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6 ELECTROMAGNETISM

P60 014

In an evacuated chamber, a pair of parallel metal plates are separated by 75.75´10^{-3 m} and have a potential difference of 1.222k V applied to them. A beam of electrons is fired with a velocity of 6.35´10⁶ m.s^-1 between the plates.

(a) Calculate the magnitude of the electric field strength between the plates.

(b) Calculate the magnitude of the electrostatic force acting on an electron between the plates.

(c) A magnetic field is applied between the plates, sufficient to cancel the force on the electron beam due to the electric field. Calculate the magnitude of the magnetic field required between the plates to cancel the deflection of the electron beam due to the electric field.

(d) Draw a diagram of the plates showing

· The charge on the plates.

· The electric field and the electric force on an electron.

· The magnetic field and the magnetic force on an electron.

· The trajectory of the electron passing through the plates with zero fields applied.

· The trajectory of the electron if the magnetic field only was set to zero.

· The trajectory of the electron if the electric field only was set to zero.

View solution below only after you have completed the answering the question.

Solution

d = 75.75´10^-3 m V = 1222 V q = e = 1.602´10^-19 C v = 6.35´10⁶ m.s^-1

(a) E = V / d = 1.61´10⁴ V.m^-1

(b) F_E = E q = 2.58´10^-15 N

(c) F_B = F_E = B q v B = F_E / q v = 2.54´10^-3 T

(d)

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Ian Cooper School of Physics University of Sydney