Question

A mass $m=20\text{g}$ has a charge $q=3.0\text{mC}$. It moves with a velocity of $20\text{m/s}$ and enters a region of electric field of $80\text{N/C}$ in the same direction as the velocity of the mass. The velocity of the mass after $3\text{sec}$ in this region is–

• A. $80\text{m/s}$
• B. $56\text{m/s}$
• C. $44\text{m/s}$
• D. $40\text{m/s}$

Answer 1

The correct option is B $56\text{m/s}$

Given:-
$m=20\text{g}=20\times {10}^{-3}\text{kg}$
$q=3\text{mC}=3\times {10}^{-3}\text{C}$
$E=80\text{N/C}$
$u=20\text{m/s}$

In an electrostatic equilibrium,

Electric force $=ma$

So, $qE=ma$

or, $a=\frac{qE}{m}=12{\text{m/s}}^2$

From first equation of the motion we get,

$v=u+at$

$=20+12\times 3$

$=56\text{m/s}$