Question

A pair of charged conducting plates produces a uniform field of E_o = 10,838 N/C directed to the right, between the plates. The separation of the plates is L = 42mm. 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 v_o = 2

Answer 1

Force on charge q in an electric field E is

F = qE.

So.

Force on electron(F) = -1.6 * 10^-19 x 10,838 = -1.734 * 10^-15 N

The field is positive (to the right) so the force on the electron is negative (to the left).

F = ma,

a = F/m.

Acceleration of electron(a) = (-1.734 * 10^-15) / (9.1 x 10^-31) = -1.905 * 10^15 m/s^2

The electron has:

initial velocity, vo = 2 * 10^7 m/s

acceleration, a = - 1.905 * 10^15 m/s2

displacement s = 42 mm = 0.042 m

We know that,

Vf^2 = Vo^2 +2as

Vf^2 = (2 * 10^7)^2 + 2 * (-1.905 * 10^15 ) x 0.042

Vf^2 = 2.3998 * 10^14

Vf = sqrt(2.3998 * 10^14 )

Vf = 1.55 * 10^7 m/s