Question.
A charge 4.59 x 10-6 C is held fixed at origin. A second charge 3.40 x 10-6 C is released from rest at the position (1.42 m, 0.746 m). If the mass of the second charge is 6.35 g, what is its speed when it moves infinitely far from the origin?
Answer.
Distance between the charges,
r = sqrt ( 1.42^2 + 0.746^2 )
= 1.60 m
Potential energy of the second charge w.r.t. the first charge,
kqq' / r gets fully converted into kinetic energy,
(1/2) mv^2 at an infinite distance.
=> kqq'/r = (1/2) mv^2
=> v = sqrt (2kqq' / ( r*m )
= √[2 (9*10^9) (4.59*10^-6) (3.40*10^-6) / (1.6) (0.00635)
= 3.72 m/s
A charge 4.59 x 10-6 C is held fixed at origin. A second charge 3.40 x 10-6 C is released from rest at the position (1.42 m, 0.746 m). If the mass of the second charge is 6.35 g, what is its speed when it moves infinitely far from the origin?
Answer.
Distance between the charges,
r = sqrt ( 1.42^2 + 0.746^2 )
= 1.60 m
Potential energy of the second charge w.r.t. the first charge,
kqq' / r gets fully converted into kinetic energy,
(1/2) mv^2 at an infinite distance.
=> kqq'/r = (1/2) mv^2
=> v = sqrt (2kqq' / ( r*m )
= √[2 (9*10^9) (4.59*10^-6) (3.40*10^-6) / (1.6) (0.00635)
= 3.72 m/s
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