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An electron enters a parallel plate capa...

An electron enters a parallel plate capacitor with horizontal speed u and is fond to deflect by angle `theta` on leaving the capacitor as shown. It is found that `tan theta = 0.4` and gravity is negligible
If the initial horizontal speed is doubled, then tan` theta` will be

A

`0.1`

B

`0.2`

C

`0.8`

D

`1.6`

Text Solution

Verified by Experts

The correct Answer is:
A
Horizontal displacement = l
Time `(t) =l/u`
Now, `v_y=u_y+at`
`implies v_y =0+(eE)/mxxl/u`
`v_y =(eEl)/("mu")`
`v_x` remain same `implies v_x =u`
`tan theta=v_y/v_x =(eEl)/("mu") xx1/u =(eEl)/("mu"^2)implies tan theta prop 1/u^2`
When speed u is doubled then `tantheta` will become `1/4th`
`tan theta = (0.4)/4 = 0.1`
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