Question

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

• A. $1.6m/s^2$
• 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
$F=eE$ ... (i)
From Newtons law $F=ma$ ... (ii)
From Eqs. (i) and (ii),
we get $a=\frac{eE}{m}$
Given, $E=9.1\times 10^{6}N/C,$
$e=1.6\times 10^{-19}C$,
$m=9.1\times 10^{-31}kg$
$a=\frac{1.6\times 10^{-19}\times 9.1\times 10^6}{9.1\times 10^{-31}}$
$\Rightarrow a=1.6\times 10^{18}m/s^2$