Gujarati
1. Electric Charges and Fields
medium

There is a uniform electric field of strength ${10^3}\,V/m$ along $y$-axis. A body of mass $1\,g$ and charge $10^{-6}\,C$ is projected into the field from origin along the positive $x$-axis with a velocity $10\,m/s$. Its speed in $m/s$ after $10\,s$ is (Neglect gravitation)

A

$10$

B

$5\sqrt 2 $

C

$10\sqrt 2 $

D

$20$

Solution

(c) Body moves along the parabolic path.
For vertical motion : By using $v = u + at$
$==> $ ${v_y} = 0 + \frac{{QE}}{m}.\,t = \frac{{{{10}^{ – 6}} \times {{10}^3}}}{{{{10}^{ – 3}}}} \times 10 = 10\,m/{\rm{sec}}$
For horizontal motion -It’s horizontal velocity remains the same i.e. after $10$ $sec$, horizontal velocity of body $v_x = 10\, m/sec$.
Velocity after $10$ $sec$ $v = \sqrt {v_x^2 + v_y^2} $$ = 10\sqrt 2 \,m/sec$

Standard 12
Physics

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