# A particle of mass 2g and charge  is held at a distance of 1 metre from a fixed charge of .  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

We have,
Mass of the particle = 2g
($Q_1$)=
$Q_2=1\mu C$
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,
$\\\Rightarrow v^2=2(4.5\times (10-1))\\$
$\\=2(4.5\times (10-1))\\ =2(405)\\ =810$
$\\v=\sqrt{810}\\ v=28.46\ m/s$

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-