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Question

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{

Vedclass · Accepted Answer

$1.8 	imes 10^{11}$

Vedclass · Answer

Let be the time taken by the electron to move from $A$ and $B$. For the motion along $X$ - axis.

$l = {{m{v}}_{m{x}}}{m{t}} \Rightarrow {m{t}} = \frac{l}{{m{v}}}$

or $\quad \mathrm{t}=\frac{3 	imes 10^{-2} \mathrm{m}}{3 	imes 10^{7} \mathrm{m} / \mathrm{s}}=10^{-9} \mathrm{s}$

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 ${\frac{d}{2}=\frac{1}{2}\left(\frac{e V}{m d}ight) t^{2}}$

or $\frac{{m{e}}}{{m{m}}} = \frac{{{{m{d}}^2}}}{{{m{V}}{{m{t}}^2}}}$

$ = \frac{{{{10}^{ - 4}}}}{{550 	imes {{10}^{ - 18}}}} = 1.8 	imes {10^{11}}{m{C}}/{m{kg}}$

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