Question

Two large vertical and parallel metal plates having a separation of 1 cm 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

• A. $1\times {10}^{-5}V$
• B. $1\times {10}^{-7}V$
• C. $1\times {10}^{-9}V$
• D. $1\times {10}^{-10}V$

Answer 1

Answer: (C)

$1\times {10}^{-9}V$

As we know that F = qE also f = mg due to mass and gravitation

there fore $qE=mg$

$q\left(V/d\right)=mg$

$V=mgd/q$

now putting the values $V=\frac{1.67\times {10}^{-27}\times 10\times {10}^{-2}}{1.6\times {10}^{-19}}$

we get V = ${10}^{-9}$ v hence the option (c) is correct