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

Field Lines, Null Points and Uniform Fields

IKey Formulas

Formula	Name	What it says
$\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\,\hat{r}$	Field of a point charge	$\hat{r}$ points from $q$ to the field point; away from $q > 0$ , toward $q < 0$
$\vec{E} = \sum_i \vec{E}_i = \sum_i \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_i^2}\,\hat{r}_i$	Superposition	Each charge has its own $r_i$ and $\hat{r}_i$ ; the arrows add tip to tail
$\vec{a} = \dfrac{q\vec{E}}{m} = -\dfrac{qE_y}{m}\,\hat{j}$	Charge between plates	Uniform field $\vec{E} = -E_y\,\hat{j}$ , so the acceleration is constant
$v = \sqrt{\dfrac{2yqE_y}{m}}$ , $\ K = qE_y\,y$	Released from rest across a gap $y$	Kinetic energy = force × distance = work done by the field

Key Insight: Field lines begin on positive charges (or at infinity) and end on negative charges (or at infinity), their number at a charge is proportional to its size, and no line passes through a null point, where $\vec{E} = 0$ .

IIVisualization 1 — Three Fields Added Tip to Tail

The field at $P$ is the vector sum of three point-charge fields, so their magnitudes do not simply add.

IIIVisualization 2 — Field Lines of Two Charges

Symmetry, the near field, the far field and null points fix the pattern; line counts are proportional to charge.

Charges

IVVisualization 3 — A Charge Driven Across a Gap

Released from rest, a charge in the uniform field between the plates gains kinetic energy $K = qE_y\,y$ .

Particle Ey (N/C) 1000 Gap y (cm) 2.0 t / T 1.00

VQuiz Questions

Problem 1 · Field of a Negative Point Charge

Given: a point charge $q = -4.0\ \text{nC}$ at the origin, with $\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ — find the electric field at the point $0.20\ \text{m}$ due east of the charge.

✅ Correct!

|\vec{E}| = (8.99 \times 10^{9})(4.0 \times 10^{-9})/(0.20)^2 = 899\ \text{N/C}

, and for

q < 0

the field points toward the charge: west.

❌ Close, but… the magnitude is right, but

q

is negative, so

\vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{r}

points opposite to

\hat{r}

(east): toward the charge.

❌ Not quite. The direction is right, but the field falls off as

1/r^2

: divide by

(0.20)^2 = 0.040

, not by

0.20

❌ Not quite. Two slips: the field goes as

1/r^2

, not

1/r

, and a negative charge's field points toward it.

Show solution

The unit vector $\hat{r}$ points from the charge to the field point, so $\hat{r} = \hat{\imath}$ (east). With $q = -4.0 \times 10^{-9}\ \text{C}$ and $r = 0.20\ \text{m}$ :

\vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\,\hat{r} = (8.99 \times 10^{9})\,\frac{-4.0 \times 10^{-9}}{(0.20)^2}\,\hat{\imath} = -899\,\hat{\imath}\ \text{N/C}

The magnitude is $899\ \text{N/C}$ , and the minus sign makes the field point west, toward the negative charge.

Problem 2 · The Dipole's Midpoint

Given: charges $+2.0\ \text{nC}$ and $-2.0\ \text{nC}$ a distance $d = 0.10\ \text{m}$ apart, with $\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ — find the magnitude of the electric field at the midpoint between them.

✅ Correct! Both fields point from

+Q

toward

-Q

, so they add:

8k_eQ/d^2 = 1.44 \times 10^{4}\ \text{N/C}

❌ Not quite. The fields cancel at the midpoint of two equal charges. Here the field of

+Q

points away from it and the field of

-Q

points toward it, which is the same direction.

❌ Close, but…

k_eQ/(d/2)^2 = 7.19 \times 10^{3}\ \text{N/C}

is the field of one charge. The second charge's field points the same way and doubles it.

❌ Not quite. Each charge is

d/2 = 0.050\ \text{m}

from the midpoint, not

d

Show solution

Each charge is $d/2 = 0.050\ \text{m}$ from the midpoint, so each field has magnitude

E_1 = E_2 = k_e\frac{Q}{(d/2)^2} = (8.99 \times 10^{9})\,\frac{2.0 \times 10^{-9}}{(0.050)^2} = 7.19 \times 10^{3}\ \text{N/C}

The field of $+Q$ points away from $+Q$ , toward $-Q$ ; the field of $-Q$ points toward $-Q$ . Both point the same way, so

E = 2k_e\frac{Q}{(d/2)^2} = \frac{8k_eQ}{d^2} = 1.44 \times 10^{4}\ \text{N/C}

The dipole has no null point between its charges.

Problem 3 · Locate a Null Point

Given: $+4.0\ \text{nC}$ at $x = 0$ and $+1.0\ \text{nC}$ at $x = 0.30\ \text{m}$ — find the point between them where $\vec{E} = 0$ .

✅ Correct! At

x = 0.20\ \text{m}

4.0/(0.20)^2 = 100 = 1.0/(0.10)^2

, so the two equal and opposite fields cancel.

❌ Not quite. The midpoint is the null point only when the charges are equal. The larger charge pushes the null point toward the smaller one.

❌ Close, but… that balances

q/r

, not

q/r^2

. Each field falls off as the inverse square of the distance.

❌ Not quite.

x = 0.10\ \text{m}

0.10\ \text{m}

from the larger charge. The null point must be closer to the smaller charge:

0.10\ \text{m}

from the

+1.0\ \text{nC}

charge.

Show solution

Between two positive charges the fields point in opposite directions, so the null point is where their magnitudes are equal. At position $x$ :

k_e\frac{4.0\ \text{nC}}{x^2} = k_e\frac{1.0\ \text{nC}}{(0.30 - x)^2} \quad\Longrightarrow\quad \frac{x}{0.30 - x} = \sqrt{4} = 2

x = 2(0.30 - x) \quad\Longrightarrow\quad 3x = 0.60 \quad\Longrightarrow\quad x = 0.20\ \text{m}

Check: $4.0/(0.20)^2 = 100$ and $1.0/(0.10)^2 = 100$ . ✓ The null point is $0.20\ \text{m}$ from the $+4.0\ \text{nC}$ charge and $0.10\ \text{m}$ from the $+1.0\ \text{nC}$ charge.

Problem 4 · Proton Versus α Particle

Given: a proton (charge $e$ , mass $m_p$ ) and an α particle (charge $2e$ , mass $4m_p$ ) are each released from rest at the positive plate and cross the same gap $y$ in the same uniform field $E_y$ — find the ratios of their kinetic energies and speeds on arrival.

What is $K_\alpha / K_p$ ?

What is $v_\alpha / v_p$ ?

✅ Correct!

K = qE_y\,y

doubles with the charge, while

v = \sqrt{2yqE_y/m}

scales as

\sqrt{q/m} = \sqrt{2/4}

❌ Not quite. The mass cancels from

K = \tfrac{1}{2}mv^2 = qE_y\,y

: the kinetic energy is the work

qE_y\,y

, which depends only on the charge.

❌ Check the energy.

K = qE_y\,y

with the same

E_y

and

y

for both.

❌ Close, but…

q/m

halves, but the speed is the square root of

2yqE_y/m

❌ Not quite. The speed depends on

q/m

, and

q/m

2e/4m_p = \tfrac{1}{2}(e/m_p)

for the α particle.

❌ Check the speed. Use

v = \sqrt{2yqE_y/m}

and compare

q/m

for the two particles.

Show solution

Energy. The field does work $qE_y\,y$ on each particle, and the mass cancels:

K = \tfrac{1}{2}mv^2 = qE_y\,y \quad\Longrightarrow\quad \frac{K_\alpha}{K_p} = \frac{(2e)E_y\,y}{e\,E_y\,y} = 2

Speed. From $v^2 = 2|a_y|y$ with $|a_y| = qE_y/m$ :

\frac{v_\alpha}{v_p} = \sqrt{\frac{2y(2e)E_y/(4m_p)}{2y\,e\,E_y/m_p}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}} \approx 0.71

The α particle arrives with twice the energy but about $71\%$ of the speed: its larger mass more than offsets its larger charge.

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