KVPY SX SB Exam 20141 Question Paper

Question : 33 of 120

Marks: +1, -0

An electron enters a parallel plate capacitor with horizontal speed u and is found to deflect by angle θ on leaving the capacitor as shown. It is found that tanθ = 0.4 and gravity is negligible

If the initial horizontal speed is doubled, then tanθ will be

Validate

Solution:

Horizontal displacement = l

t=l/u

v_y = u_y + at

=0+{eE}/m×l/u

vy ={eE}/m×l/u

v_x

remain same and it is equal to u

\tan θ =v_y/v_x={eE}/m \ l/u×1/u={eEl}/{mu^2}

\tan θ ∝1/u^2

When speed u is doubled then tan θ will become

1/4

th.

∴ \tan θ = 0.4/4=0.1

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