Question

A pair of charged conducting plates produces a uniform field of Eo = 10,859 N/C directed to the right, between the plates. The separation of the plates is L = 37 mm. In Figure, an electron (e = - 1.6 x 10-19 C; m = 9.1 x 10-31 kg) is projected from plate A, directly toward plate B, with an initial velocity of Vo = 2.1×107 m/s. The velocity of the electron (expressed in general form as a whole number) as it strikes plate B is what?

Answer 1

Force on charge placed in external electric field

$F = qE$

$F = 1.6*10^{-19} * 10859$

$F =17374.4* 10^{-19}$

Now by work energy theorem

Work done by all force = change in kinetic energy

$W = K_f - K_i$

$F.d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$17374.4* 10^{-19}*37*10^{-3} = \frac{1}{2}*9.1*10^{-31}*(v^2 - (2.1*10^7)^2)$

$v = 2.41 * 10^7 m/s$

so above is the speed of electron on reaching other plate

Answer 2

from Newton's law of motion

$v^2 - v_0^2 = 2as </p><p>acceleration is given by [tex]a = \frac{eE}{m}$

then

v^2 - v_0^2 = 2 \times (\frac{eE}{m}) \times L

putting in values

v^2 - (2.1 \times 10^7)^2 = 2 \times (\frac{1.6\times 10^{-19} \times 10859}{9.1 \times 10^{-31}}) \times 0.037[/tex]

we get

$v = 1.731 \times 10^7 m/s$