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Question

As shown in the figure, a dust particle with mass m=5.0×109 kg and charge q0=2.0 nC starts from the rest at point a and moves in a straight line to point b. What is its speed v at point b?

A
26 ms1
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B
34 ms1
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C
46 ms1
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D
14 ms1
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Solution

The correct option is C 46 ms1
Let Va and Vb be velocity of particle at point a and b respectively.
As the particle starts from rest
Va=0
Let Ua be the total energy of the particle at point a
Ua=P.E+K.E
Ua=(k×3×109×q00.01k×3×109×q00.02) + 0
Let Ub be the total energy of the particle at point b
Ub=(k×3×109×q00.02k×3×109×q00.01) + 12mV2b
By total energy conservation
Ua=Ub
k×3×109×q00.01k×3×109×q00.02 + 0 = k×3×109×q00.02k×3×109×q00.01) + 12mV2b

putting all the values (k=9×109,qo=2×109 and m=5×109)

Upon rearranging the terms and putting the values of q0 and m we get
Vb46 ms1

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