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A uniform electric field of strength E e...

A uniform electric field of strength E exists in region. A electron enters a point A with velocity v as shown. It moves through the electric field and reaches at point B. Velocity particle at B is `2v` at `30^@` with x-axis .
Then

A

electric field `E=(3mv^2)/(2ea) hati`

B

rate of work done by electric field at B is `(3mv^2)/(2ea)`

C

Bota a and b are correct

D

both a and b are wrong

Text Solution

Verified by Experts

The correct Answer is:
A
`(v_A)y=vimplies(v_B)_y=2vsin30^@v`
Since y-component of velocity remains unchanged. Hence electric field is along `(-hati)` direction. Work done by electrostatic force in moving from A and B =change in its kinetic energy
`:. (eE)(2a-a)1/2m(4v^2-v^2)`
`E=(mv^2)/(2ea)`
`or E=-(3mv^2)/(2ea)hati`
Rate of doing work done =power
`=Fvcostheta`
`((3mv^2)/(2ea))^2(e)(2v)cos30^@`
`=3sqrt3/2(mv^2)/a`
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