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Question

Two equal charges A and B each of 1/3×106C are placed 200 cm apart in air. A particle carrying a charge of 1/3×106C is projected along the perpendicular bisector from the point O midway between A and B with a kinetic energy of 103J. Before the particle starts to return it will cover a distance

A
1m
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B
2 m
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C
3 m
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D
1/3 m
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Solution

The correct option is A 3 m

Initial potential energy: 2kq1q2r=(2×9×109×13×106×13×106)/(1)=2×103

Initial kinetic energy is 10^(-3) J.

Let the body is at r distance from both the charges (will be at equal distance from both as both charges are same)

Final Kinetic energy is zero (as body stopped).

By equating Initial Total Energy & Final Total Energy:

2×103+103=2kq1q2r=(2×9×109×13×106×13×106)/(r)1×103=2×103/(r)r=2

Therefore distance traveled:

(r2fr2i)=41=3


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