Q: The deflecting plates in Thomson’s setup are ‘ x ‘ meters long. Intensity of electric field applied between
the plates is E. the plates are maintained at a p.d of V volts. Electrons accelerated through a p.d of V
volts enter from one edge of the plated midway in a direction parallel to the plates. Find the deflection
at the other edge of the plates.
Sol: Motion of a charged particle projected perpendicular to a uniform electric field is a parabola. Horizontal
distance travelled by the electron in times t is
$\large x = v t \Rightarrow t = \frac{x}{v} $
Deflection of the electron in y – direction is
$\large y = 0 + \frac{1}{2} a t^2 $
$\large y = \frac{1}{2} (\frac{e E}{m}) (\frac{x}{v})^2 $
(since uy = 0)
$\large y = \frac{1}{4} (\frac{e E}{\frac{1}{2}m v^2})x^2 $ …(i)
We known that, $\large \frac{1}{2} m v^2 = e V $ …….(ii)
From (i) and (ii)
$\large y = \frac{E}{4 V} x^2 $