Two large vertical and parallel metal plates having a separation of 1cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at `45^@` to the vertical JUST after release. Then X is nearly

Options

A. `1xx10^-5V`
B. `1xx10^-7V`
C. `1xx10^-9V`
D. `1xx10^(-10)V`

Answer

C

Explanation

X=potential difference applied between two large vertical plates held distance `d=1cm=10^-2m` apart.
Electric field intensity in a region between the two plates, `E=(pot. di ff.)/(distance)=(X)/(10^-2)V//m`
When proton is released at O, the force on it due to electric field `=qE` acting along the direction of field, i.e., along OA.

Weight of proton=mg acting vertically downwards, i.e., along OB.
Since proton moves at `45^@` to the vertical just after the released, so `tan 45^@=qE//mg`
or `qE=mg` or `q(X)/(d)=mg`
or `X=(mgd)/(q)=((1*6xx10^(-27))xx10xx10^-2)/(1*6xx10^(-19))`
`=1xx10^-9V`