An electron is accelerated (from rest) through a potential difference of 20,000 V...

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a)

To find the magnitude of the electric field we apply the relationship between the field and the potential for a parallel plates capacitor:

∆V = E*d

∆V is the potential difference between the plates.

E is the intensity of the electric field.

d is the distance between the plates.

E = ∆V/d = 20000/0.04

E = 500000 V/m

As the electron moves from the left plate to the right, the direction of the field is from the right plate to the left.

b)

To find the speed we apply the conservation of energy. The kinetic energy of the electron when it enter into the magnetic field is equal to the work done by the electric field:

∆V*q= (mv^2)/2

q is the charge of the electron.

m is the mass of the electron.

v is the speed of the electron when it enters in the magnetic field.

v^2 = 2(∆V*q)/m

v = sqrt [2(∆V*q)/m] = sqrt [2(2*10^4*1.602*10^-19)/9.109*10^-31]

v = 8.3*10^7 m/s

c)

To find the magnetic force on the electron we apply the following equation:

Fm = qvB sen θ

B is the induction of the magnetic field.

θ is the angle between the speed and the field.

As the velocity is perpendicular to the magnetic field, we have:

Fm = q*v*B = (1.602*10^-19)(8.3*10^7)(2)

Fm = 2.66*10^-11 N

In accordance with the right-hand rule, we discover that the force is directed down because the charge is negative.

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