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In free space, a particle A of charge 1μC is held fixed at a point P. Another particle B of the same charge and mass 4μg is kept at a distance of 1mm from P. If B is released, then it's velocity at a distance of 9mm from P is:-
Asked by Harshvardhansinghr10 | 26 Apr, 2022, 10:55: PM
answered-by-expert Expert Answer
Force F between charges A and B when distance between them is 1 mm is calculated as follows
 
F = K × ( q2 / d2 ) = 9 × 109 × 10-12 / 10-6 = 9000 N
 
where K = 1/ (4πεo ) = 9×109 N m2 C-2  is Coulomb's constant , q is magnitude of either charge
and d is distance between them
 
acceleration = Force / mass = 9000 / ( 4 ×10-9 ) = 2.25 × 1012 m/s2
 
Velocity at a point Q  which is at a distance 9 mm from charge B is determined from the following
equation of motion
 
v2 = u2 + ( 2 a S )
 
where v is final velocity , u is initial velocity which is zero because charge B is released from rest ,
a is acceleration and S is distance travelled
 
begin mathsize 14px style v space equals space square root of 2 cross times a cross times S end root space equals space square root of 2 cross times 2.25 space cross times 10 to the power of 12 cross times 9 space cross times 10 to the power of negative 3 end exponent end root space equals space 2.012 space cross times 10 to the power of 5 space m divided by s end style
Answered by Thiyagarajan K | 27 Apr, 2022, 08:14: AM
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