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Physics-

General

Easy

Question

A particle of mass $2 cross times 10 to the power of negative 5 end exponent K g$ and charge $4 cross times 10 to the power of negative 3 end exponent C$ moves from rest in a uniform electric field of magnitude $5 blank V divided by m.$ its kinetic energy after 10 seconds is

${2 \times 10}^{3} J$
$10^{3} J$
$2 \times 10^{- 3} J$
$10^{2} J$

The correct answer is: $10 to the power of 3 end exponent blank J$

Here, mass of particle, $m equals 2 cross times 10 to the power of negative 5 end exponent K g$
Charge of a particle, $q equals 4 cross times 10 to the power of negative 3 end exponent C$
Electric field, $E equals 5 blank V blank m to the power of negative 1 end exponent$
Force on a charged particle in a uniform electric field is
$F equals q E$
$therefore$ Acceleration of the particle, $a equals fraction numerator q E over denominator m end fraction$
$equals fraction numerator 4 cross times 10 to the power of negative 3 end exponent cross times 5 over denominator 2 cross times 10 to the power of negative 5 end exponent end fraction equals 10 to the power of 3 end exponent m s to the power of negative 2 end exponent$
Let $v$ be velocity of particle after $10 blank s$
As $v equals u plus a t$
$therefore v equals a t equals open parentheses 10 to the power of 3 end exponent m s to the power of negative 2 end exponent close parentheses open parentheses 10 blank s close parentheses equals 10 to the power of 4 end exponent m s to the power of negative 1 end exponent blank left square bracket u equals 0 right square bracket$
Kinetic energy of a particle after 10 s is
$K equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 to the power of negative 5 end exponent cross times 10 to the power of 4 end exponent cross times 10 to the power of 4 end exponent equals 10 to the power of 3 end exponent blank J$

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