Question

What is the acceleration of an electric field of magnitude $50\, V - cm ^{-1}$ ? [Given that $\frac{e}{m}=1.76 x 10^{11} C - kg ^{-1}$]

• A. $ 8.8\times 10^{16}m/s^{2}$
• B. $ 8.8\times 10^{14}m/s^{2}$
• C. $ 8.8\,m/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 $F=e E$, and its acceleration is.
$a=\frac{F}{m}=\frac{e E}{m}$
Given, $\frac{e}{m}=1.76 \times 10^{11} C - kg ^{-1}$,
$E=50 \times 10^{2} V / m$
$a=50 \times 10^{2} \times 1.76 \times 10^{11}$
$\Rightarrow a=8.8 \times 10^{14} m / s ^{2}$