6. An electron moving with a constant horizontal velocity enters a region of uni

Question

6. An electron moving with a constant horizontal velocity enters a region of uniform electric field of 2 x 103 N/c directed upward between two parallel plates. a. If the E-field is in the direction stated, b. What is the magnitude and direction of the force on the electron c. What is the shape of the electron's path as it moves through the d. Suppose we set up a magnetic fi?ld in the region between the plates what must be the sign of the charge on each plate? due to the E-field? (0.32 fN, down) E-field? Hint: Is its acceleration uniform? so that we now have both E and B fields in the region through which he electron travels. We also find that if the magnitude of the -field is 0.2.T, the electron passes between the plates undeflected. What must be the direction of the F-field for this to occur? (out of the page) e. Under the conditions stated in (d), what is the speed of the electron? (10 km/s)

Explanation / Answer

(a)

Direction of electric field is from +ve to -ve charge.

so, lower plate must be of positive charge and upper plate must be of negative charge.

(b)

Magnitute of force on the electron,

F = qE

F = 1.6*10(-19) * 2*103

F = 0.32*10(-15) = 0.32 fN

Direction -

electron is negatively charged particle . so, direction of force is opposite of direction of electric field.

so, the direction is downward.

(c)

As the acceleration of electron is not uniform.

Shape of electron's path is parabolic as it moves through the E-field.

(d)

lf we also have B field in the region,

The direction of magnetic field is out of the page.

(e)

Let speed of electron = v

By balancing forces on electron,

Fe = Fm

qE = qvB

v = 2*103 / 0.2

v = 10*103 m/s

v = 10 km / s

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