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# A BALL OF MASS 100g and having a charge of 40*10^-5 is relesed from rest in aregion where horizontal E.F of 2*10^4 N/C exists. a)find the resulting force acting on the ball. b) whay will be the path of the ball C) where will be the ball at the end of 2seconds

Apoorva Arora IIT Roorkee
7 years ago
There will ne a vertical force acting downwards = mg=1 newton
and one horizontal force towards the right, say, = Eq= 8 newton
so the resultant force is
$F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos\theta }=\sqrt{65}newton$
and the angle made by the resultant force will be$cos^{-1}\frac{1}{8}$
and hence it will move along a straight line with an angle of 82.8 degrees with the vertical
So after 2 seconds
$x=\frac{1}{2}\frac{F}{m}2^{2}$= 160 m
and
$y=\frac{1}{2}\frac{F}{m}2^{2}$=20 m
assuming that the origin is the position of ball at t=0 and the x axis increases towards the right and y axis increases towards the downward direction.

Thanks and Regards
Apoorva Arora
IIT Roorkee