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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)

$1.5\times 10^{2}m/s$

Views

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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