Discussion Forum

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 $45^{0}$ to the vertical just after release. Then X is nearly

Answer:

Explanation:

net force  is at $45^{0}$  from vertical

$\therefore$   qE=mg

or   $\frac{qX}{d}=mg$      ($\because E=\frac{X}{d}$)

 or      $X= \frac{mgd}{q}$

  =  $\frac{(1.67 \times 10^{-27})(9.8)(10^{-2})}{(1.6 \times 10^{-19})}$

    =$1 \times 10^{-9} V$