JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 10)
A uniform electric field $$\mathrm{E}=(8 \mathrm{~m} / \mathrm{e}) \,\mathrm{V} / \mathrm{m}$$ is created between two parallel plates of length $$1 \mathrm{~m}$$ as shown in figure, (where $$\mathrm{m}=$$ mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of $$2 \mathrm{~m} / \mathrm{s}$$. The angle of the deviation $$(\theta)$$ of the path of the electron as it comes out of the field will be _________.
_28th_July_Evening_Shift_en_10_1.png)
Explanation
$$E = {{8\,m} \over e}$$ V/m
$$l = 1$$ m
$${v_x} = 2$$ m/s
$${a_y} = - 8$$ m/s2
$$t = {l \over {{v_x}}} = {1 \over 2}s$$
$$ \Rightarrow |{v_y}| = 4$$ m/s
$$\Rightarrow$$ angle of deviation = $$\theta$$
$$\tan \theta = {{{v_y}} \over {{v_x}}}$$
$$\theta = {\tan ^{ - 1}}\left( {{4 \over 2}} \right) = {\tan ^{ - 1}}(2)$$
Comments (0)
