Question 2. The X-ray tube shown at right is used to accelerate electrons which

Question

Question 2. The X-ray tube shown at right is used to accelerate electrons which hit a target and produce X-rays. The device consists of two parallel, metal plates of very large area separated by a gap of length d. The X-rays plate on the left (''cathode'') is connected to ground, while the plate on the right (?anode?) is set to a potential of V. Electrons arc produced (at rest) at the cathode plate. They accelerate across the gap, and strike the target area where they slow down rapidly and produce X-rays. (a) Calculate the final speed y achieved by electrons of mass m and charge e. (b) When the tube is operating, the surface charge densities on the cathode and anode are equal and opposite. What is their magnitude, sigma? (e) If the charge densities on the cathode and anode are equal and opposite as above, outside the tube the electric field will point i. to the right on the right side of the tube, and to the left on the left side of the tube. ii. to the left on the right side of the tube, and to the right on the left side of the tube. iii. the field will be zero everywhere to the left of the cathode and to the right of the anode.

Explanation / Answer

By conservation of energy,

(Ui - Uf) = Kf - Ki

Here,

Ui - Uf = eV

Also, as Ki = 0

Kf - Ki = Kf = 1/2m v^2

Thus,

1/2 m v^2 = eV

Solving for v,

v = sqrt(2eV/m)   [ANSWER, PART A]

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Note that

V = Ed

However, for a parallel plate capacitor like this,

E = sigma/Eo

where Eo = 8.85E-12 C^2/N m^2

Thus,

V = (sigma/Eo)d

Solving for sigma,

sigma = Vd/Eo    [ANSWER, PART B]

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PART C: Choice iii is correct: It will be zero everywhere outside, as the fields cancel.

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