Question

Consider the parallel-plate capacitor shown in the figure. The plate separation is 1.8 mm and the the electric field inside is 23 N/C. An electron is positioned halfway between the plates and is given some initial velocity, vi.
(a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate?

(b) If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case?

Answer 1

We have that the Velocities is mathematically given as

a)v=2.8e5m/s

b)v_2=3.11e5m/s

From the question we are told

• Consider the parallel-plate capacitor shown in the figure.
• The plate separation is 1.8 mm and the the electric field inside is 23 N/C.
• An electron is positioned halfway between the plates and is given some initial velocity, vi.

Velocity

Generally the equation for the workdone   is mathematically given as

a)

$-qEr=0.5m(u/2)^2\\\\Therefore\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31*v^2/4$

v=2.8e5m/s

b)

applying work

$-qEr=0.5m(u/2)^2\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31(v_2^2-7.75e10)$

v_2=3.11e5m/s

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