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An electron enters a parallel plate capa...

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

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

A

`0.1`

B

`0.2`

C

`0.8`

D

`1.6`

Text Solution

Verified by Experts

The correct Answer is:
A

`V_(y)=ay.t`
`V_(y)=(eE)/(m).(l)/(u)`
`V_(x)=u`
`tan^(theta) =(eEl)/(m u^(2))`
`(tan theta_(1))/(tan theta_(2))= 4`
`tan theta_(2)=(tan theta_(1))/(4)=0.1 `
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