Question

What is the acceleration of an electric field of magnitude $50V-c{m}^{-1}$ ? [Given that $\frac{e}{m}=1.76x10^{11}C-k{g}^{-1}$]

• A. $8.8\times 10^{16}m/s^2$
• B. $8.8\times 10^{14}m/s^2$
• C. $8.8m/s^2$
• D. $8.8\times 10^{20}m/s^2$

Answer 1

Answer: (B)

The force experienced by an electron $(e)$ in an electric field $(E)$ is $F=eE$, and its acceleration is.
$a=\frac{F}{m}=\frac{eE}{m}$
Given, $\frac{e}{m}=1.76\times 10^{11}C-k{g}^{-1}$,
$E=50\times 10^{2}V/m$
$a=50\times 10^2\times 1.76\times 10^{11}$
$\Rightarrow a=8.8\times 10^{14}m/s^2$