Question

Suppose that the particle 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? $(q_e=1.6\times {10}^{-19}C,m_e=9.1\times {10}^{-31}kg.)$

Answer 1

electron

$charge=1.6\times {10}^{-19}C$

mass = $9.1\times {10}^{-31}kg$

and

$V_x=2\times {10}^{6}m/s$

$d=0.005m$

Let the electron strike the up upper plate at the end of plate 'L' When deflection is 'S' therefore

$S=\frac{qE{L}^2}{2m{v}_x^2}$

$L=\sqrt{\frac{2dm{v}_x^2}{qE}}$

=$\sqrt{\frac{2\times 0.005\times 9.1\times {10}^{-31}\times 2\times {10}^6\times 2\times {10}^6}{1.6\times {10}^{-10}\times 9.1\times {10}^2}}$

$=1.6\times {10}^{-2}m$

=1.6cm