First slide

Motion of charged particle in uniform electric field

Question

An electron is released with a velocity of $5 \times 10^{6} {ms}^{- 1}$ in an electric field of $10^{3} N C^{- 1}$ which has been applied so as to oppose its motion. How much time could it take before it is brought to rest?

Moderate

A

$2 .8 \times 10^{- 8} s$
B

$1 .8 \times 10^{- 8} s$
C

$6 .4 \times 10^{- 8} s$
D

$4 .2 \times 10^{- 8} s$

Solution

Since electric field is applied so as to oppose the motion of electron,
$\begin{array}{l} a = - \frac{eE}{m} (retardation) = - \frac{1 .6 \times 10^{- 19} \times 10^{3}}{9 .1 \times 10^{- 31}} = - 1 .758 \times 10^{14} {ms}^{- 2} \\ Now, u = 5 \times 10^{6} {ms}^{- 1}, v = 0 \\ Using the relation : v^{2} - u^{2} = 2 as, we get \\ s = 7 .11 \times 10^{- 2} m \\ Using the relation : v = u + at, we get \\ t = 2.844 \times 10^{- 8} s \end{array}$

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