ElectrostaticsHard
Question
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 :
Options
A.
B.
C.0
D.cannot be determined
Solution
Applying conservation of energy, we get
= 
+ 
mv2
∴
mv2 = 
or v = 
∴ Force required to keep Q fixed = Coulombic repulsion between the charges.
= force on charge q
∴ Impulse = change in momentum of q. = mv =
= 
= + mv2∴
mv2 = or v = ∴ Force required to keep Q fixed = Coulombic repulsion between the charges.
= force on charge q
∴ Impulse = change in momentum of q. = mv =
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