Electricity-And-Magnetism · Unit 5 · Video 3 · Interactive Practice

Electron Deflected by an Electric Field Between Two Plates

IKey Formulas

Formula	Name	What it says
$\vec{\mathbf{F}} = (-e)(-E_y\hat{\mathbf{j}}) = eE_y\hat{\mathbf{j}},\qquad a_y = \dfrac{eE_y}{m}$	Force and acceleration	Constant and upward, toward the positive plate; $v_x = v_0$ never changes
$t_1 = \dfrac{L_1}{v_0},\qquad y_1 = \tfrac{1}{2}\dfrac{eE_y}{m}\left(\dfrac{L_1}{v_0}\right)^2$	Transit through the plates	A parabola, like a ball thrown horizontally, turned upside down
$\tan\theta_1 = \dfrac{v_y}{v_x} = \dfrac{eE_yL_1}{mv_0^2}$	Exit angle	After the plates the path is a straight line at $\theta_1$
$y = \dfrac{eE_yL_1}{mv_0^2}\left(\tfrac{1}{2}L_1 + L_2\right) = \dfrac{eE_yL_1\left(\tfrac{1}{2}L_1 + L_2\right)}{2K}$	Deflection at the screen	Proportional to the field, inversely proportional to the kinetic energy

Key Insight: Since $y_1 = \tfrac{1}{2}L_1\tan\theta_1$ , the straight exit path traced backward crosses the original line at the middle of the plates: the electron lands as if it had turned through $\theta_1$ at $x = L_1/2$ .

IIVisualization 1 — Field Up, Energy Down

The deflection $y = eE_yL_1(\tfrac{1}{2}L_1 + L_2)/(2K)$ doubles with the field and halves when $K$ doubles.

E_y (×10⁴ N/C) 1.00 K (keV) 2.0

IIIVisualization 2 — Predict the Spot

With $L_1 = 4\ \text{cm}$ and $L_2 = 10\ \text{cm}$ , where on the screen does each beam land?

IVQuiz Questions

Problem 1 · The Electron's Acceleration

Given: Between the plates $\vec{\mathbf{E}} = -E_y\hat{\mathbf{j}}$ with $E_y = 2.0\times10^{3}\ \text{N/C}$ . Use $e = 1.6\times10^{-19}\ \text{C}$ , $m = 9.1\times10^{-31}\ \text{kg}$ . Find the electron's acceleration.

✅ Correct! Two minus signs cancel: the force on the electron points up, against the field.

❌ Not quite. The size is right, but a negative charge is pushed opposite to the field:

(-e)(-E_y\hat{\mathbf{j}}) = +eE_y\hat{\mathbf{j}}

❌ Not quite. That divides by the proton mass; the electron is about

1800

times lighter.

❌ Not quite.

eE_y = 3.2\times10^{-16}\ \text{N}

is the force; divide by

m

❌ Not quite. Use

\vec{\mathbf{a}} = q\vec{\mathbf{E}}/m

with

q = -e

Show solution

\vec{\mathbf{a}} = \frac{(-e)(-E_y\hat{\mathbf{j}})}{m} = \frac{(1.6\times10^{-19})(2.0\times10^{3})}{9.1\times10^{-31}}\,\hat{\mathbf{j}} = 3.5\times10^{14}\ \text{m/s}^2\ \hat{\mathbf{j}}.

Problem 2 · Deflection at the Screen

Given: The electron of Problem 1 enters with $v_0 = 2.0\times10^{7}\ \text{m/s}$ . The plates are $L_1 = 2.0\ \text{cm}$ long and the screen is $L_2 = 20\ \text{cm}$ beyond them. Find the total deflection $y$ at the screen.

✅ Correct!

\tan\theta_1 = 0.0176

times

\tfrac{1}{2}L_1 + L_2 = 21\ \text{cm}

❌ Not quite. The lever arm is

\tfrac{1}{2}L_1 + L_2

, not

L_1 + L_2

: the exit path points back to the middle of the plates.

❌ Not quite.

L_2\tan\theta_1

is only the rise after the plates; add

y_1 = \tfrac{1}{2}L_1\tan\theta_1

from inside them.

❌ Not quite. That is

y_1

, the rise inside the plates; the straight flight to the screen adds far more.

❌ Not quite. Find

\tan\theta_1 = eE_yL_1/(mv_0^2)

, then

y = \tan\theta_1(\tfrac{1}{2}L_1 + L_2)

Show solution

\tan\theta_1 = \frac{eE_yL_1}{mv_0^2} = \frac{(1.6\times10^{-19})(2.0\times10^{3})(0.020)}{(9.1\times10^{-31})(2.0\times10^{7})^2} = \frac{6.4\times10^{-18}}{3.64\times10^{-16}} = 0.0176.

y = \tan\theta_1\left(\tfrac{1}{2}L_1 + L_2\right) = 0.0176 \times 0.21\ \text{m} = 3.7\times10^{-3}\ \text{m} = 3.7\ \text{mm}.

Problem 3 · An Oscilloscope Beam from Its Energy

Given: An electron with kinetic energy $K = 2.0\ \text{keV} = 3.2\times10^{-16}\ \text{J}$ passes plates of length $L_1 = 3.0\ \text{cm}$ with $E_y = 1.0\times10^{4}\ \text{N/C}$ , and the screen is $L_2 = 15\ \text{cm}$ beyond them. Find the deflection at the screen.

✅ Correct!

y = eE_yL_1(\tfrac{1}{2}L_1 + L_2)/(2K) = 1.24\ \text{cm}

❌ Not quite.

mv_0^2 = 2K

, not

K

; dividing by

K

doubles the answer.

❌ Not quite. Use

\tfrac{1}{2}L_1 + L_2 = 16.5\ \text{cm}

, not

L_1 + L_2

❌ Not quite. Keep the

\tfrac{1}{2}L_1

: the deflection inside the plates counts too.

❌ Not quite. Replace

mv_0^2

2K

in the deflection formula.

Show solution

y = \frac{eE_yL_1\left(\tfrac{1}{2}L_1 + L_2\right)}{2K} = \frac{(1.6\times10^{-19})(1.0\times10^{4})(0.030)(0.165)}{2(3.2\times10^{-16})} = \frac{7.92\times10^{-18}}{6.4\times10^{-16}}\ \text{m}.

y = 1.24\times10^{-2}\ \text{m} \approx 1.2\ \text{cm}.

Problem 4 · A Proton with the Same Energy

Given: The electron is replaced by a proton with the same kinetic energy, entering the same plates on the same line. Find how its deflection at the screen compares with the electron's.

✅ Correct!

y = qE_yL_1(\tfrac{1}{2}L_1 + L_2)/(2K)

has no mass in it, and

q

changes sign.

❌ Not quite. The heavier proton is slower, so it spends longer in the plates; at equal

K

the mass cancels exactly.

❌ Not quite. The proton's charge is

+e

, so the force

e\vec{\mathbf{E}}

points down, along the field.

❌ Not quite. Write the deflection in terms of

K

and look for

m

Show solution

With $mv_0^2 = 2K$ , the deflection is $y = \dfrac{qE_yL_1\left(\tfrac{1}{2}L_1 + L_2\right)}{2K}$ (taking up as positive for $q = -e$ in a downward field).

The mass is gone, so equal $K$ and equal $|q|$ give an equal size. The proton's charge $+e$ is pushed along the downward field, so it is deflected down by the same distance.

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