
As the other charge particle exerts force on the charge Q, the external agent has to apply equal and opposite force to keep it stationary. This force is thus equal to the force exerted on q by Q.
Net impulse is I=∫
0
t
Fdt=mΔ
v
q
Initial Potential energy q is U
i
=
4πϵ
0
1
r
Final Potential energy of q is U
f
=
4πϵ
0
1
2r
As the charge q is only under the influence of electrostatic forces, mechanical energy is conserved.
Final Kinetic energy id K
f
=U
i
−U
f
=
4πϵ
0
1
2r
Thus, v
q
=
m
2K
f
⇒I=mv
q
=
2mK
f
=
4πϵ
0
r
Qqm