A stream of electrons travelling with speed \mathrm{vm} / \mathrm{s} at right angles to a uniform electric field E is deflected in a circular path of radius r . Prove that \frac{e}{m}=\frac{v^{2}}{r E} ?
A stream of electrons travelling with speed \mathrm{vm} / \mathrm{s} at right angles to a uniform electric field E is deflected in a circular path of radius r . Prove that \frac{e}{m}=\frac{v^{2}}{r E} ?

Ans: The path of the electron that is travelling with velocity \mathrm{vm} / \mathrm{s} at right angle of \bar{E} is of circular shape.
It requires a centripetal in nature, F=\frac{m v^{2}}{r}
It is provided by an electrostatic force F=\mathrm{eE}.

    \[\begin{array}{l} \Rightarrow \mathrm{eE}=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \\ \Rightarrow \frac{\mathrm{e}}{\mathrm{m}}=\frac{\mathrm{v}^{2}}{\mathrm{rE}} \end{array}\]