Electricity-And-Magnetism · Unit 4 · Video 5 · Interactive Practice

Electric Pressure and Tension: The Stress ε₀E²/2

IKey Formulas

Formula	Name	What it says
$\dfrac{\varepsilon_0 E^2}{2}$	Electric stress	Tension (pull along the outward normal) where $\vec E$ runs along $\pm\hat n$ ; pressure (push into the surface) where $\vec E$ lies along the face
$\dfrac{\text{C}^2}{\text{N}\cdot\text{m}^2}\cdot\dfrac{\text{N}^2}{\text{C}^2} = \dfrac{\text{N}}{\text{m}^2} = \dfrac{\text{J}}{\text{m}^3}$	Units of $\varepsilon_0 E^2$	Force per area, and also energy per volume
$F \sim \dfrac{\varepsilon_0 E_0^2}{2}\cdot\dfrac{q}{\varepsilon_0 E_0} = \dfrac{qE_0}{2}$	Sphere estimate	Stress times the area $4\pi r_0^2 = q/(\varepsilon_0 E_0)$ of the sphere through the null point, where $E_0 = \dfrac{q}{4\pi\varepsilon_0 r_0^2}$
$\dfrac{F}{A} = \dfrac{\varepsilon_0 E^2}{2} - 0$	Pull on a capacitor plate	Tension on the face in the gap, nothing on the face inside the plate

Key Insight: The same number $\varepsilon_0E^2/2$ is the energy density $u_E$ stored in the field (given here; derived later in the course) — the field's stress and its stored energy per cubic metre have the same value.

IIVisualization 1 — A Charge in a Growing Field

As $\vec E_{\text{ext}} = -E_0\sin^4(2\pi t/T)\,\hat k$ grows, the total field's lines crowd above $q$ and stretch taut below it.

t / T 0.250

IIIVisualization 2 — Sizing the Force from the Stress

On the sphere through the null point, every patch carries a stress $\varepsilon_0E^2/2$ pointing straight down.

Angle from the top θ 135°

IVVisualization 3 — Faraday's Box and the Capacitor Plate

Each face feels $\varepsilon_0E^2/2$ ; only a face where the field stops leaves a net force on the box.

Gap field E (10⁶ N/C) 1.0

💡 The same rule explains the pendulums: the lines of like charges bend away from each other and leave a gap between them, and the pressure the field transmits across that gap, perpendicular to the lines, pushes the bobs apart; unlike charges are joined by lines whose tension pulls the bobs together.

VQuiz Questions

Problem 1 · How Big Is the Stress?

Given: air breaks down at about $E = 3.0\times10^{6}\ \text{N}/\text{C}$ , and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ — find the stress $\varepsilon_0E^2/2$ the field transmits.

✅ Correct! About

40\ \text{N}/\text{m}^2

, which is also

40\ \text{J}/\text{m}^3

of stored field energy — tiny next to atmospheric pressure,

1.0\times10^5\ \text{N}/\text{m}^2

❌ Not quite. The field enters squared:

E^2 = 9.0\times10^{12}\ \text{N}^2/\text{C}^2

, not

3.0\times10^{6}

❌ Close, but… that is

\varepsilon_0E^2

. The stress carries a factor

\tfrac{1}{2}

❌ Not quite. That used

k = 1/(4\pi\varepsilon_0) = 8.99\times10^{9}

in place of

\varepsilon_0 = 8.85\times10^{-12}

Show solution

\frac{\varepsilon_0E^2}{2} = \frac{(8.85\times10^{-12})(3.0\times10^{6})^2}{2} = \frac{(8.85\times10^{-12})(9.0\times10^{12})}{2} = \frac{79.7}{2} \approx 40\ \text{N}/\text{m}^2

Units: $\dfrac{\text{C}^2}{\text{N}\cdot\text{m}^2}\cdot\dfrac{\text{N}^2}{\text{C}^2} = \dfrac{\text{N}}{\text{m}^2}$ , a pressure. Multiplying top and bottom by a metre gives $\text{N}\cdot\text{m}/\text{m}^3 = \text{J}/\text{m}^3$ , so the field also stores about $40\ \text{J}$ in every cubic metre.

Problem 2 · The Bottom Face of Faraday's Box

Given: an imaginary closed box sits in a uniform electric field $\vec E$ pointing up. The bottom face has outward normal $\hat n$ pointing down, so $\vec E$ enters the box there — what does the field transmit across the bottom face?

✅ Correct! A tension pulls along the outward normal whether

\vec E

points out of the surface or into it, so the bottom face is pulled down — and it exactly balances the upward pull on the top face.

❌ Not quite. A pressure acts where

\vec E

lies along the face. Here

\vec E

runs along

\pm\hat n

, so the stress is a tension: a pull outward.

❌ Not quite. The rule depends only on the angle between

\vec E

and

\hat n

, not on which way

\vec E

crosses:

\vec E

along

\pm\hat n

means tension.

❌ Close, but… the size is right, the direction is not. Tension pulls along the face's outward normal (down here), not along

\vec E

; an upward pull on both top and bottom would leave a net force in a uniform field.

Show solution

Faraday's rule names the stress by the angle between $\vec E$ and the face's outward normal:

$\vec E$ along $\pm\hat n$ → tension, a pull along the outward normal $\hat n$ ;
$\vec E \perp \hat n$ → pressure, a push along $-\hat n$ , into the box.

On the bottom face $\vec E$ is along $-\hat n$ , so it is a tension: a pull along $\hat n$ , i.e. downward, of size $\varepsilon_0E^2/2$ . The top face is pulled up by the same amount and the side pressures cancel in pairs, so the net force on the inside of the box is zero in a uniform field.

Problem 3 · The Sphere Estimate in Numbers

Given: a charge $q = 1.0\ \text{nC}$ is held at rest in a uniform downward external field $E_0 = 1.0\times10^{4}\ \text{N}/\text{C}$ ; $1/(4\pi\varepsilon_0) = 8.99\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ — find the distance $r_0$ of the null point and the stress-times-area estimate of the force.

How far above the charge is the null point?

What force does the estimate $F \sim (\varepsilon_0E_0^2/2)\cdot 4\pi r_0^2$ give?

✅ Correct!

r_0 \approx 3.0\ \text{cm}

and

F \sim qE_0/2 = 5.0\times10^{-6}\ \text{N}

— within a factor of two of the exact force

qE_0 = 1.0\times10^{-5}\ \text{N}

❌ Not quite.

8.99\times10^{9}\cdot 10^{-9}/10^{4} = 9.0\times10^{-4}\ \text{m}^2

r_0^2

; take its square root.

❌ Not quite. That is

\sqrt{q/(\varepsilon_0E_0)}

, the square root of the whole area

4\pi r_0^2

. Divide the area by

4\pi

before taking the root.

❌ Check the power of ten.

r_0^2 = 9.0\times10^{-4}\ \text{m}^2

, so

r_0 = 3.0\times10^{-2}\ \text{m}

❌ Close, but…

qE_0

is the exact force. The estimate multiplies one magnitude,

\varepsilon_0E_0^2/2

, by the whole area, and falls short by a factor of two.

❌ Not quite. That uses the bottom-point stress

2\varepsilon_0E_0^2

over the whole sphere. The estimate takes

|\vec E| \sim E_0

everywhere.

❌ Not quite.

4.4\times10^{-4}

is the stress in

\text{N}/\text{m}^2

; multiply by the area

q/(\varepsilon_0E_0) = 0.0113\ \text{m}^2

to get a force.

Show solution

Null point: the charge's upward field cancels the downward external field where

E_0 = \frac{q}{4\pi\varepsilon_0 r_0^2} \;\Rightarrow\; r_0^2 = \frac{(8.99\times10^{9})(1.0\times10^{-9})}{1.0\times10^{4}} = 9.0\times10^{-4}\ \text{m}^2,\quad r_0 \approx 3.0\ \text{cm}

Area: $A = 4\pi r_0^2 = \dfrac{q}{\varepsilon_0E_0} = \dfrac{1.0\times10^{-9}}{(8.85\times10^{-12})(1.0\times10^{4})} \approx 0.0113\ \text{m}^2$ .

Stress: $\dfrac{\varepsilon_0E_0^2}{2} = \dfrac{(8.85\times10^{-12})(1.0\times10^{8})}{2} \approx 4.4\times10^{-4}\ \text{N}/\text{m}^2$ .

F \sim \frac{\varepsilon_0E_0^2}{2}\cdot\frac{q}{\varepsilon_0E_0} = \frac{qE_0}{2} = \frac{(1.0\times10^{-9})(1.0\times10^{4})}{2} = 5.0\times10^{-6}\ \text{N}

The exact force is $qE_0 = 1.0\times10^{-5}\ \text{N}$ . The factor of two is the averaging: on the sphere $E^2$ runs from $0$ at the top to $4E_0^2$ at the bottom and averages $2E_0^2$ , so the average stress is $\varepsilon_0E_0^2$ — twice what the estimate assumed.

Problem 4 · Pull Between Capacitor Plates

Given: two large parallel plates carry surface charge densities $+\sigma$ and $-\sigma$ with $\sigma = 2.0\times10^{-6}\ \text{C}/\text{m}^2$ ; each plate has area $A = 0.010\ \text{m}^2$ . Each plate alone makes a field $\sigma/(2\varepsilon_0)$ on both sides — find the force the field exerts on one plate.

What is the size of the force?

Which way does it act on the lower ( $+$ ) plate?

✅ Correct!

F/A = \varepsilon_0E^2/2 = \sigma^2/(2\varepsilon_0) \approx 0.23\ \text{N}/\text{m}^2

, so

F \approx 2.3\times10^{-3}\ \text{N}

, pulling the plates together.

❌ Close, but… that is

\varepsilon_0E^2 A

. The stress is

\varepsilon_0E^2/2

❌ Not quite. That used one sheet's field

\sigma/(2\varepsilon_0)

in the stress. The stress needs the total field in the gap, where the two sheets add to

\sigma/\varepsilon_0

❌ Not quite.

0.23

is the force per area in

\text{N}/\text{m}^2

; multiply by

A = 0.010\ \text{m}^2

❌ Not quite. The face of the plate in the gap has

\vec E

along its normal — a tension, which pulls toward the other plate. Pressure acts across the lines, not along them.

❌ Not quite. A box around the plate has its top face in the gap and its bottom face inside the plate, where

E = 0

: the pulls do not balance.

Show solution

Total field: in the gap the two sheets' fields add, $E = \dfrac{\sigma}{2\varepsilon_0} + \dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0}$ ; outside the pair (and inside the conducting plate) they cancel.

Stress on a box around the plate's surface: top face in the gap, bottom face where $E = 0$ :

\frac{F}{A} = \frac{\varepsilon_0E^2}{2} - 0 = \frac{\varepsilon_0}{2}\cdot\frac{\sigma^2}{\varepsilon_0^2} = \frac{\sigma^2}{2\varepsilon_0} = \frac{(2.0\times10^{-6})^2}{2(8.85\times10^{-12})} \approx 0.226\ \text{N}/\text{m}^2

F \approx (0.226)(0.010) \approx 2.3\times10^{-3}\ \text{N}

The face in the gap is pulled along its outward normal, toward the other plate: the plates attract, as opposite charges should.

Check with Coulomb forces: the charge $\sigma A$ on one plate sits in the other plate's field $\sigma/(2\varepsilon_0)$ , so $F = \sigma A\cdot\dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma^2 A}{2\varepsilon_0}$ — the same answer, obtained from the field's shape alone.

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