In free space, a particle A of charge 1   μ C is held fixed at point P . Another particle B of the same charge and mass 4   μ g is kept at a distance of 1   m m from P . If B is released, then its velocity at a distance of 9   m m from P is: [Take   1 4 π…

Question

In free space, a particle A of charge 1 &#160; &#956; C is held fixed at point P . Another particle B of the same charge and mass 4 &#160; &#956; g is kept at a distance of 1 &#160; m m from P . If B is released, then its velocity at a distance of 9 &#160; m m from P is: [Take &#160; 1 4 &#960; &#1013; 0 = 9 &#215; 10 9 &#160; N &#160; m 2 &#160; C - 2 ]

Aicharya · Accepted Answer

From conservation in mechanical energy:

- Δ P . E . = Δ K . E .

\frac{1}{4 π ε_{0}} {(1 \times 10^{6})}^{2} [\frac{1}{10^{- 3}} - \frac{1}{9 \times 10^{- 3}}] = \frac{1}{2} \times 4 \times 10^{- 6} \times V^{2}

9 \times 10^{9} \times 10^{- 12} [\frac{8}{9 \times 10^{- 3}}] = 2 \times 10^{- 6} \times V^{2}

V^{2} = 4 \times 10^{6}

V = 2 \times 10^{3} m s^{- 1}

In free space, a particle A of charge 1   μ C is held fixed at point P . Another particle B of the…

Solution

In free space, a particle A of charge 1 &#160; &#956; C is held fixed at point P . Another particle B of the…

Solution

In free space, a particle A of charge 1 μ C is held fixed at point P . Another particle B of the…