Q. 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
IIT JEEIIT JEE 2012Electrostatic Potential and Capacitance
Solution:
Net force is at $45^0 $ from vertical.
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, qE=mg $
or $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac {qX}{d}=mg \, \, \, \, \, \bigg (\because E= \frac {X}{d}\bigg ) $
or $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, X= \frac {mgd}{q} $
$ = \frac {(1.67 \times10^{-27})(9.8)(10^{-2})}{(1.6 \times10^{-19})} $
$ \approx 1 \times 10^{-9}V $
