A uniform electric field of 10N C ^ - 1 is created between two...

Given here, kinetic energy $\dfrac{1}{2}e = \dfrac{1}{2}m{u}_{x}^{2} ⇒ {u}_{x} = \sqrt{\dfrac{e}{m}}$

Acceleration of electron due to electric field is $a = \dfrac{eE}{m}$, where, $e$ is charge on electron.

Using equation of motion, $s = {u}_{x}t + \dfrac{1}{2}{a}_{x}{t}^{2}$ along $x$ direction, we get

$L = {u}_{x}t + 0\left(∵ {a}_{x} = 0\right) ⇒ 10 \times {10}^{- 2} = {u}_{x}t ⇒ t = \dfrac{{10}^{- 1}}{{u}_{x}} . . . . \left(1\right)$

Using equation of motion, ${v}_{y} = {u}_{y} + {a}_{y}t$ along y direction, we get

${v}_{y} = 0 + \dfrac{eE}{m}t = \dfrac{eE}{m}t . . . \left(2\right)$

From equation \left(1\right)and\left(2\right)$, we have</p><p>${v}_{y} = \dfrac{eE}{m}\left(\dfrac{{10}^{- 1}}{{u}_{x}}\right)$</p><p>Now,$tan\theta = \dfrac{{v}_{y}}{{v}_{x}} = \dfrac{eE}{m}\left({10}^{- 1}{u}_{x}^{2}\right) = \dfrac{e \times 10 \times {10}^{- 1}}{m \times \dfrac{e}{m}} = 1$</p><p>The angle of deviation of the path of electron as it comes out of the field is$\theta = 45 °.