A uniform electric field of $20 N / C$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5 e V$ .The length of each plate is 5 cm. The angle $θ$ of deviation of the path of electron as it comes out of the field is……. (in degree).

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answer is 45.

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Detailed Solution

(45) kinetic energy $= \frac{1}{2} m v_{x}^{2}$
$0.5 eV = \frac{1}{2} m v_{x}^{2} \Rightarrow v_{x} = \sqrt{\frac{e V}{m}}$
Along $x, L = v_{x} t = \sqrt{\frac{e V}{m}} t$
Along $y, v_{y} = \frac{eE}{m} t$
Divide equation (ii) by (i)
$\begin{array}{r} \frac{v_{y}}{L} = E \sqrt{\frac{e V}{m}} = E v_{x} \\ \Rightarrow \tan θ = \frac{v_{y}}{v_{x}} = E \times L = 20 \times 0.05 = 1 \\ θ = 45^{\circ} \end{array}$

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