A electric field of $1.5 \times 10^{4} N C^{- 1}$ exists between two parallel plates of length 2 cm. An electron enters the region between the plates at right angles to the field with a kinetic energy of $E_{k} = 2000 e V .$ The deflection that the electron experiences at the deflecting plates is:

0.34 mm

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0.57 mm

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7.5 mm

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0.75 mm

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Solution

The correct option is D 0.75 mm
Initial energy of $e = 2000 e v$
$\Rightarrow \frac{1}{2} m_{e} v^{2} = 2000 e v$
$\Rightarrow v = \sqrt{\frac{2 E_{k}}{m_{e}}}$
time taken to travel $2 c m = \frac{2 \times 10^{- 2} \sqrt{m_{e}}}{\sqrt{2 E_{k}}}$
Now displacement $S = \frac{1}{2} a E^{2}$
$= \frac{1}{2} (\frac{E υ}{m}) \frac{4 \times 10^{- 4} m}{2 E_{k}}$
$\frac{1}{2} \times \frac{1.5 \times 10^{4} \times e \times 4 \times 10^{- 4}}{2 \times 2000 \times e J}$
$S = 0.75 m m$

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A electric field of 1.5×104NC−1 exists between two parallel plates of length 2 cm. An electron enters the region between the plates at right angles to the field with a kinetic energy of Ek=2000eV. The deflection that the electron experiences at the deflecting plates is:

The correct option is D 0.75 mmInitial energy of e=2000 ev⇒12mev2=2000 ev⇒v=√2 Ekmetime taken to travel 2 cm=2×10−2√me√2EkNow displacement S=12a E2=12(Eυm)4×10−4m2Ek12×1.5×104×e×4×10−42×2000×e JS=0.75 mm

A electric field of $1.5 \times 10^{4} N C^{- 1}$ exists between two parallel plates of length 2 cm. An electron enters the region between the plates at right angles to the field with a kinetic energy of $E_{k} = 2000 e V .$ The deflection that the electron experiences at the deflecting plates is: