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A cathode ray tube contains a pair of parallel metal plates $1.0\, cm$ apart and $3.0\, cm$ long. A narrow horizontal beam of electron with a velocity $3 \times 10^7\, m/s$ passed down the tube midway between the plates. When a potential difference of $550\, V$ is maintained across the plates, it is found that the electron beam is so deflected that it just strikes the end of one of the plates. Then the specific charge of the electron in $C/kg$ is
$1.8 \times 10^9$
$1.8 \times 10^{10}$
$1.8 \times 10^{11}$
$1.8 \times 10^{12}$
Solution
Let be the time taken by the electron to move from $A$ and $B$. For the motion along $X$ – axis.
$l = {{\rm{v}}_{\rm{x}}}{\rm{t}} \Rightarrow {\rm{t}} = \frac{l}{{\rm{v}}}$
or $\quad \mathrm{t}=\frac{3 \times 10^{-2} \mathrm{m}}{3 \times 10^{7} \mathrm{m} / \mathrm{s}}=10^{-9} \mathrm{s}$
The force on the electron along $+y$ direction
$\mathrm{F}_{y}=\mathrm{eE}=\frac{\mathrm{eV}}{\mathrm{d}}$
where $\mathrm{V}=550$ volts and $\mathrm{d}=10^{-2} \mathrm{m}$
The acceleration along ty divection is
$a_{y}=\frac{F_{y}}{m}=\frac{e V}{m d}$
For the motion along $+y$ -axis.
${y=0+\frac{1}{2} a_{y} t^{2}}$
or ${\frac{d}{2}=\frac{1}{2}\left(\frac{e V}{m d}\right) t^{2}}$
or $\frac{{\rm{e}}}{{\rm{m}}} = \frac{{{{\rm{d}}^2}}}{{{\rm{V}}{{\rm{t}}^2}}}$
$ = \frac{{{{10}^{ – 4}}}}{{550 \times {{10}^{ – 18}}}} = 1.8 \times {10^{11}}{\rm{C}}/{\rm{kg}}$
