EG1216 · Physics 2 — Electricity & Magnetism

Theme 1 · Electrostatics — charge & the electric field

Motion of a charge in a uniform field

Drop a charge into the steady field between two plates and it accelerates — speeding up in a straight line, or curving into a parabola just like a thrown ball.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §23.7.

Before you start

What you need first

Topic 3 — force from a field — $F = qE$, and that a $+$ charge is pushed along $E$, a $-$ charge against it.
Topic 4 — the uniform field — between two oppositely charged parallel plates the field is the same everywhere.
Physics 1 — constant-acceleration motion — the equations $v = v_0 + at$, $v^2 = v_0^2 + 2a\,\Delta x$, and projectile (parabolic) motion.

What you'll be able to do

Get the acceleration of a charge in a uniform field: $a = qE/m$.
Find the speed a charge gains moving along the field.
Find the parabolic deflection of a charge fired across the field between two plates.
Find the exit deflection $y$ and the exit angle $\theta$.

The setting

A steady field between two plates

From Topic 4, the field between two oppositely charged parallel plates is uniform — it has the same size and the same direction at every point between the plates. It points from the $+$ plate to the $-$ plate.

Because $E$ is the same everywhere, any charge sitting between the plates feels the same force wherever it is. That single fact is what makes the motion easy.

Uniform downward electric field between a positive top plate and a negative bottom plate

A uniform field $E$: same size, same direction everywhere between the plates.

The key step

A constant force gives a constant acceleration

The force on the charge is $F = qE$, and $E$ is constant — so $F$ is constant. Newton's second law $F = ma$ then gives a constant acceleration:

$$ a = \frac{F}{m} = \frac{qE}{m} $$

where:
Symbol	Meaning	SI unit
a	acceleration of the charge	m/s²
q	the charge (use its size; the sign sets the direction)	C
E	the uniform field	N/C
m	mass of the charged particle	kg

Direction: a $+$ charge accelerates along $E$; a $-$ charge (like an electron) accelerates opposite $E$.

It's just constant-acceleration motion. Once you know $a$ is constant, every Physics-1 tool applies unchanged — $v=v_0+at$, $v^2=v_0^2+2a\,\Delta x$, and the projectile picture. Gravity is utterly negligible here: for an electron $a$ comes out near $10^{14}\ \text{m/s}^2$, about $10^{13}$ times $g$.

Case A — straight line

Moving along the field: a straight-line speed-up

If the charge starts from rest (or moves straight along $E$), it just speeds up in a straight line. Using $v^2 = v_0^2 + 2a\,\Delta x$ with $v_0=0$ and $a=qE/m$, after moving a distance $d$:

$$ v = \sqrt{\frac{2qEd}{m}} $$

where:
Symbol	Meaning	SI unit
v	speed after moving the distance $d$ (from rest)	m/s
d	distance travelled along the field	m
q, E, m	charge size, field, mass (as above)	C, N/C, kg

Same idea as work–energy: the field does work $qEd$ on the charge, which becomes kinetic energy $\tfrac12 mv^2$. Setting them equal gives the same result.

📐 Worked example 1

An electron speeding up

An electron is released from rest in a uniform field $E = 2.0\times10^{3}$ N/C. ($m_e = 9.11\times10^{-31}$ kg, $e = 1.60\times10^{-19}$ C.)

(a) Find its acceleration. (b) Find its speed after it has moved $d = 0.020$ m.

1Acceleration from $a = eE/m_e$:

$$ a = \frac{(1.60\times10^{-19})(2.0\times10^{3})}{9.11\times10^{-31}} \approx 3.5\times10^{14}\ \text{m/s}^2 $$

2Speed from rest after $d$, using $v=\sqrt{2ad}$:

$$ v = \sqrt{2(3.5\times10^{14})(0.020)} \approx 3.7\times10^{6}\ \text{m/s} $$

The electron is negative, so it accelerates opposite to $E$ (toward the $+$ plate). The acceleration dwarfs $g$, so we ignore gravity completely.

Case B — the parabola

Firing a charge across the field

Now send the charge into the gap moving across the field, with speed $v_0$ along the plates. This is exactly a projectile:

Along the plates there is no force, so that speed stays $v_0$: $x = v_0 t$.
Across the gap the force $qE$ gives constant acceleration, so the sideways drift grows as $y = \tfrac12 a t^2$.

Eliminating $t$ gives $y = \dfrac{a}{2}\left(\dfrac{x}{v_0}\right)^2$ — a parabola, just like a ball thrown horizontally under gravity.

A $+$ charge enters at $v_0$. The constant force $F=qE$ bends it into a parabola, deflecting it by $y$ over the plate length $L$ and sending it out at angle $\theta$. (An electron would curve the other way.)

1Time spent between the plates (set by the steady sideways speed $v_0$ and the plate length $L$):

$$ t = \frac{L}{v_0} $$

2The deflection at the far end, from $y=\tfrac12 a t^2$ with $a=qE/m$:

$$ y = \frac{1}{2}\,\frac{qE}{m}\left(\frac{L}{v_0}\right)^{2} $$

3The exit angle, from the sideways speed gained $v_y=at$ compared with $v_0$:

$$ \tan\theta = \frac{v_y}{v_0} = \frac{qEL}{m\,v_0^{2}} $$

where (new symbols):
Symbol	Meaning	SI unit
v₀	entry speed, along the plates (stays constant)	m/s
L	length of the plates (how far the charge travels across)	m
y	sideways deflection as the charge leaves the plates	m
θ	angle of the path to its original direction, at exit	—
v_y	sideways speed gained across the field, $v_y=at$	m/s

📐 Worked example 2

Deflecting a proton between plates

A proton enters along the centre line of two parallel plates at $v_0 = 1.0\times10^{6}$ m/s. The field between the plates is $E = 1.0\times10^{4}$ N/C and the plates are $L = 0.10$ m long. ($m_p = 1.67\times10^{-27}$ kg, $e = 1.60\times10^{-19}$ C.)

(a) Find the deflection $y$ as it leaves. (b) Find the exit angle $\theta$.

1Acceleration across the gap, $a = eE/m_p$:

$$ a = \frac{(1.60\times10^{-19})(1.0\times10^{4})}{1.67\times10^{-27}} \approx 9.6\times10^{11}\ \text{m/s}^2 $$

2Time between the plates, $t = L/v_0 = 0.10/(1.0\times10^{6}) = 1.0\times10^{-7}$ s. Then $y=\tfrac12 a t^2$:

$$ y = \tfrac12(9.6\times10^{11})(1.0\times10^{-7})^{2} \approx 4.8\times10^{-3}\ \text{m} \;(\approx 4.8\ \text{mm}) $$

3Exit angle from $\tan\theta = v_y/v_0$, with $v_y = at = (9.6\times10^{11})(1.0\times10^{-7}) = 9.6\times10^{4}$ m/s:

$$ \tan\theta = \frac{9.6\times10^{4}}{1.0\times10^{6}} = 0.096 \;\Rightarrow\; \theta \approx 5.5^\circ $$

The proton is positive, so it deflects toward the $-$ plate (along $E$) — exactly the path drawn above.

✏️ Try it yourself

An electron enters along the centre line of two plates at $v_0 = 1.0\times10^{7}$ m/s. The field is $E = 1.0\times10^{3}$ N/C and the plates are $L = 0.050$ m long. ($m_e = 9.11\times10^{-31}$ kg.)

(a) Find the acceleration. (b) Find the deflection $y$ at exit. (c) Which plate does it bend toward?

(a) $a = \dfrac{eE}{m_e} = \dfrac{(1.60\times10^{-19})(1.0\times10^{3})}{9.11\times10^{-31}} \approx 1.8\times10^{14}\ \text{m/s}^2.$ (b) Step 1. $t = L/v_0 = 0.050/(1.0\times10^{7}) = 5.0\times10^{-9}$ s. Step 2. $y = \tfrac12 a t^2 = \tfrac12(1.8\times10^{14})(5.0\times10^{-9})^2 \approx 2.2\times10^{-3}\ \text{m}\;(\approx 2.2\ \text{mm}).$ (c) The electron is negative, so the force is opposite $E$: it bends toward the $+$ plate — the opposite way to the proton above.

Common mistakes

Mistake	Fix
Including gravity.	For charged particles $a=qE/m$ is enormous; $g$ is negligible — leave it out.
Letting the sideways speed $v_0$ change inside the plates.	There is no force along the plates, so $v_0$ stays constant — only the across-field speed grows.
Using $d$ (along the field) and $L$ (across) interchangeably.	$L$ sets the time $t=L/v_0$; the deflection $y$ is the across-field distance.
Sending a $-$ charge the same way as a $+$ charge.	An electron's force is opposite $E$; it deflects toward the $+$ plate.

Recap — the whole topic on one screen

Idea	What you own now
Acceleration	$a = qE/m$, constant (force is constant in a uniform field).
Direction	$+$ along $E$; $-$ opposite $E$. Gravity ignored.
Along the field	Straight-line speed-up; from rest $v=\sqrt{2qEd/m}$ (work–energy).
Across the field	Parabola: $t=L/v_0$, $y=\tfrac12\dfrac{qE}{m}\left(\dfrac{L}{v_0}\right)^2$.
Exit angle	$\tan\theta = \dfrac{qEL}{m v_0^2}$.

Next topic

Electric flux & Gauss's law

So far we add up fields piece by piece. Next we count how much field "flows through" a surface — the flux — and meet a law that ties that flux directly to the charge enclosed, often replacing a hard integral with one line of algebra.

→ Topic 6 · Electric flux & Gauss's law

Mistake	Fix
Including gravity.	For charged particles \(a=qE/m\) is enormous; \(g\) is negligible — leave it out.
Letting the sideways speed \(v_0\) change inside the plates.	There is no force along the plates, so \(v_0\) stays constant — only the across-field speed grows.
Using \(d\) (along the field) and \(L\) (across) interchangeably.	\(L\) sets the time \(t=L/v_0\); the deflection \(y\) is the across-field distance.
Sending a \(-\) charge the same way as a \(+\) charge.	An electron's force is opposite \(E\); it deflects toward the \(+\) plate.

Symbol	Meaning	SI unit
v	speed after moving the distance \(d\) (from rest)	m/s
d	distance travelled along the field	m
q, E, m	charge size, field, mass (as above)	C, N/C, kg

Idea	What you own now
Acceleration	\(a = qE/m\), constant (force is constant in a uniform field).
Direction	\(+\) along \(E\); \(-\) opposite \(E\). Gravity ignored.
Along the field	Straight-line speed-up; from rest \(v=\sqrt{2qEd/m}\) (work–energy).
Across the field	Parabola: \(t=L/v_0\), \(y=\tfrac12\dfrac{qE}{m}\left(\dfrac{L}{v_0}\right)^2\).
Exit angle	\(\tan\theta = \dfrac{qEL}{m v_0^2}\).