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Please help! - Electrostatics - JEE Main-8

In free space , a particle A of charge 1\mu C is held fixed at a point P. Another particle B of the same charge and mass of 4\mu g is kept at a distance of 1mm from P. If B is relesed .then its velocity at a distance of 9mm from  p is:

\left [ Take \, \, \frac{1}{4\pi \epsilon _{0}}=9\times 10^{9} Nm^{2}C^{-2}\right ]

 

 

 

 

  • Option 1)

    1.0m/s

  • Option 2)

    3.0\times 10^{4}m/s

  • Option 3)

    2.0\times 10^{3}m/s

  • Option 4)

    1.5\times 10^{2}m/s

 
Answers (1)
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V Vakul

As we know

loss in potential energy =gain in kinetic energy

K\times 10^{-6}\times 10^{-6}\left [ \frac{1}{10^{-3}}-\frac{1}{9\times 10^{-3}} \right ]=\frac{1}{2}mv^{2}

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

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

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


Option 1)

1.0m/s

Option 2)

3.0\times 10^{4}m/s

Option 3)

2.0\times 10^{3}m/s

Option 4)

1.5\times 10^{2}m/s

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