Question

Electrons are accelerated from rest through a potential difference V_o and then pass through a region between two parallel metal plates, as shown above. The region between the plates can contain a uniform electric field E and a uniform magnetic field B. With only the electric field present, the electrons follow path 1. With only the magnetic field present, the electrons follow path 3. As drawn, the curved paths between the plates show the correct direction of deflection for each field, but not necessarily the correct path shape. With both fields present, the electrons pass undeflected along the straight path 2.

(a)

i. Which of the following describes the shape of the portion of path 1 between the plates?

____Circular ____Parabolic ____Hyperbolic ____Exponential

Justify your answer.

ii. What is the direction of the electric field?

____To the left ____To the top of the page ____Into the page ____To the right ____To the bottom of the page ____Out of the page

Justify your answer.

(b)

i. Which of the following describes the shape of the portion of path 3 between the plates?

____Circular ____Parabolic ____Hyperbolic ____Exponential Justify your answer.

ii. What is the direction of the magnetic field?

____To the left ____To the top of the page ____Into the page____To the right ____To the bottom of the page ____Out of the page

Justify your answer.

Between the plates the magnitude of the electric field is 3.4 x 10⁴ V/m, and that of the magnetic field is 2.0 x 10^–3 T .

(c) Calculate the speed of the electrons given that they are undeflected when both fields are present.

(d) Calculate the potential difference V_o required to accelerate the electrons to the speed determined in part (c).

Answer 1

(a) i) Parabolic. The electrons have constant speed to the right. The constant electric force provides a constant acceleration toward the top of the page. This is similar to a projectile under the influence of gravity, so the shape is parabolic.

ii) Down. To create path 1, the electric force must be toward the top of the page. The electron is negatively charged, so the field must point in the opposite direction to the electric force.

(b) i) Circular. The magnetic force is always perpendicular to the velocity of the electrons and has constant magnitude. Thus it acts as a centripetal force making the electrons follow a circular path.

ii) Into the page. To create path 3, the initial magnetic force must be toward the bottom of the page. With the initial velocity to the right, the right-hand rule gives a field pointing out of the page. But the electron is negatively charged, so the field must point in the opposite direction.

(c) F_e = F_b … Eq = qvB … v = E/B … v = (3.4x10^–4) / (2x10^–3) = 1.7x10⁷ m/s

(d) W = K … Vq = ½ mv² … V = mv²/2Q … V = (9.11x10^–31)(1.7x10⁷)² / 2(1.6x10^–19) = 823 V