ElectrostaticsHard
Question
A particle of mass 2g and charge 1μC is held at a distance of 1mC. if the particle is released it will be repelled. The speed of particle when it is at a distance of 10 metre from the fixed charge is
Options
A.90m/s
B.100m/s
C.45m/s
D.55m/s
Solution
Potential at 1 m from the charge
VA =
= K × 10-6
Potential at 10m from the charge
VB = = K × 10-7
Potential diff.= VA - VB = K(10-6 - 10-7)
Its velocity at 10 m is then
× 2 × 10-3 × v2 = K × 10-6
× 10-3
v2 =
K × 
× 10-6
= 9 × 10-9 × × 10-6 = 81 × 100
v = 90 m/sec
VA =
= K × 10-6Potential at 10m from the charge
VB =
= K × 10-7Potential diff.= VA - VB = K(10-6 - 10-7)
Its velocity at 10 m is then
× 2 × 10-3 × v2 = K × 10-6 × 10-3v2 =
K × × 10-6= 9 × 10-9 ×
× 10-6 = 81 × 100v = 90 m/sec
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