Question

An electron moving along positive X-axis with a $3\times {10}^6$m/s speed, enters the region of a uniform electric field. The electron stops after travelling a distance of $90mm$ in the field. The electric field strength is (charge on electron $=1.6\times {10}^{-19}$, mass of electron$=9.1\times {10}^{-31}kg$) :

• A. $2.84KV{m}^{-1}$ along -ve X-axis
• B. $0.284KV{m}^{-1}$ along -ve X-axis
• C. $0.284KV{m}^{-1}$ along +ve X-axis
• D. $28.4KV{m}^{-1}$ along +ve X-axis

Answer 1

Answer: (C)

$0.284KV{m}^{-1}$ along +ve X-axis

e stops after travelling a distance d

Now work done by electric field must be equal

to change in kinetic rengy of $e^{-}$

$\therefore Eed=\frac{1}{2}m{v}^2$

$\Rightarrow E=\frac{1}{2}\left(\frac{m}{e}\right)\frac{v^2}{d}$

$\Rightarrow E=\frac{1}{2}\left(\frac{9.1\times {10}^{-31}}{1.6\times {10}^{-19}}\right)$ $\frac{(9\times {10}^{12})}{90\times {10}^{-3}}$

$\Rightarrow E=0.284\times {10}^3$

$\therefore E=0.284\frac{KV}{m}\left(\hat{x}\right)$

$\to$ as $e^{-}$ is deaccelerating, force on it must be in opposite direction of velocity

$\therefore$ E is in $\hat{x}$ direction