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

Faraday's Field Lines: How Electric Fields Carry Force, Energy and Momentum

IKey Formulas

FormulaNameWhat it tells you
Fβƒ—=14πΡ0q1q2r2 r^\vec{F} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}\,\hat{r}Coulomb's lawThe size of the force the field stress carries
E⃗=E⃗q+E⃗0\vec{E} = \vec{E}_q + \vec{E}_0Total field (superposition)The field Faraday reads every force from
q4πΡ0r02=E0β€…β€Šβ‡’β€…β€Šr0=q4πΡ0E0\dfrac{q}{4\pi\varepsilon_0 r_0^2} = E_0 \;\Rightarrow\; r_0 = \sqrt{\dfrac{q}{4\pi\varepsilon_0 E_0}}Cancellation radiusWhere the charge's own field cancels the background (E=0E = 0 at the top of the sphere)
Fβƒ—e=qEβƒ—0=βˆ’qE0 k^\vec{F}_e = q\vec{E}_0 = -qE_0\,\hat{k}Force on qq in Eβƒ—0=βˆ’E0 k^\vec{E}_0 = -E_0\,\hat{k}The net result of tension below and pressure above

Key Insight: Field lines carry a tension along their length and a pressure across one another. The field stores the charge's energy and gives it back, but momentum only passes through it, to the charges that produce the field.

IIVisualization 1 β€” Pressure Apart, Tension Together

Between like charges no line crosses the gap; between unlike charges lines join the spheres and pull.

IIIVisualization 2 β€” The Force Shows Only in the Sum

A charge at rest in a downward field: tension below and pressure above both push it down.

πŸ’‘ How large the tension and pressure are, per unit area, is the next video's subject; here only their directions matter.

IVVisualization 3 β€” Energy Banked, Momentum Passed On

As the charge climbs and falls back, the field banks its kinetic energy but hands its momentum to the source charges.

VQuiz Questions

Problem 1 Β· What Pulls Unlike Charges Together?

Given: The small sphere now carries charge of the opposite sign to the van de Graaff dome. Find: in Faraday's picture, which stress pulls the small sphere toward the dome?

βœ… Correct! For unlike charges lines run from one sphere to the other, and like stretched rubber bands their tension pulls the spheres together.
❌ That is the like-charge mechanism. Pressure across the lines pushes like charges apart; with unlike charges lines join the spheres and their tension does the pulling.
❌ Not quite. A single charge's own field is perfectly symmetric, so it points to no direction of force. The force shows only in the total field.
❌ Not quite. Faraday's picture rejects action at a distance: each sphere touches only the field filling the gap, and the stress in that field carries the force at the speed of light.
Show solution

Draw the total field of both spheres. For opposite signs, some lines leave the dome and end on the small sphere. A field line carries a tension along its length and pulls on whatever it ends on, so these connecting lines pull the small sphere toward the dome (and the dome toward the small sphere).

For like signs, no line runs from one sphere to the other; instead, the lines lying side by side in the gap press sideways on each other and push the spheres apart. Either way, Coulomb's law gives the size of the force, and the field stress is how it gets across.

Problem 2 Β· The Sphere Where the Fields Cancel

Given: A charge q=4.0Β nCq = 4.0\ \text{nC} is momentarily at rest in a uniform field Eβƒ—0=βˆ’E0 k^\vec{E}_0 = -E_0\,\hat{k} with E0=1.0Γ—104Β N/CE_0 = 1.0 \times 10^{4}\ \text{N/C}. Use 14πΡ0=9.0Γ—109Β Nβ‹…m2/C2\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2. Find the radius r0r_0 of the sphere at whose top the total field is zero, and the total field at the bottom of that sphere.

What is r0r_0?

Total field at the bottom of the sphere?

βœ… Correct! r0=6.0Β cmr_0 = 6.0\ \text{cm}; below the charge both fields point down and add to 2E02E_0, which is why the lines through the lower half are packed closer.
❌ Missing the square root. q4πΡ0E0=3.6Γ—10βˆ’3Β m2\frac{q}{4\pi\varepsilon_0 E_0} = 3.6 \times 10^{-3}\ \text{m}^2 is r02r_0^2, not r0r_0.
❌ Stray factor of 2. 2q/(4πΡ0E0)\sqrt{2q/(4\pi\varepsilon_0 E_0)} is where the own field is only E0/2E_0/2; cancellation needs q4πΡ0r02=E0\frac{q}{4\pi\varepsilon_0 r_0^2} = E_0.
❌ Stray factor of Β½. The own field must equal the full background E0E_0 at the top: q4πΡ0r02=E0\frac{q}{4\pi\varepsilon_0 r_0^2} = E_0.
❌ Check r0r_0. Solve q4πΡ0r02=E0\frac{q}{4\pi\varepsilon_0 r_0^2} = E_0 for r0r_0.
❌ That is the top. At the top the charge's own field points up and cancels the background; at the bottom it points down.
❌ Only the background. At distance r0r_0 the charge's own field also has size E0E_0, and below the charge it points down too, so the two add.
❌ Check the direction. Below a positive charge its own field points away from it β€” downward β€” and the background points down as well.
❌ Check the field. Add the charge's own field and the background at the bottom point.
Show solution

At the top of the sphere the charge's own field points up (+k^+\hat{k}) and the background points down. They cancel when

q4πΡ0r02=E0β‡’r0=(9.0Γ—109)(4.0Γ—10βˆ’9)1.0Γ—104=3.6Γ—10βˆ’3Β m2=0.060Β m.\frac{q}{4\pi\varepsilon_0 r_0^2} = E_0 \quad\Rightarrow\quad r_0 = \sqrt{\frac{(9.0\times10^{9})(4.0\times10^{-9})}{1.0\times10^{4}}} = \sqrt{3.6\times10^{-3}\ \text{m}^2} = 0.060\ \text{m}.

At the bottom of the sphere the own field has the same size E0E_0 but points down (βˆ’k^-\hat{k}), so by superposition

Eβƒ—=βˆ’E0 k^βˆ’E0 k^=βˆ’2E0 k^=βˆ’2.0Γ—104Β N/CΒ k^.\vec{E} = -E_0\,\hat{k} - E_0\,\hat{k} = -2E_0\,\hat{k} = -2.0\times10^{4}\ \text{N/C}\ \hat{k}.

The lines part at the empty top point and drape over the upper flanks, and run straight down, packed closer, through the lower half.

Problem 3 Β· How High, and How Much Energy Banked?

Given: A charge q=3.0Β ΞΌCq = 3.0\ \mu\text{C} of mass m=2.0Β gm = 2.0\ \text{g} moves up the zz axis at v0=3.0Β m/sv_0 = 3.0\ \text{m/s} through Eβƒ—0=βˆ’E0 k^\vec{E}_0 = -E_0\,\hat{k} with E0=2.0Γ—104Β N/CE_0 = 2.0\times10^{4}\ \text{N/C} (gravity neglected). Find how far it rises before stopping, and how much energy the field has taken in at that moment.

Height risen?

Energy stored in the field when the charge is at rest?

βœ… Correct! It rises 0.15Β m0.15\ \text{m}, and all 9.0Β mJ9.0\ \text{mJ} of kinetic energy now sits in the field, ready to be handed back on the way down.
❌ Missing the Β½. From v2=v02βˆ’2ahv^2 = v_0^2 - 2ah with v=0v = 0: h=v02/(2a)h = v_0^2/(2a), not v02/av_0^2/a.
❌ That is a time, not a height. v0/a=0.10 sv_0/a = 0.10\ \text{s} is how long the climb takes; the height is v02/(2a)v_0^2/(2a).
❌ Check the factor. The deceleration is a=qE0/m=30 m/s2a = qE_0/m = 30\ \text{m/s}^2, and h=v02/(2a)h = v_0^2/(2a).
❌ Check the height. Use a=qE0/ma = qE_0/m and h=v02/(2a)h = v_0^2/(2a).
❌ The energy has not vanished. Kinetic energy plus field energy stays constant, so the kinetic energy the charge lost is now stored in the field.
❌ Missing the ½. Kinetic energy is 12mv02\tfrac12 m v_0^2, not mv02m v_0^2.
❌ Too small. At rest the charge has no kinetic energy left, so the field holds all of 12mv02\tfrac12 m v_0^2, not half of it.
❌ Check the energy. Kinetic energy plus field energy is constant.
Show solution

The force is constant: Fe=qE0=(3.0Γ—10βˆ’6)(2.0Γ—104)=0.060Β NF_e = qE_0 = (3.0\times10^{-6})(2.0\times10^{4}) = 0.060\ \text{N}, downward, so the charge decelerates at

a=qE0m=0.0602.0Γ—10βˆ’3=30Β m/s2,h=v022a=9.060=0.15Β m.a = \frac{qE_0}{m} = \frac{0.060}{2.0\times10^{-3}} = 30\ \text{m/s}^2, \qquad h = \frac{v_0^2}{2a} = \frac{9.0}{60} = 0.15\ \text{m}.

Kinetic energy plus field energy stays constant, so at rest the field holds all the kinetic energy the charge started with:

12mv02=12(2.0Γ—10βˆ’3)(3.0)2=9.0Γ—10βˆ’3Β J=qE0h=(0.060)(0.15)Β J.\tfrac12 m v_0^2 = \tfrac12 (2.0\times10^{-3})(3.0)^2 = 9.0\times10^{-3}\ \text{J} = qE_0 h = (0.060)(0.15)\ \text{J}.

Problem 4 Β· Where Does the Momentum Go?

Given: The charge of Problem 3 (m=2.0Β gm = 2.0\ \text{g}, launched up at 3.0Β m/s3.0\ \text{m/s}) returns to its starting height moving down at 3.0Β m/s3.0\ \text{m/s}. Find the momentum passed to the charges that produce the background field over the whole trip.

βœ… Correct! The charge's momentum changes by βˆ’2mv0 k^-2mv_0\,\hat{k}, and the source charges take up the equal and opposite +2mv0 k^+2mv_0\,\hat{k}; the field holds none.
❌ Energy and momentum take different routes. The field stores energy and returns it, but no momentum is held in the field: it passes through to the source charges.
❌ Check the sign. That is the charge's own change. The stress that pushes down on the charge pushes up on the source charges: Ξ”pβƒ—sources=βˆ’Ξ”pβƒ—charge\Delta\vec{p}_{\text{sources}} = -\Delta\vec{p}_{\text{charge}}.
❌ Only half the change. The momentum reverses, from +mv0+mv_0 to βˆ’mv0-mv_0, so its change is 2mv02mv_0, not mv0mv_0.
Show solution

The charge's momentum goes from +mv0 k^+mv_0\,\hat{k} to βˆ’mv0 k^-mv_0\,\hat{k}:

Ξ”pβƒ—charge=βˆ’2mv0 k^=βˆ’2(2.0Γ—10βˆ’3)(3.0) k^=βˆ’1.2Γ—10βˆ’2Β kgβ‹…m/sΒ k^.\Delta\vec{p}_{\text{charge}} = -2mv_0\,\hat{k} = -2(2.0\times10^{-3})(3.0)\,\hat{k} = -1.2\times10^{-2}\ \text{kg}\cdot\text{m/s}\ \hat{k}.

No momentum is held in the field, so the source charges take up the opposite change, Ξ”pβƒ—sources=βˆ’Ξ”pβƒ—charge=+1.2Γ—10βˆ’2Β kgβ‹…m/sΒ k^\Delta\vec{p}_{\text{sources}} = -\Delta\vec{p}_{\text{charge}} = +1.2\times10^{-2}\ \text{kg}\cdot\text{m/s}\ \hat{k}.

Check: the trip lasts 2v0/a=0.20Β s2v_0/a = 0.20\ \text{s} under the constant 0.060Β N0.060\ \text{N} force, an impulse of (0.060)(0.20)=1.2Γ—10βˆ’2Β Nβ‹…s(0.060)(0.20) = 1.2\times10^{-2}\ \text{N}\cdot\text{s}.

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