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?

• A. $2.8 \times 10^{-8} s$
• B. $1.8 \times 10^{-8} s$
• C. $6.4 \times 10^{-8} s$
• D. $4.2 \times 10^{-8} s$

Answer 1

Answer: (A)

Since electric field is applied so as to oppose the motion of electron,
$a=-\frac{e E}{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: $v^{2}-u^{2}=2 a s$, we get
$s=7.11 \times 10^{-2} m$
Using the relation: $v=u+a t$, we get
$t=2.844 \times 10^{-8} s$