QJEEM19Ph1ef2: JEE Main 2019 Physics - Class 12 JEEM19Ph1ef2 Solution: In free space, a particle A of charge 1 μ C is held fixed at a…

Question

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In free space, a particle A of charge 1

μ

C is held fixed at a point P. Another particle B of the same charge and mass 4

μ

g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is :

[T a k e \frac{1}{4 π \in_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}]

Accepted Answer

q_A = 1

μ

c ; q_B = 1

μ

c, m_B = 4 × 10^–9 kg, r_AB = 10^–3 m

\frac{1}{2} M_{B} V^{2} = k q_{A} q_{B} {\frac{1}{10^{- 13}} - \frac{1}{9 \times 10^{- 3}}}

\frac{1}{2} 4 \times 10^{- 9} V^{2} = 9 \times 10^{9} \times 10^{- 6} \times \frac{8}{9} \times 10^{3}

V^{2} = \frac{8}{2} \times 10^{9} = 4 \times 10^{9}

V =

\sqrt{40} \times 10^{4} m / s

= 6.32

\times

10⁴ m/s

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