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  4. 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: [ textTake (1/4πε0 ) = 9×109 Nm2 C-2]

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Q. 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 $4\, \mu \,g$ 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 :
$\left[\text{Take} \frac{1}{4\pi\varepsilon_{0} } = 9\times10^{9} Nm^{2} C^{-2}\right] $

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Solution:

$W_{E} = -\left[\Delta U\right] =U_{i} -U_{F} = \frac{1}{2} mv^{2} $
$ U = \frac{kq_{1}q_{2}}{r} $
$ \frac{\left(9\times10^{9}\right)\times10^{-12}}{10^{-3}} - \frac{\left(9\times10^{9} \right)\times10^{-12}}{9\times10^{-3}} = \frac{1}{2} \times\left(4 \times10^{-6}\right)v^{2} $
$v^{2} = 4\times10^{6} $
$v = 2 \times10^{3} \frac{m}{s}$