A uniform electric field E = (8m/e) V/m is created between two parallel plate of length 1m as shown in figure, (where m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s. The angle of the deviation ( θ) of the path of the electron as it comes out of the field will be________.

Option 1 -

tan^-1(4)

Option 2 -

tan^-1(2)

Option 3 -

tan^-1 ^{$(\frac{1}{3})$}

Option 4 -

$t a n^{- 1} (3)$

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$E = \frac{8 m}{e} v / m$

Time to come out, t = $\frac{1}{2} s e c$

$V_{y} = a t = (\frac{e E}{m}) t = (\frac{e}{m} * \frac{8 m}{e}) * \frac{1}{2} = 4 m / s$

So, $t a n ? = \frac{V_{y}}{V_{x}} = \frac{4}{2}$

$? ? = t a n^{? 1} 2$

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