phy (electrostatics)

2011-02-13 8:37 pm
Two positive charged particles are kept at a certain distance apart in vacuum. If the particles are released, their subsequent accelerations will
A. increase with time.
B. decrease with time.
C. remain unchanged.
D. be undeterminable

THE ANS IS B 但我唔明點解...

回答 (2)

2011-02-13 8:51 pm
✔ 最佳答案
The electrostatic force F between the two charged particles is given by Coulomb's Law,F = k(Q1).(Q2)/r^2where k is a constant
Q1 and Q2 are the charges respectively on the particles
r is their distance of separation

Since the two charges are both positive, the force is repulsive.
acceleration a = F/m, where m is the mass of each particle.

thus, a = k(Q1).(Q2)/m(r^2)
It is thus clear that because of the repulsive force, the two particles will be moving away from one another. In doing so, r increases with time, and hence a decreases with time.
2011-02-13 8:51 pm
因為佢地都係 +ve charge, 即係佢地係like charges,佢地released 咗之後,就會repel, 走相反方向。最初acceleration 會大D, 因為佢地distance 較近,當他們一直 repel, distance就長咗,影響大家的程度少咗,repel 的程度就會細咗,所以acceleration 都會細咗,所以就decrease with time。


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