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Question

A uniform electric field E is created between two parallel and oppositely charged plates as shown in figure. An electron enters the field symmetrically between the plates with a speed v0. The length of each plate is l. Find the angle of deviation (which is given by θ in the figure) of the path of the electron as it comes out of the field interms of mass of electron m and charge of the electron e. (Acceleration due to gravity can be neglected)


A
θ=tan1(eElmv20)
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B
θ=cot1(2eElmv20)
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C
θ=tan1(2eElmv20)
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D
θ=tan1(eEmv20)
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Solution

The correct option is A θ=tan1(eElmv20)
Here, the only force on the electron is due to the electric field. Hence, by using Newton's second law of motion.

F=eE=ma

a=eEm

So, the acceleration of the electron is a=eEm in the upward direction.

The horizontal velocity remains v0 as there is no acceleration in this direction. Thus, the time taken to cross the plates is

t=lv0.....(1)

The upward component of the velocity of the electron as it emerges from the field region is

vy=uy+at=0+eElmv0

The horizontal component of the velocity remains

vx=v0

The angle θ made by the resultant velocity with the original direction is given by

tanθ=vyvx=eElmv20

Thus, the electron deviates by an angle

θ=tan1eElmv20

Hence, option (b) is correct answer.

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