Question

An electron entering field normally with a velocity $4\times 10^{7}m{s}^{-1}$ travels a distance of $0.10m$ in an electric field of intensity $3200V{m}^{-1}$ . What is the deviation from its path?

• A. $1.76mm$
• B. $17.6mm$
• C. $176mm$
• D. $0.176mm$

Answer 1

Answer: (A)

Time to cross $0.10\text{m},$ $t=\frac{0.10}{4\times 10^7}s$
Distance deviated in this time due to electric field $=\frac{1}{2}(\frac{eE}{m}).t^2$
$=\frac{1}{2}\times \frac{1.6\times 10^{-19}\times 3200}{9.1\times 10^{-31}}$ $\times (\frac{0.1}{4\times {\left(10\right)}^7}{)}^2\text{m}$
$=01.76mm$