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A particle of mass 2g and charge 1\mu C is held at a distance of 1 metre from a fixed charge of 1\mu C.  If the particle is released it will be repelled. The speed of the particle when it is at a distance of 10 metres from the fixed charge is

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We have,
Mass of the particle = 2g
(Q_1)= 1\mu C
Q_2=1\mu C
and distance between them (r) = 1m
Therefore, Force between them is given as
F= k\frac{Q_1.Q_2}{r^2}

     =9\times 10^9(\frac{1\times 10^{-6}\times 1\times 10^{-6}}{1^2})
      =9\times 10^{-3}

So, acceleration of a particle is calculated as
a=\frac{F}{m}
\\a=\frac{9\times 10^{-3}}{2\times 10^{-3}}\\ a=4.5\ m/s^2

We know that,
v^2 = u^2-2as

Putting the values in the above equation (initially, the particle is at rest so, u = 0)
Now, 
\\\Rightarrow v^2=2(4.5\times (10-1))\\
             \\=2(4.5\times (10-1))\\ =2(405)\\ =810
\\v=\sqrt{810}\\ v=28.46\ m/s

 

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manish

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