Question

A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed $v_x$ (like particle 1 in Fig.). The length of plate is L and an uniform electric field E is maintained between the plates.

Suppose that the particle in Exercise is an electron projected with velocity $v_x=2.0\times {10}^{6}m{s}^{-1}$. If E between the plates separated by 0.5 cm is $9.1\times {10}^2$ N/C, where will the electron strike the upper plate? $(|e|=1.6\times {10}^{-19}C,m_e=9.1\times {10}^{-31}kg.)$

Answer 1

Velocity of the electron $v_x=2\times {10}^6$ m/s
Distance covered $d=0.005m$
Electric field between plates $E=9.1\times {10}^{2}N/C$
charge of electron is $q$ and the mass is $m$.

Acceleration of the electron $a=\frac{qE}{m}$

Time taken by the electron $t=L/v_x$
Using $d=ut+\frac{1}{2}a{t}^2$

where initial speed of the electron along y-axis $u=0m/s$

$\therefore$ $d=0+\frac{1}{2}\times \frac{qE}{m}(\frac{L}{v_x}{)}^2$

so, we get $d=\frac{qE{L}^2}{2m{v}_x^2}$

$⟹L=\sqrt{\frac{2dm{v}_x^2}{qE}}$

$=\sqrt{\frac{2\left(0.005\right)(9.1\times {10}^{-31})(2\times {10}^6{)}^2}{(1.6\times {10}^{-19})(9.1\times {10}^2)}}$ $=1.6\times {10}^{-2}m=1.6cm$