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 $2cm$. 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

Answer: 8

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 $=eE$
$\therefore$ Acceleration $a=\frac{eE}{m}$, 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.,
$0=u_{\perp }^2-2ay$, with $y\le d$
where $d$, distance between the plates.
$\therefore$ Maximum velocity for the particle not to hit the upper plate
(for this $y=d=2cm$ )
$u_{\perp }=\sqrt{2ay}$
$=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 10^3\times 2\times 10^{-2}}{1.6\times 10^{-30}}}=2\times 10^{6}m/s$
$u_{\max }=u_{\perp }/\cos 45^{\circ }=2\sqrt{2}\times 10^{6}m/s$