Question

A charged particle of mass 100 gm and charge +1.5 C is fixed at a position 2 m from the ground. Two charged particles of the same mass are in equilibrium 50 cm below and 60 cm above the 1st particle respectively. What is the nature and quantity of their charge?

Answer 1

Answer:

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Explanation:

Loss of potential energy = gain in kinetic energy

U

1

−U

2

=ΔK.

kQq[

r

1

1

−

r

2

1

]=1/2mv

2

−0

v=

m

2lQq

[

r

1

r

2

−

r

2

r

1

]

=

0.1×3×0.5

2×9×10

9

×5×10

−6

×2×10

−6

×2.5

=

3

m/s.