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(B) The electric field between the plates is $E = \frac{V}{d}$.The force on an electron of charge $e$ is $F = eE = \frac{eV}{d}$.The acceleration of t

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{eVL}{mdv_{0}^{2}}ight)$

Vedclass · Answer

(B) The electric field between the plates is $E = \frac{V}{d}$.The force on an electron of charge $e$ is $F = eE = \frac{eV}{d}$.The acceleration of the electron in the transverse direction ($y$-axis) is $a_y = \frac{F}{m} = \frac{eV}{md}$.The time taken to travel an axial distance $L$ with constant horizontal velocity $v_0$ is $t = \frac{L}{v_0}$.The transverse velocity $v_y$ acquired after time $t$ is $v_y = a_y t = \left(\frac{eV}{md}ight) \left(\frac{L}{v_0}ight) = \frac{eVL}{mdv_0}$.The horizontal velocity remains constant at $v_x = v_0$.The angle $	heta$ that the beam makes with the plates is given by $	an 	heta = \frac{v_y}{v_x}$.Substituting the values,$	an 	heta = \frac{eVL / mdv_0}{v_0} = \frac{eVL}{mdv_0^2}$.Therefore,$	heta = 	an ^{-1}\left(\frac{eVL}{mdv_0^2}ight)$.

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