Question

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

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

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 $= \frac{1}{2} \left(\right. \frac{e E}{m} \left.\right) . \, t^{2}$
$= \frac{1}{2} \times \frac{1.6 \times 10^{- 19} \times 3200}{9.1 \times 10^{- 31}}$ $\times \left(\right. \frac{0.1}{4 \times \left(10\right)^{7}} \left.\right)^{2} \text{m}$
$= 01.76 \, mm$