In free space, a particle A of charge 1 μC is held fixed at a point P. Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P. If B is released, then its velcoity at a diatnce of 9 mm from P is [Take 14πε0=9×109 Nm2C−2]

In free space, a particle A of charge 1 μC is held fixed at a point

1.	In free space, a particle A of charge 1 μC is held fixed at a point P. Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P. If B is released, then its velcoity at a diatnce of 9 mm from P is [Take 14πε0=9×109 Nm2C−2]
Answer» 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 velcoity at a diatnce of $9 mm from P$ is $[Take \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2}]$

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In free space, a particle A of charge 1 μC is held fixed at a point P. Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P. If B is released, then its velcoity at a diatnce of 9 mm from P is [Take 14πε0=9×109 Nm2C−2]

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