Question

A particle of mass $9\times {10}^{-31}$kg having a negative charge of $1.6\times {10}^{-19}$C is projected horizontally with a velocity of ${10}^{6}m{s}^{-1}$ into a region between two infinite horizontal parallel plates of metal. The distance between the plates is $d=03$cm, and the particle enters 0.1 cm below the top plate. The top and bottom plates are connected, respectively, to the positive and negative terminals of a 30 V battery. Find the components of the velocity of the particle just before it hits one of the plates

Answer 1

An electron is a negatively charged particle, so it will be attracted by the positive plate with force $F=eE$. Hence, acceleration of electron along y-axis will be

$a=\frac{F}{m}=\frac{eE}{m}=\frac{eV}{md}$ $\{asE=\frac{V}{d}\}$ (i)

So, from equation of motion, $v^2=u^2+2as$ along the x-axis,

$v_x=v_0={10}^6\left(m{s}^{-1}\right)\{as{a}_x=0\}$ (ii)

And along the y-axis, $v_y^2=2a{y}_0|\{asu=0ands=y_0\}$

Now, as $y_0=1cm$ (given) and is given by equation (i), the electron will hit the top plate with the velocity

$v_y=\sqrt{\frac{2y_{0}eV}{md}}=\sqrt{\frac{2\times 1.6\times {10}^{-19}\times 20\times 1\times {10}^{-3}}{9\times {10}^{-31}\times 3\times {10}^{-3}}}$

$=\left(4\sqrt{2}/3\right)\times {10}^6=1.885\times {10}^{6}m{s}^{-1}$ (iii)

So, the electron will hit the upper plate with the velocity

$v_x={10}^6\left(m{s}^{-1}\right)$ and $v_y=1.885\times {10}^{6}m{s}^{-1}$

Note

In this problem, time taken by the electron to hit the plate

$t=\sqrt{\frac{2y_0}{a}}=\sqrt{\frac{2y_{0}m}{eE}}=\sqrt{\frac{2y_{0}md}{eV}}$

$=\sqrt{\frac{2\times (2\times {10}^{-3})\times (9\times {10}^{-31})\times (3\times {10}^{-3})}{1.6\times {10}^{-19}\times 30}}$

$=\frac{3}{2\sqrt{2}}\times {10}^{-9}s=1.06\times {10}^{-6}s$

And in this time, electron will travel a horizontal distance

$s=v_{0}t={10}^6\times 1.06\times {10}^{-9}=1.06\times {10}^{-3}m$