Unit 2

Electric Force, Field, and Potential

The whole unit is one move made twice: take a quantity that belongs to a pair of charges and divide out the test charge. Force becomes field. Potential energy becomes potential. Once a charge’s surroundings are described by E and V, you can predict what any charge placed there will do — without ever looking at the charges that made the field.

0 · Foundations: one pattern, four quantities

Every equation in the unit sits in a two-by-two grid. Learn the grid and the formula sheet stops looking like a list of unrelated facts.

Charge and the inverse-square force

Charge comes in whole multiples of e = 1.60×10⁻¹⁹ C and is conserved in every process. In solids it is the electrons that move; a positive object is one that has lost some.

Field and potential: divide out the test charge

The uniform field: where everything gets linear

Between two large parallel plates the field is the same everywhere, which turns the whole unit into kinematics and bookkeeping:

1 · Electric charge & charging

Ordinary matter is electrically neutral because it carries exactly as many protons as electrons. Every charged object you will meet in this unit got that way by moving electrons around — never by creating or destroying charge. Keep asking “where did the electrons go?” and this whole topic is bookkeeping.

What charge is

Charge is a property of particles, like mass, that determines how strongly they push and pull on each other electrically. Unlike mass it comes in two kinds, which Benjamin Franklin labelled positive and negative:

• Like charges repel, unlike charges attract. That single rule is all the direction information you will ever need.
• Protons carry +e, electrons carry −e, neutrons carry none.
• The SI unit is the coulomb (C). One coulomb is enormous: the charges in this unit are usually microcoulombs (μC, 10⁻⁶ C) or nanocoulombs (nC, 10⁻⁹ C).

“Net charge” means the imbalance. A negatively charged balloon still contains roughly 10²⁴ protons; it just has a few billion extra electrons on top. That imbalance is a fantastically small fraction of the total, which is why charged objects do not weigh noticeably more.

Conservation of charge

The net charge of an isolated system never changes. Charge can be transferred between objects, and particle pairs can be created or annihilated (an electron and a positron together carry zero net charge), but the total is fixed.

Sharing charge between identical conductors

When two identical conducting spheres touch, electrons move until the two are at the same potential. By symmetry that means each ends up with half the total:

If the spheres are different sizes they do not split evenly: the larger sphere takes the larger share. AP problems almost always say “identical” when they want you to halve.

Conductors, insulators & grounding

What separates the two kinds of material is whether some electrons are free to wander through the whole object.

Grounding

The Earth is a conductor so large that it can give or absorb practically any number of electrons without its own charge changing noticeably. Connecting an object to it (“grounding”) gives the object’s charges a path to escape or arrive. A grounded conductor with no other charges nearby ends up neutral; a grounded conductor next to a charged object does not — which is the whole trick behind charging by induction.

Three ways to charge an object

1. Friction (triboelectric charging)

Rub two different materials together and electrons transfer to whichever material holds them more tightly. Rubber rubbed with fur becomes negative (it gains electrons; the fur becomes positive). Glass rubbed with silk becomes positive. The two objects end with equal and opposite charges — conservation in action.

2. Conduction (contact)

Touch a charged object to a neutral conductor and some of the excess charge flows across. The neutral object ends with the same sign as the charged one, and the total is shared.

3. Induction (no contact)

Bring a charged rod near a conductor, ground the conductor while the rod is still there, break the ground, then remove the rod. The conductor ends with the opposite sign to the rod, and the rod loses nothing.

Polarization: why neutral things get attracted

A charged balloon sticks to a neutral wall. A charged comb picks up neutral bits of paper. Neither target has any net charge, so why is there a force?

Because the nearby charge rearranges the neutral object. In a conductor, free electrons migrate across the whole object. In an insulator, each molecule’s electron cloud shifts a tiny distance, turning every molecule into a small dipole. Either way, the side facing the rod acquires charge of the opposite sign, and the far side the same sign.

The electroscope

An electroscope is a metal knob connected by a metal stem to two light metal leaves, all insulated from its case. Because the whole thing is one conductor, any charge on it spreads to the leaves, which repel each other and spread apart. The angle shows how much charge is on the leaves — not its sign.

Using a charged electroscope to identify a sign

Charge the electroscope with a known sign first. Then bring the mystery object near the knob:

• Leaves spread further ⇒ more like charge was pushed into the leaves ⇒ the object has the same sign as the electroscope.
• Leaves fall together ⇒ like charge was pulled up toward the knob ⇒ the object is opposite in sign or simply neutral (a neutral conductor polarizes and has a weaker version of the same effect).

As with attraction, only the “spread further” result is conclusive.

Worked examples

1. Counting the extra electrons

A balloon rubbed on hair acquires a charge of −3.2 nC. How many excess electrons does it carry, and by how much did its mass increase?

n = |q|e = 3.2×10⁻⁹1.60×10⁻¹⁹ = 2.0×10¹⁰ electrons

Δm = nm_e = (2.0×10¹⁰)(9.11×10⁻³¹) = 1.8×10⁻²⁰ kg Twenty billion electrons, and the mass change is still unmeasurable. Electric effects are huge compared with the amount of matter moved.

The hair, by conservation, is left at +3.2 nC.

2. Touching identical spheres in sequence

Identical conducting spheres A (+6.0 μC) and B (−2.0 μC) are touched together and separated. B is then touched to an identical neutral sphere C and separated. Find the final charges, and the number of electrons that moved in the first contact (and which way).

First contact. Total = +6.0 + (−2.0) = +4.0 μC, shared equally:

q_A = q_B = +2.0 μC

Second contact. B (+2.0 μC) and C (0) share +2.0 μC:

q_B = q_C = +1.0 μC,    q_A = +2.0 μC Check: 2.0 + 1.0 + 1.0 = +4.0 μC, the same total we started with. Charge conserved.

Electrons in the first contact. B went from −2.0 μC to +2.0 μC, a change of +4.0 μC. Only electrons move, so B lost electrons and A gained them:

n = 4.0×10⁻⁶1.60×10⁻¹⁹ = 2.5×10¹³ electrons, from B to A

3. Induction, step by step

A positive rod is held near (not touching) a neutral metal sphere on an insulating stand. The far side of the sphere is touched with a grounded wire, the wire is removed, and then the rod is taken away. What is the final charge on the sphere? Explain in terms of electron motion.

1. The positive rod attracts free electrons to the near side of the sphere, leaving the far side positive.
2. Grounding gives electrons a new path: they are attracted up from the Earth into the sphere by the rod.
3. Removing the ground wire traps those extra electrons on the sphere.
4. Removing the rod lets the extra electrons spread over the whole surface.

Final charge: negative — opposite to the rod, as induction always gives. Note that it is not “positive charge flowing to the ground”; electrons flow in from the ground.

4. Reading an electroscope

An electroscope is charged positively, so its leaves are spread. Three objects are brought near the knob in turn, without touching. Object X makes the leaves spread further; Y makes them partly collapse; Z makes them partly collapse. What can you conclude about each?

• X is positive. A positive object attracts free electrons up from the leaves toward the knob, leaving the leaves even more positive, so they spread further. Neither a negative nor a neutral object can do that.
• Y and Z: negative or neutral — can’t tell which. A negative object pushes electrons down into the leaves, partly cancelling their positive charge. A neutral conductor polarizes, and its near side (negative) has a weaker version of the same effect.

To distinguish Y from Z, bring each near a negatively charged electroscope as well: a negative object will make those leaves spread further; a neutral one will not.

5. Which charges can exist?

Which of these could be the net charge on an oil drop: (a) 4.8×10⁻¹⁹ C, (b) 8.0×10⁻²⁰ C, (c) −6.4×10⁻¹⁹ C, (d) 2.4×10⁻¹⁹ C?

Divide each by e = 1.60×10⁻¹⁹ C and look for an integer:

Charge	q/e	Possible?
(a) 4.8×10−19 C	3	Yes — missing 3 electrons
(b) 8.0×10−20 C	0.5	No
(c) −6.4×10−19 C	−4	Yes — 4 extra electrons
(d) 2.4×10−19 C	1.5	No

This is exactly the logic of Millikan’s oil-drop experiment: every measured charge was a whole-number multiple of one smallest value, which is how e was discovered.

Checklist

• I can explain any charged object in terms of electrons gained or lost — never protons moving in a solid.
• I can test whether a charge is possible by checking q/e is an integer.
• I share charge between identical spheres by averaging, signs included, and check that the total is conserved.
• I know conduction gives the same sign and induction the opposite sign, and I can narrate the four induction steps with electrons.
• I can explain why a charged object attracts a neutral one, and why only repulsion proves an object is charged.
• I can predict what an electroscope’s leaves do, and know which outcome is conclusive.

2 · Coulomb’s law & superposition

Coulomb’s law tells you how hard two charges push or pull on each other. It looks exactly like Newton’s law of gravitation, and the algebra is the easy part. The marks are in two places: getting the direction from a sketch, and adding forces as vectors when more than two charges are involved.

The law

Three facts about it, each one tested:

• It is a magnitude. Use absolute values. The direction is along the line joining the charges: away from each other if the signs are alike, toward each other if they differ.
• It is inverse-square. The force falls off as 1/r², the same geometry as gravity and light intensity — the influence spreads over a sphere of area 4πr².
• It applies to point charges, and to uniformly charged spheres measured from their centers (as long as you are outside them).

Third-law pairs & proportional reasoning

The force of charge 1 on charge 2 and the force of charge 2 on charge 1 are a Newton’s-third-law pair: equal in magnitude, opposite in direction, always. The formula even shows it — the product q₁q₂ is symmetric. A +3 μC charge does not push harder on a +1 μC charge than the +1 μC pushes back.

What can differ is the response. If the two charges have different masses, a = F/m gives them different accelerations.

Most multiple-choice questions never need a calculator. Write the new force as a ratio of the old one and cancel everything that did not change:

Electric vs. gravitational force

The two laws have the same shape, F ∝ (product)/r², and that is where the resemblance ends.

How lopsided is it? In a hydrogen atom the electron and proton are 5.29×10⁻¹¹ m apart:

So for charged particles — electrons, protons, ions — ignore gravity unless a problem tells you otherwise. Gravity dominates the everyday world only because ordinary matter is so precisely neutral that the enormous electric forces cancel.

Superposition in one dimension

With three or more charges, the force on any one of them is the vector sum of the forces from each of the others, each computed as if the rest were not there:

On a line, “vector sum” just means assigning each force a sign by its direction (+ for right, − for left) and adding. The procedure:

1. Draw the charges and pick a positive direction.
2. For each other charge, compute the magnitude with Coulomb’s law.
3. Decide each direction from attract/repel, and attach the sign.
4. Add.

Superposition in two dimensions

When the charges are not on one line, the forces point in different directions and you must add them by components. Never add magnitudes of forces that are not parallel.

1. Find each force’s magnitude from Coulomb’s law.
2. Find each force’s direction from the geometry — usually a right triangle, so you can read the cosine and sine as side ratios.
3. Split each force into x and y components, with signs.
4. Add the x-components; add the y-components.
5. Recombine: |F| = √(F_x² + F_y²), θ = tan⁻¹(|F_y|/|F_x|), then state the quadrant in words.

Where is the net force zero?

A classic question: two fixed charges sit on a line; where could a third charge be placed so that it feels no net force? The answer depends only on the two fixed charges — the third charge’s sign and size cancel out, because it appears in both forces.

For the opposite-sign case, try +9.0 nC at x = 0 and −4.0 nC at x = 0.20 m. The balance point is beyond the −4.0 nC charge, at distance d from the origin: 9/d² = 4/(d − 0.20)² ⇒ 3/d = 2/(d − 0.20) ⇒ d = 0.60 m. Check: 9/0.36 = 25 and 4/0.16 = 25. ✓

Worked examples

1. Magnitude and direction for two charges

A +3.0 μC charge and a −5.0 μC charge are 0.20 m apart. Find the force each exerts on the other.

F = (8.99×10⁹)(3.0×10⁻⁶)(5.0×10⁻⁶)(0.20)² = 0.1350.040 = 3.4 N Absolute values only — the − sign on the second charge tells us attractive, nothing more.

Each charge is pulled toward the other with 3.4 N. Same magnitude on both, even though the charges differ (third law).

2. Touch, separate, compare

Two identical small conducting spheres carry +6.0 μC and −2.0 μC and attract with force F when their centers are 0.30 m apart. They are touched together and returned to 0.30 m. Find the new force in terms of F, and its direction.

From charge sharing, each ends with (6.0 − 2.0)/2 = +2.0 μC. Distance is unchanged, so only the charge product matters:

F′F = (2.0)(2.0)(6.0)(2.0) = 412  ⇒  F′ = F/3, now repulsive

Numerically: F = (8.99×10⁹)(12×10⁻¹²)/(0.30)² = 1.2 N before, and F′ = (8.99×10⁹)(4.0×10⁻¹²)/(0.30)² = 0.40 N after. The direction flipped because both spheres now carry the same sign.

3. Superposition on a line

q₁ = +2.0 μC sits at x = 0 and q₂ = −3.0 μC at x = 0.30 m. Find the net force on q₃ = +1.0 μC at x = 0.10 m.

From q₁ (0.10 m away, both positive ⇒ repel ⇒ pushed in +x):

F₁₃ = (8.99×10⁹)(2.0×10⁻⁶)(1.0×10⁻⁶)(0.10)² = 1.80 N,  +x

From q₂ (0.20 m away, opposite signs ⇒ attract ⇒ pulled toward q₂, i.e. +x):

F₂₃ = (8.99×10⁹)(3.0×10⁻⁶)(1.0×10⁻⁶)(0.20)² = 0.674 N,  +x

F_net = 1.80 + 0.674 = 2.47 N in the +x direction The trap: students see one positive and one negative source charge and subtract. Here both forces point right — pushed by one, pulled by the other.

4. Superposition in two dimensions (components)

q₀ = +2.0 μC is at the origin. q₁ = +3.0 μC is at (0, 0.30 m) and q₂ = −8.0 μC is at (0.40 m, 0.30 m). Find the net force on q₀. (This is the figure above.)

Distances. r₁ = 0.30 m. r₂ = √(0.40² + 0.30²) = 0.50 m — a 3-4-5 triangle.

Magnitudes.

F₁ = (8.99×10⁹)(3.0×10⁻⁶)(2.0×10⁻⁶)(0.30)² = 0.05390.090 = 0.599 N

F₂ = (8.99×10⁹)(8.0×10⁻⁶)(2.0×10⁻⁶)(0.50)² = 0.1440.25 = 0.575 N

Directions. q₁ repels q₀, pushing it straight away from q₁: the −y direction. q₂ attracts q₀ toward (0.40, 0.30), so along the diagonal with cosθ = 0.40/0.50 = 0.80 and sinθ = 0.30/0.50 = 0.60.

Force	Fx (N)	Fy (N)
F1 (repulsion, down)	0	−0.599
F2 (attraction, up-right)	0.575 × 0.80 = +0.460	0.575 × 0.60 = +0.345
Net	+0.460	−0.254

|F_net| = √(0.460² + 0.254²) = √0.276 = 0.53 N

θ = tan⁻¹(0.2540.460) = 29° below the +x-axis Fourth quadrant: positive x, negative y. Always say which quadrant in words — a bare angle from tan⁻¹ is ambiguous.

Sanity check: the two individual forces are about the same size, but one points straight down and the other mostly right and partly up. The y-parts partly cancel, the x-part survives, so the net force should be mostly rightward, a bit downward, and smaller than the sum 0.599 + 0.575. It is.

5. Three charges on a line in equilibrium

q₁ = +4.0 μC is fixed at x = 0 and q₂ = +1.0 μC at x = 0.60 m. (a) Where can a third charge q₃ be placed so the net force on it is zero? (b) If q₁ and q₂ are released too, what must q₃ be so that all three charges feel zero net force?

(a) Like charges, so the point is between them (see the table). Let q₃ be at x. Setting the two magnitudes equal, k and q₃ cancel:

4.0x² = 1.0(0.60 − x)²  ⇒  2x = 10.60 − x  ⇒  x = 0.40 m Taking the positive square root keeps us between the charges. True for any q₃, positive or negative.

(b) Now look at q₁. q₂ repels it to the left, so q₃ must pull it to the right: q₃ must be negative. Balance on q₁:

k(4.0)|q₃|(0.40)² = k(4.0)(1.0)(0.60)²  ⇒  |q₃| = (1.0)0.160.36 = 0.44 μC  ⇒  q₃ = −0.44 μC (exactly −⁴⁄₉ μC)

Check on q₂. Push from q₁: (8.99×10⁹)(4.0×10⁻⁶)(1.0×10⁻⁶)/(0.60)² = 0.0999 N to the right. Pull from q₃: (8.99×10⁹)(0.444×10⁻⁶)(1.0×10⁻⁶)/(0.20)² = 0.0999 N to the left. Balanced ✓ Once two of the three are balanced, the third must be too: the internal forces of the system sum to zero by the third law.

This equilibrium is unstable: nudge the negative q₃ toward q₁ and its attraction to q₁ grows while its attraction to q₂ shrinks, so it keeps going.

Checklist

• I compute Coulomb force magnitudes with absolute values and take directions from a sketch.
• I know the force pair between two charges is equal and opposite even when the charges differ.
• I can answer “double this, triple that” questions as a ratio without a calculator.
• I add forces on a line with signs from direction, and in 2-D by components — never by adding magnitudes.
• I state a 2-D answer as a magnitude plus an angle and a quadrant in words.
• I can find the zero-force point for two charges and say whether it lies between or outside them.

3 · The electric field

Coulomb’s law tells you the force between two charges you already have. The field idea splits that in half: a charge first sets up a condition in space, and any other charge that wanders in simply responds to the condition where it sits. Once you know the field, you never need to know what made it.

What a field is

Put a small test charge q₀ at some point and measure the electric force on it. Double the test charge and the force doubles; flip its sign and the force flips. The ratio, though, stays put — it belongs to the point in space, not to the test charge. That ratio is the electric field:

The second form is the one you use. It says the force on any charge is its charge times the field at its location. Two consequences follow straight from the sign of q:

• A positive charge feels a force along E⃗.
• A negative charge feels a force opposite to E⃗ — same field, same line, reversed arrow.

The field of a point charge

Divide Coulomb’s law by the test charge and what is left is the field of the source charge q alone:

Superposition — fields add as vectors

With several source charges, each produces its own field as if the others were not there, and the net field is the vector sum:

The procedure never changes:

1. Draw the point and every source. Mark each distance r_i.
2. For each source, compute the magnitude k|q_i|/r_i² and draw its arrow at the point (away from +, toward −).
3. Resolve each arrow into x and y components, using geometry (often a 3-4-5 triangle) for the sines and cosines.
4. Sum the components; recombine with Pythagoras and an inverse tangent.
5. Symmetry check first — components that cancel by symmetry need no arithmetic at all.

Field lines

Field lines are a picture of the field, not a physical thing. They are drawn so that the field’s direction and strength can be read off at a glance, and they obey a short list of rules that exam questions test directly:

1. Start on positive charges, end on negative charges (or at infinity).
2. The field at any point is tangent to the line through it, pointing along the arrow.
3. The number of lines leaving or entering a charge is proportional to its magnitude: a −2q charge gets twice the lines of a +q.
4. Line density shows strength. Crowded lines, strong field; sparse lines, weak field.
5. Lines never cross. At a crossing the field would have two directions at one point, which is meaningless.

The uniform field between parallel plates

Two large, flat, parallel conducting plates carrying equal and opposite charges produce the most useful field in the unit: the same magnitude and direction at every point between them, pointing from the + plate to the − plate, and (ideally) zero outside.

Two expressions for the strength, which you will meet from different directions:

The remarkable thing about E = σ/ε₀ is what is missing: the distance from the plate. Unlike a point charge, a large sheet of charge does not get weaker as you back away from it, because as you move away more and more of the sheet “comes into view” at a shallow angle, exactly compensating. That is why the field is uniform.

Charges moving in a uniform field

In a uniform field the force qE is constant, so the acceleration is constant — and every constant-acceleration tool from AP Physics 1 comes back:

Entering parallel to the field (or released from rest): straight-line motion with constant acceleration — speed it up or slow it down, like a ball thrown straight up.

Entering perpendicular to the field: this is a projectile. The velocity component perpendicular to E⃗ never changes, and the component along E⃗ grows steadily. The path between the plates is a parabola; after leaving the plates the particle moves in a straight line.

Worked examples

1. Field of a point charge, then the force on something placed there

A +4.0 μC charge is fixed in place. (a) Find the electric field 30 cm away. (b) A −2.0 nC charge is placed at that point. Find the force on it.

E = k|q|r² = (8.99×10⁹)(4.0×10⁻⁶)(0.30)² = 4.0×10⁵ N/C, directed away from the charge

F = |q|E = (2.0×10⁻⁹)(4.0×10⁵) = 8.0×10⁻⁴ N, toward the +4.0 μC charge Negative charge ⇒ force opposite to E⃗. The field pointed away from the source, so the force points back toward it — attraction, as it must be for opposite charges.

If the −2.0 nC charge were swapped for −4.0 nC, the force would double and the field would not change at all.

2. Two-dimensional superposition

A +6.0 nC charge sits at the origin and a −8.0 nC charge at (0.40 m, 0). Find the net electric field at P = (0, 0.30 m). (This is the figure in the superposition section.)

Distances. P is 0.30 m from the + charge and √(0.30² + 0.40²) = 0.50 m from the − charge.

E₁ = (8.99×10⁹)(6.0×10⁻⁹)(0.30)² = 599 N/C, straight up (+y)

E₂ = (8.99×10⁹)(8.0×10⁻⁹)(0.50)² = 288 N/C, from P toward (0.40, 0) The direction from P to the negative charge is (+0.40, −0.30)/0.50 = (+0.80, −0.60). A 3-4-5 triangle hands you the cosines.

Field	Ex (N/C)	Ey (N/C)
E1	0	+599
E2	+288(0.80) = +230	−288(0.60) = −173
Net	+230	+427

E_net = √(230² + 427²) = 485 N/C, at tan⁻¹(427/230) = 61.7° above +x

Sanity check against the figure: the net arrow leans right of vertical, toward the negative charge, and is shorter than E₁ because E₂ partly opposes it. ✓

3. Where is the field zero?

A charge +q is at x = 0 and a charge +4q at x = 0.90 m. Where on the x-axis is the net field zero?

Which region? Outside the pair, both fields point the same way (both away from two positives), so they cannot cancel. Between them they point in opposite directions — the null point is between, and closer to the smaller charge.

kqx² = k(4q)(0.90 − x)²  ⇒  0.90 − x = 2x  ⇒  x = 0.30 m Take the positive square root of both sides — the region was already chosen, so the other root is not needed. Check: 0.30 m from q, 0.60 m from 4q; four times the charge at twice the distance gives the same field. ✓

Opposite charges

For a +q and a −4q, the fields point the same way between the charges, so the zero must lie outside, on the side of the smaller-magnitude charge. Pick the region first; the algebra is easy after that.

4. An electron accelerated between plates

An electron is released from rest at the negative plate of a pair of parallel plates 2.0 cm apart, where the field is 2.5×10⁴ N/C. Find its acceleration, its speed on reaching the positive plate, and the time it takes.

F = eE = (1.60×10⁻¹⁹)(2.5×10⁴) = 4.0×10⁻¹⁵ N

a = Fm_e = 4.0×10⁻¹⁵9.11×10⁻³¹ = 4.4×10¹⁵ m/s², toward the + plate

v = √(2ad) = √(2 × 4.39×10¹⁵ × 0.020) = 1.3×10⁷ m/s About 4% of the speed of light, reached in t = v/a = 3.0×10⁻⁹ s. Classical mechanics is still fine at this speed.

Gravity: m_eg = 8.9×10⁻³⁰ N, about 10⁻¹⁵ of the electric force. Ignoring it was safe. The potential section (4) gets the same speed in one line using energy instead of kinematics.

5. Projectile-like deflection

An electron enters the region between two horizontal plates moving horizontally at v₀ = 3.0×10⁷ m/s, midway between them. The plates are 5.0 cm long and the field between them is 2.0×10⁴ N/C, pointing down. Find the vertical deflection at the exit, and the angle at which the electron leaves.

Direction first. E⃗ is down and the electron is negative, so the force — and the deflection — is up.

Time between the plates (horizontal motion, no horizontal force):

t = Lv₀ = 0.0503.0×10⁷ = 1.67×10⁻⁹ s

Vertical acceleration:

a = eEm_e = (1.60×10⁻¹⁹)(2.0×10⁴)9.11×10⁻³¹ = 3.51×10¹⁵ m/s²

Deflection and exit velocity:

y = ½at² = ½(3.51×10¹⁵)(1.67×10⁻⁹)² = 4.9×10⁻³ m ≈ 4.9 mm upward

v_y = at = 5.85×10⁶ m/s  ⇒  θ = tan⁻¹(5.85×10⁶3.0×10⁷) = 11° above the horizontal For the electron to get out at all, the plate separation must be more than 2 × 4.9 mm ≈ 1 cm, since it started midway.

Scaling check (the kind of reasoning the free-response section loves): y = qEL²/(2mv₀²). Halve the entry speed and the deflection quadruples; double the field and it doubles. A proton at the same speed deflects the other way by only about 1/1800 as much.

Checklist

• I can explain why E at a point does not depend on the test charge placed there.
• I get the direction of a point-charge field from the sign of the source, and the direction of the force from the sign of the charge feeling it.
• I can add fields from two or three charges in 2-D by components, using symmetry first.
• I can state the five field-line rules and sketch a dipole and two like charges.
• I know the field between parallel plates is uniform, points + to −, and equals σ/ε₀ = |ΔV|/d.
• I can treat a charge entering a uniform field as a projectile with a = qE/m.

4 · Electric potential energy & potential

Forces and fields are vectors, and vectors are work. Energy is a single number with a sign. This page rebuilds electrostatics in the language of energy — and in exchange for one new idea, potential, most problems drop from two dimensions of trigonometry to one line of arithmetic.

Electric potential energy

The electric force is conservative, exactly like gravity: the work it does depends only on where a charge starts and ends, not the route. So it has a potential energy. For two point charges a distance r apart:

Note the r, not r². The energy falls off more slowly than the force — the force is the slope of the energy graph.

Systems of several charges

Energy is a scalar, so a group of charges has a total potential energy equal to the sum over every pair, each counted once:

The physical meaning is the cleanest way to remember it: U_total is the external work needed to assemble the charges, starting infinitely far apart and at rest, ending at rest in place. Bring in charge 1 — free, nothing there yet. Bring in charge 2 — costs U₁₂. Bring in charge 3 — costs U₁₃ + U₂₃. The order you choose does not change the total.

Electric potential

Potential energy depends on two charges. Just as the field is force per coulomb, the electric potential is potential energy per coulomb — a property of the location alone, set up by the source charges:

For a single point charge, divide kqq₀/r by q₀:

Equipotentials

An equipotential is a surface (a line, in a 2-D drawing) on which V has the same value everywhere. Two facts carry the whole idea:

1. Moving a charge along an equipotential takes no work, since ΔV = 0 means ΔU = qΔV = 0.
2. Equipotentials are everywhere perpendicular to field lines. If the field had a component along the surface, moving along it would take work — contradicting fact 1.

And the field always points from higher to lower potential, crossing the equipotentials at right angles — like water running straight downhill across the contour lines of a topographic map.

Uniform fields: ΔV = −EΔx

Between parallel plates the field is uniform, so the potential changes at a steady rate. Move a distance Δx along the field and the potential drops by EΔx:

This is also where the unit V/m comes from: a field of 3000 V/m means the potential changes by 3000 volts per metre along it. Check that 1 V/m = 1 (J/C)/m = 1 N/C — the two units for E are the same thing.

Work, energy conservation & the electron-volt

Three relations, three different questions:

That last line is the reason this topic exists. In a non-uniform field the force changes from point to point, so constant-acceleration kinematics fails — but ΔK = −qΔV needs only the potential at the two endpoints.

High to low, or low to high?

The electron-volt

The eV makes particle problems almost free: anything with charge ±e accelerated from rest through 500 V gains exactly 500 eV of kinetic energy, whatever the field looked like in between. For an electron that is 500 × 1.60×10⁻¹⁹ = 8.0×10⁻¹⁷ J, so v = √(2K/m) = 1.3×10⁷ m/s — the same answer Section 3 got with forces and kinematics, in a fraction of the work. Convert to joules before using ½mv².

Worked examples

1. Potential energy of a pair, and what its sign means

A +3.0 μC charge and a −2.0 μC charge are 20 cm apart. Find the potential energy of the pair, and the work needed to separate them completely.

U = (8.99×10⁹)(+3.0×10⁻⁶)(−2.0×10⁻⁶)0.20 = −0.27 J

Negative: the pair is bound. Pulling them to infinite separation (from rest to rest) takes W_ext = ΔU = 0 − (−0.27) = +0.27 J of work from outside.

2. Potential energy of a system of three charges

Charges q₁ = +4.0 nC at (0, 0), q₂ = +2.0 nC at (0.30 m, 0) and q₃ = −3.0 nC at (0, 0.40 m). Find the total electric potential energy of the system.

Distances: r₁₂ = 0.30 m, r₁₃ = 0.40 m, r₂₃ = 0.50 m (3-4-5 again).

Pair	kqiqj (J·m)	r (m)	U (J)
1–2	+7.19×10−8	0.30	+2.40×10−7
1–3	−1.08×10−7	0.40	−2.70×10−7
2–3	−5.39×10−8	0.50	−1.08×10−7
Total			−1.4×10−7

Negative overall: the two attractive pairs outweigh the one repulsive pair. It would take 1.4×10⁻⁷ J of external work to pull all three charges apart to infinity, and that is also how much work the field did on them while they were being assembled.

3. Potential from two charges, then the work to bring in a third

q₁ = +4.0 nC is at the origin and q₂ = −6.0 nC at (0.30 m, 0). (a) Find the potential at P = (0.30 m, 0.40 m). (b) How much work must an external agent do to bring a +2.0 nC charge slowly from very far away to P?

P is 0.50 m from q₁ and 0.40 m from q₂.

V_P = (8.99×10⁹)(4.0×10⁻⁹)0.50 + (8.99×10⁹)(−6.0×10⁻⁹)0.40 = 71.9 − 134.9 = −63 V Compare with the 2-D field example: no components, no angles. That is what a scalar buys you.

W_ext = qΔV = (2.0×10⁻⁹)(−63 − 0) = −1.3×10⁻⁷ J

Negative work by the agent: the net pull of the charges draws the +2.0 nC charge in, and the agent has to hold it back to keep it from speeding up.

4. Parallel plates: field, potential, and a proton’s energy

Two parallel plates 4.0 mm apart are held at 12 V and 0 V (the figure in the uniform-field section). Find (a) the field, (b) the potential 1.0 mm from the 0 V plate, and (c) the kinetic energy of a proton released from rest at the 12 V plate when it reaches the 0 V plate.

(a) E = |ΔV|d = 124.0×10⁻³ = 3.0×10³ V/m, from the 12 V plate toward the 0 V plate

(b) V = 0 + E(1.0×10⁻³) = 3.0 V Moving from the 0 V plate toward the 12 V plate is moving against the field, so the potential rises.

(c) ΔK = −qΔV = −(+e)(0 − 12 V) = 12 eV = 1.9×10⁻¹⁸ J

An electron released from rest at the 0 V plate would gain the same 12 eV moving the other way, toward higher potential: −(−e)(+12 V) = +12 eV.

5. Energy conservation in a non-uniform field

A proton is released from rest 2.0 cm from a fixed +2.0 μC charge. How fast is it moving when it is 6.0 cm from the charge?

No kinematics allowed — the force weakens as the proton moves out, so the acceleration is not constant. Use energy.

V_i = (8.99×10⁹)(2.0×10⁻⁶)0.020 = 8.99×10⁵ V     V_f = (8.99×10⁹)(2.0×10⁻⁶)0.060 = 3.00×10⁵ V

ΔK = −eΔV = −e(3.00×10⁵ − 8.99×10⁵) = 5.99×10⁵ eV = 9.59×10⁻¹⁴ J About 0.60 MeV. The proton moved from high to low potential, as a positive charge released from rest must.

v = √2ΔK/m_p = √2(9.59×10⁻¹⁴)/(1.67×10⁻²⁷) = 1.07×10⁷ m/s

Checklist

• I keep the signs of the charges in U = kq₁q₂/r and V = kq/r, and explain what a negative U means.
• I can find the potential energy of three charges by summing over all three pairs.
• I add potentials as plain numbers, and I know zero field and zero potential are different conditions.
• I can sketch equipotentials perpendicular to field lines, with the field pointing from high to low V.
• I use ΔV = −EΔx and |E| = |ΔV|/Δx in uniform fields.
• I use ΔK = −qΔV whenever the field is not uniform, and I can convert between eV and J.

5 · Conductors & capacitors

In a conductor, charge is free to move — so it keeps moving until the field inside has nothing left to push on. Everything on this page follows from that one sentence: why metal boxes block fields, why charge piles up on sharp points, and how two metal plates with a gap between them store energy.

Conductors in electrostatic equilibrium

“Electrostatic equilibrium” means the charges have stopped moving. Put excess charge on a metal object, or hold it in an external field, and within a tiny fraction of a second the free electrons rearrange themselves and come to rest. Once they have, four things are true — and each has a one-line reason:

1. E = 0 everywhere inside the conducting material. If it were not, the free electrons there would feel a force and move — so it would not be equilibrium.
2. Any excess charge sits on the outer surface. Like charges push each other as far apart as they can get; and with E = 0 inside, there can be no net charge buried in the interior.
3. Just outside, the field is perpendicular to the surface. A component along the surface would push surface charges sideways, and they would move.
4. The whole conductor is one equipotential. With E = 0 inside, no work is done moving a charge between any two points of it, so ΔV = 0. That is also why the surface field must be perpendicular — field lines always cross equipotentials at right angles.

Shielding and the Faraday cage

Put a hollow conductor in an external field. Its free charges shift: one side becomes negative, the other positive, and they arrange themselves so that their own field exactly cancels the external field everywhere inside the metal — and therefore throughout the empty cavity too. The cavity is shielded.

This is a Faraday cage, and it does not need to be solid: a mesh with holes much smaller than the wavelength involved works almost as well. Examples worth knowing: a car struck by lightning protects the people inside (the charge stays on the outer metal skin); the metal mesh in a microwave door keeps the microwaves in; your phone loses signal in a metal elevator; the braided outer conductor of a coaxial cable keeps stray fields off the signal wire.

Capacitance and the parallel-plate capacitor

A capacitor is any two conductors separated by an insulator. Move charge Q from one to the other — leaving +Q and −Q — and a potential difference ΔV appears between them. The two are always proportional, and the constant of proportionality is the capacitance:

Deriving C = ε₀A/d

Four steps, each one a formula you already have:

1. Charge per area on each plate: σ = Q/A.
2. Field between large plates (from Section 3): E = σ/ε₀ = Q/(ε₀A).
3. Uniform field, so ΔV = Ed = Qd/(ε₀A).
4. Divide: C = Q/ΔV = ε₀A/d. The Q cancels — as promised, geometry only.

Why does a smaller gap help? For a given charge, the field σ/ε₀ is fixed, so the voltage Ed shrinks with d. Less voltage for the same charge means more charge per volt.

Energy stored in a capacitor

Charging a capacitor means moving charge from one plate to the other against a voltage that grows as you go: the first bit of charge crosses for free, the last bit crosses the full ΔV. On a graph of voltage against charge that is a straight line, and the work done is the area under it:

Where is the energy? It is stored in the field between the plates. The same field filling three times the volume holds three times the energy — which is exactly what happens in worked example 2.

Dielectrics

A dielectric is an insulator placed between the plates. Its molecules cannot conduct, but they can polarize: in the field, each molecule’s electron cloud shifts slightly toward the + plate (or polar molecules rotate to line up). The net effect is a thin layer of negative charge on the slab face next to the + plate and positive charge on the face next to the − plate.

Weaker field for the same charge means a smaller ΔV = Ed, and so a larger C = Q/ΔV — by exactly the factor κ. Dielectrics also let the plates sit very close without touching, and they withstand a larger field before breaking down (sparking through) than air does, which for dry air is about 3×10⁶ V/m.

Connected or disconnected?

The single most-tested capacitor question: something about the capacitor changes — plates pulled apart, a dielectric slid in — and you are asked what happens to Q, ΔV, C, E and U. The whole answer hinges on one decision, made before any algebra:

• Still connected to the battery ⇒ ΔV is fixed. The battery holds it there, adding or removing charge as needed.
• Charged, then disconnected ⇒ Q is fixed. The charge has nowhere to go.

Then: find the new C from κε₀A/d; get the other of Q or ΔV from Q = CΔV; get E from ΔV/d; get U from the energy form that uses the fixed quantity.

Worked examples

1. A parallel-plate capacitor on a battery

Two plates, each 0.020 m² in area, are 1.0 mm apart in air and connected to a 12 V battery. Find C, Q, E and U.

C = ε₀Ad = (8.85×10⁻¹²)(0.020)1.0×10⁻³ = 1.77×10⁻¹⁰ F = 177 pF

Q = CΔV = (1.77×10⁻¹⁰)(12) = 2.1×10⁻⁹ C     E = ΔVd = 121.0×10⁻³ = 1.2×10⁴ V/m

U = ½C(ΔV)² = ½(1.77×10⁻¹⁰)(12)² = 1.3×10⁻⁸ J Cross-check the field another way: σ/ε₀ = Q/(ε₀A) = 2.12×10⁻⁹/(1.77×10⁻¹³) = 1.2×10⁴ V/m ✓

A tabletop-sized capacitor holds only nanocoulombs and stores about a hundredth of a microjoule. A farad is enormous.

2. Disconnect, then pull the plates apart

The capacitor from example 1 is disconnected from the battery, and the plates are pulled apart from 1.0 mm to 3.0 mm. Find the new C, Q, ΔV, E and U, and the work you did.

Decide first: disconnected ⇒ Q = 2.1 nC is fixed.

C′ = 177 pF3 = 59 pF     ΔV′ = QC′ = 2.12×10⁻⁹5.9×10⁻¹¹ = 36 V

E′ = 363.0×10⁻³ = 1.2×10⁴ V/m (unchanged)     U′ = Q²2C′ = 3 × 1.27×10⁻⁸ = 3.8×10⁻⁸ J

Work done by you = ΔU = 3.82×10⁻⁸ − 1.27×10⁻⁸ = 2.5×10⁻⁸ J. Same field, three times the volume of field, three times the energy.

3. Stay connected, slide in a dielectric

Start again from example 1 (1.0 mm gap, connected to 12 V). A slab with κ = 3.0 is slid in to fill the gap completely. Find the new values.

Decide first: connected ⇒ ΔV = 12 V is fixed.

C′ = κC = 3.0 × 177 pF = 531 pF     Q′ = C′ΔV = (5.31×10⁻¹⁰)(12) = 6.4×10⁻⁹ C

E′ = 12/(1.0×10⁻³) = 1.2×10⁴ V/m (unchanged)     U′ = ½(5.31×10⁻¹⁰)(12)² = 3.8×10⁻⁸ J

The battery pushed an extra 6.37 − 2.12 = 4.2 nC onto the plates. Compare with the disconnected case in the table: there, the same slab would have cut the field and the energy by a factor of 3.

4. Camera flash

A camera flash charges a 150 μF capacitor to 300 V, then dumps it through the flash tube in about 1.0 ms. Find the stored energy, the charge, and the average power during the flash.

U = ½C(ΔV)² = ½(150×10⁻⁶)(300)² = 6.8 J     Q = CΔV = 0.045 C

P_avg = 6.75 J1.0×10⁻³ s ≈ 6.8 kW That is the point of a capacitor here: a small battery charges it slowly over a few seconds, and the capacitor releases the energy thousands of times faster than the battery could.

Scaling: to store twice the energy in the same capacitor, the voltage must rise by √2, to about 420 V — energy goes as (ΔV)².

5. A charge inside a hollow conductor

A +5.0 nC point charge sits at the center of a neutral, hollow metal sphere with inner radius 0.10 m and outer radius 0.15 m. Find the charge on each surface of the sphere, and the field at r = 0.05 m, 0.12 m and 0.20 m from the center.

Surfaces. The field inside the metal must be zero, so every field line leaving the +5.0 nC must end on the inner surface: the inner surface carries −5.0 nC. The sphere is neutral, so the outer surface carries +5.0 nC.

r	Region	E
0.05 m	Cavity	kq/r2 = (8.99×109)(5.0×10−9)/(0.05)2 = 1.8×104 N/C
0.12 m	Inside the metal	0
0.20 m	Outside	(8.99×109)(5.0×10−9)/(0.20)2 = 1.1×103 N/C

In the cavity and outside, the field is just that of the +5.0 nC point charge: a uniformly charged spherical shell makes no field inside itself and acts like a point charge at its center outside. Outside, the shell’s inner −5.0 nC and outer +5.0 nC cancel, leaving only the central charge. The metal did not hide the charge from the outside world — exactly the one-way shielding from the shielding section. Ground the sphere and the outer +5.0 nC drains away; the field outside then drops to zero.

Checklist

• I can state the four properties of a conductor in electrostatic equilibrium and give the reason for each.
• I can explain shielding by induced charges, and why a charge inside a shell is still felt outside unless the shell is grounded.
• I know C = Q/ΔV depends only on geometry and material, and can derive C = ε₀A/d from E = σ/ε₀.
• I can get U = ½QΔV from the area under a ΔV–Q graph, and pick the energy form that uses the fixed quantity.
• I can explain how a dielectric weakens the field and raises C by κ.
• For any change, I decide “connected (ΔV fixed) or disconnected (Q fixed)?” before doing anything else.

6 · Formula sheet & constants

Everything the unit uses, with the condition each one needs to be legal.

Constants worth memorizing

7 · Unit-wide traps

• Plugging signs into a magnitude formula. F = k|q₁q₂|/r² and E = k|q|/r² give sizes. Direction comes from the diagram. A negative answer for a force magnitude is a red flag.
• Adding fields as numbers, or potentials as arrows. Fields and forces are vectors; potentials and energies are scalars. Two equal and opposite charges give E ≠ 0 but V = 0 at their midpoint.
• r vs r². Force and field go as 1/r²; potential and potential energy go as 1/r. Doubling the distance quarters F but only halves V.
• Forgetting the microcoulombs. Most problems quote μC or nC. An answer of 10¹² N means a prefix was dropped.
• Unequal charges, unequal forces? No. By Newton’s third law, the force of q₁ on q₂ always equals the force of q₂ on q₁ in size. Their accelerations differ if their masses do.
• Electrons go the “wrong” way. The force on a negative charge is opposite to E, and an electron released from rest moves toward higher potential.
• Connected vs disconnected capacitor. Connected to a battery: ΔV fixed. Disconnected: Q fixed. Decide which one before touching any formula — see Section 5.

8 · Labs for this unit

Every lab for this unit lives in Interactive Laboratories, each one paired with the section above that explains it.

Interactive lab Coulomb force & superposition Open it in Interactive Laboratories → Interactive lab Field lines & equipotentials Open it in Interactive Laboratories → Interactive lab Charged particle between plates Open it in Interactive Laboratories → Interactive lab Parallel-plate capacitor Open it in Interactive Laboratories → Bench lab Mapping equipotentials Open it in Interactive Laboratories →