Question

An electron is sent in electric field of intensity $ 9.1\times 10^{6}N/C$ The acceleration produce:

• A. $ 1.6\,m/s^{2}$
• B. $ 1.6\times 10^{18}m/s^{2}$
• C. $ 3.2\times 10^{18}m/s^{2}$
• D. $ 0.8\times 10^{18}m/s^{2}$

Answer 1

Answer: (B)

The force $F$ on an electron $(e)$ due to electric field $(E)$ is given by
$F=e E$ ... (i)
From Newtons law $F=m a$ ... (ii)
From Eqs. (i) and (ii),
we get $a=\frac{e E}{m}$
Given, $E=9.1 \times 10^{6} N/C,$
$e=1.6 \times 10^{-19} C$,
$m=9.1 \times 10^{-31} kg$
$a=\frac{1.6 \times 10^{-19} \times 9.1 \times 10^{6}}{9.1 \times 10^{-31}}$
$\Rightarrow a=1.6 \times 10^{18} m / s ^{2}$