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Grade: 12

                        

a particle of mass 0.002 kg and a charge 1 mewC is held at rest on a frictionless horizontal surfaces at a distance of 1 m from a fixed charge of 1mC .if the particle is relaxed,it will be repelled.the speed of the particle when it is at a distance of 10m from the fixed charge is- 60m/s 75m/s 90m/s 100m/s

6 months ago

Answers : (2)

Arun
24742 Points
							
mass of particle, m = 2g = 2 × 10^-3 kg
 
charge on particle , q = 10^-6 C
 
another charged particle Q = 1mC = 10^-3C is fixed at r = 1m from first particle.
 
repulsion force acts between particles , F = kqQ/r²
 
= 9 × 10^9 × 10^-3 × 10^-6/1²
 
= 9N
 
from Newton's 2nd law, Acceleration of first particle , a = F/m
 
= 9/2 × 10^-3 = 9000/2 = 450 m/s²
 
Let final velocity of particle is v m/s
 
initial velocity of article is 0 m/s
 
displacement covered by particle = final position - initial position
 
= 10m - 1m = 9m
 
using formula, v² = u² + 2as
 
or, v² = 0 + 2 × 450 × 9
 
or, v² = 8100 = 90² => v = 90 m/s
 
hence, velocity of particle it is at a distance of 10 m from the fixed charge is 90m/s
 
 
6 months ago
Saurabh Koranglekar
askIITians Faculty
8407 Points
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6 months ago
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