Two particles have equal masses of 5.0g each and opposite charges of +4.0×10−5C and −4.0×10−5C. 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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The charge on two particles is q1=q2=4×10−5
The distance between particles is s=1m
The mass of each particle is m=5g=0.005kg
The electrostatic force is given by:
The acceleration 'a' of the particles will be:
Now, u=0, s=50cm=0.5m, a=2880m/s2, v=? The final speed of the particles can be obtained as: