Question

A particle having charge that of an electron and mass $1.6 \times 10^{-30} kg$ is projected with an initial speed $u$ at an angle $45^{\circ}$ to the horizontal from the lower plate of a parallel plate capacitor as shown in figure. The plates are sufficiently long and have separation $2\, cm$. The maximum value of velocity of particle not to hit the upper plate is $\sqrt{W} \times 10^{6} m / s$. Take electric field between the plates $=10^{3} V / m$ directed upward. The value of $W$ is _____.

Answer 1

Resolving the velocity of particle parallel and perpendicular to the plate.

$u_{11}=u \cos 45^{\circ}=\frac{u}{\sqrt{2}} $
and $ u_{\perp}=u \sin 45^{\circ}=\frac{u}{\sqrt{2}}$
Force on the charged particle in downward direction normal to the plate $=e E$
$\therefore$ Acceleration $a=\frac{e E}{m}$, where $m$ is the mass of charged particle.
The particle will not hit the upper plate, if the velocity component normal to plate becomes zero before reaching if, i.e.,
$0=u_{\perp}^{2}-2 a y$, with $y \leq d$
where $d$, distance between the plates.
$\therefore$ Maximum velocity for the particle not to hit the upper plate
(for this $y=d=2 \,cm$ )
$u_{\perp}=\sqrt{2 a y}$
$=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 10^{3} \times 2 \times 10^{-2}}{1.6 \times 10^{-30}}}=2 \times 10^{6} m / s $
$u_{\max }=u_{\perp} / \cos 45^{\circ}=2 \sqrt{2} \times 10^{6} \,m / s$