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Question

Two particles have equal masses of 5.0g each and opposite charges of +4.0×105C and 4.0×105C. They are released from rest with a separation of 1.0m between them. Find the speeds of the particles when the separation is reduced to 50cm.

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Solution

The charge on two particles is q1=q2=4×105
The distance between particles is s=1m
The mass of each particle is m=5g=0.005kg

The electrostatic force is given by:
F=Kq2r2

=9×109×(4×105)212=14.4N

The acceleration 'a' of the particles will be:
a=Fm

=14.40.005=2880m/s2

Now, u=0, s=50cm=0.5m,
a=2880m/s2, v=?
The final speed of the particles can be obtained as:
v2=u2+2as
v2=2×2880×0.5

v=2880=53.66 m/s54 m/s for each particle.

1565132_1720391_ans_9c4501da22574ca5837816c788c817b0.png

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