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A problem of practical interest is to make a beam of electrons turn at $90^o$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 × 10^{-17}\ J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is $y × 10^5\ N/C$ . The value of $y$ is
$4$
$8$
$10$
$1$
Solution
Range $=\mathrm{R}=\frac{\mathrm{u}^{2}}{\mathrm{a}}=\frac{\mathrm{u}^{2} \mathrm{m}}{\mathrm{qE}}$ for $\theta=45^{\circ}$
$\mathrm{E}=\frac{\mathrm{u}^{2} \mathrm{m}}{\mathrm{qR}}$
$\frac{1}{2} \mathrm{mu}^{2}=\mathrm{KE}$
$\mathrm{E}=\frac{\mathrm{m}}{\mathrm{q} \mathrm{R}} \frac{2 \mathrm{KE}}{\mathrm{m}}=\frac{2(\mathrm{KE})}{\mathrm{qR}}$
