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D.2

Electric and magnetic fields

Theme D · Fields · SL and HL, with additional HL material marked

Two field types on one page, and that is deliberate: almost everything here is either the same as D.1 with charge in place of mass, or a difference that traces back to one fact — charge comes in two signs.

🎯What you need to be able to do

  • State that charge comes in two kinds and is always conserved, and distinguish conductors from insulators.
  • Apply Coulomb’s law in both its forms, and say what it shares with Newton’s law of gravitation and where it differs.
  • Use electric field strength \( E = F/q = kQ/r^{2} \), and sketch the field patterns for one charge and for two.
  • Use the uniform field between parallel plates, \( E = V/d \).
  • Describe Millikan’s experiment and what it established about charge.
  • Sketch magnetic field patterns for a bar magnet, a straight wire, a flat coil and a solenoid, and get the directions right using the right-hand grip rule.
  • HLUse electric potential and electric potential energy, including the fact that both can be positive, and add potentials as scalars.
  • HLUse the potential gradient \( E = -\Delta V_e/\Delta r \), and work in electronvolts.
  • HLCompare gravitational and electric fields point by point.

⚖️Charge: two kinds, and never created or destroyed

There are exactly two kinds of charge, positive and negative, and equal amounts of them cancel. Matter with equal amounts of each is neutral, which is what almost all matter is. We know charge exists at all only because of the forces between charged objects: like charges repel, unlike charges attract.

The experimental fact underneath everything on this page is that charge is conserved. Rub a comb through your hair and the comb becomes negative while the hair becomes equally positive — no charge was made, electrons were simply moved from one to the other. The total before equals the total after, in every process ever measured.

Two panels. On the left, conservation of charge: a neutral comb and neutral hair before rubbing, then afterwards the comb carrying a negative charge and the hair an equal positive charge, with a note that electrons moved from the hair to the comb and that the totals before and after are identical. On the right, a comparison of conductors and insulators: conductors such as metals and graphite have charges free to move through them, drawn as electrons moving between atoms; insulators such as plastics, dry wood, glass and ceramics do not, so charge placed on them stays where it is put, which is why a charged object usually has to be an insulator to stay charged.
Charging by friction moves electrons; it never creates them. That is also why a charged object usually has to be an insulator — on a conductor the charge would simply flow away.

A conductor lets charge flow through it — all metals, and graphite. An insulator does not: plastics, dry wood, glass, ceramics. In a solid conductor the thing that actually moves is always electrons, passing from atom to atom.

⚡Coulomb’s law: Newton’s law with a sign

\[ F = \frac{kQ_1Q_2}{r^{2}} \qquad k = 8.99 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2} \]

You will meet the same law written a second way. The Coulomb constant is itself built out of a more fundamental one, the permittivity of free space \( \varepsilon_0 = 8.85\times10^{-12} \) C\(^{2}\) N\(^{-1}\) m\(^{-2}\):

\[ k = \frac{1}{4\pi\varepsilon_0} \qquad\text{so}\qquad F = \frac{Q_1Q_2}{4\pi\varepsilon_0 r^{2}} \]

The two forms are identical — check it: \( 1/(4\pi \times 8.85\times10^{-12}) = 8.99\times10^{9} \). Use whichever the question uses. If the charges sit in a material rather than a vacuum, it is \( \varepsilon \) for that material that appears, and the force is smaller.

Structurally this is Newton’s law of gravitation with charge in place of mass: the same inverse square, the same centre-to-centre \(r\), the same equal-and-opposite pair of forces on the two bodies. Everything you learned about handling \(1/r^{2}\) in D.1 transfers unchanged.

Two panels. In the first, two positive charges are shown with force arrows on each pointing directly away from the other, labelled repulsion, and the two arrows are drawn the same length. In the second, a positive and a negative charge have force arrows pointing towards each other, labelled attraction, again the same length as each other. Beneath both, a note states that the forces are always equal in magnitude and opposite in direction whatever the sizes of the two charges, exactly as in gravitation, and that the only new thing is that the pair can now push apart as well as pull together.
The one structural difference from gravity: like charges repel. Everything else on this page that differs between the two field types comes back to this.

The other difference is scale, and it is not a small one.

A comparison of the electric and gravitational forces between two protons, drawn on a logarithmic scale spanning thirty-six orders of magnitude. The gravitational attraction is marked at one end and the electric repulsion at the other, with the ratio labelled as about one times ten to the thirty-six. A note explains that this is why gravity is irrelevant inside an atom and yet dominates the universe: large bodies are electrically neutral, so their enormous positive and negative charges cancel almost exactly, leaving only the far weaker force that has no cancelling opposite.
Between two protons the electric repulsion is around \( 10^{36} \) times the gravitational attraction. Gravity runs the universe only because it is the force that cannot be cancelled.

✏️Worked example 1 — the two forces side by side

A charge of \( +2.0\ \mu\text{C} \) and one of \( -3.0\ \mu\text{C} \) are 5.0 cm apart. Find the force between them, and compare it with the gravitational force between two 1.0 g masses at the same separation.
\[ F = \frac{kQ_1Q_2}{r^{2}} = \frac{8.99\times10^{9} \times 2.0\times10^{-6} \times 3.0\times10^{-6}}{(0.050)^{2}} = 22\ \text{N} \]

The charges have opposite signs, so the force is attractive. Sub in the masses instead:

\[ F = \frac{Gm_1m_2}{r^{2}} = \frac{6.67\times10^{-11} \times (1.0\times10^{-3})^{2}}{(0.050)^{2}} = 2.7\times10^{-14}\ \text{N} \]
A factor of \( 8\times10^{14} \). 22 N is roughly the weight of a 2 kg bag of rice, from two specks of charge you could not see. The gravitational force between two paperclips at the same distance is around \( 10^{-14} \) N, which no ordinary instrument can detect. Handle the signs by working out the magnitude from the numbers and deciding attractive-or-repulsive separately, from the signs — putting negative charges into the formula and trusting the output sign is how people end up reporting a negative force.

🧭Electric field strength and field patterns

Electric field strength \( E = F/q = kQ/r^{2} \) is the force per unit positive charge, measured in N C\(^{-1}\) (or, equivalently, V m\(^{-1}\)). That definition fixes every direction on the page: field lines point away from positive charges and towards negative ones.

Four electric field patterns. First, an isolated positive charge with straight field lines radiating outwards in all directions. Second, an isolated negative charge with straight lines pointing inwards. Third, two like positive charges, whose lines curve away from each other and leave a point midway between them where the field is zero. Fourth, a dipole of one positive and one negative charge, with lines that leave the positive charge, curve across the gap and end on the negative charge, crowding together in the region between them where the field is strongest.
Lines start on positive and end on negative. Between two like charges there is a null point, exactly as there was between two masses in D.1; between unlike charges there is no null point anywhere.

With more than two charges, work out each force separately from Coulomb’s law and then add them as vectors — the same parallelogram you used for forces in Theme A. Electric field strength is a vector too, so fields from several charges add the same way. (Potential, later on this page, is the exception: it is a scalar and simply adds.)

Three rules that are marked strictly. Field lines never cross, because the field at a point has one direction. They meet a conductor’s surface at 90°, because any component along the surface would push the free charges sideways until it was cancelled. And the arrowheads are not decoration — a sketch without them scores nothing, because the direction is most of the physics.

🔢The uniform field between parallel plates

Two parallel plates at a potential difference \(V\), a distance \(d\) apart, make a field that is the same everywhere between them:

\[ E = \frac{V}{d} \]
Two horizontal parallel plates, the upper one positive and the lower one negative, with evenly spaced straight field lines running from the positive plate down to the negative one. The lines are strictly parallel and equally spaced across the central region, and curve outwards at the two ends where the field fringes. The separation d is marked between the plates and the potential difference V across them. A positive charge placed in the field has a force arrow drawn on it pointing down the field, and the note states that the force is the same size wherever in the uniform region the charge is put.
Uniform means the force on a charge is the same everywhere between the plates — unlike every other field on this page. That is exactly why this arrangement is used to accelerate and steer charged particles, which is where D.3 picks up.

✏️Worked example 2 — a uniform field

Two parallel plates 5.0 cm apart are connected to a 2000 V supply. Find the field strength between them, and the force and acceleration of an electron placed in it. (\( e = 1.60\times10^{-19} \) C, \( m_e = 9.11\times10^{-31} \) kg.)

Field strength. \( E = V/d = 2000/0.050 = 4.0\times10^{4} \) V m\(^{-1}\).

Force. \( F = qE = 1.60\times10^{-19} \times 4.0\times10^{4} = 6.4\times10^{-15} \) N, directed towards the positive plate, because the electron is negative.

Acceleration. \( a = F/m = 6.4\times10^{-15} / 9.11\times10^{-31} = 7.0\times10^{15} \) m s\(^{-2}\).

That number is not a mistake. It is \( 7\times10^{14} \) times \(g\). An electron is so light that even a modest laboratory field throws it across the gap in nanoseconds, and this is why electron beams are steered electrically rather than mechanically. Note also that the mass never entered the force calculation — unlike a gravitational field, an electric field does not produce the same acceleration for everything in it.

💧Millikan’s experiment: charge comes in lumps

The uniform field between parallel plates is what let Robert Millikan settle one of the most important questions in physics: is charge continuous, or does it come in a smallest lump?

He sprayed a fine mist of oil into the gap between two horizontal plates and watched single droplets through a microscope. A drop falling freely reaches terminal velocity, which gives its radius and hence its weight. Charge the drop with X-rays, switch on the field, and adjust the voltage until the drop hangs stationary — at which point the electric force exactly balances the weight and the drag has vanished, because nothing is moving.

On the left, the apparatus: an atomizer sprays oil droplets through a small hole into the gap between two horizontal parallel plates connected to a variable potential difference, with a microscope viewing the drops through a window and the whole assembly in a constant-temperature bath. In the middle, the force diagram for a stationary drop, with the weight acting downwards and the electric force q E acting upwards and equal to it, and the note that the drag force is zero because the drop is not moving. On the right, a number line of the charges Millikan actually measured, showing values at minus 3.2, minus 1.6, plus 1.6, plus 4.8 and plus 6.4 times ten to the minus nineteen coulombs, with nothing in between, all of them whole-number multiples of one value of 1.6 times ten to the minus nineteen coulombs.
The result that mattered was not any single measurement but the pattern of them: every charge he found was a whole-number multiple of \( 1.6\times10^{-19} \) C, and no fraction of it has ever been observed.

For a stationary drop the forces balance, so

\[ qE = mg \qquad\Longrightarrow\qquad q = \frac{mg}{E} = \frac{mgd}{V} \]

Repeating this for droplet after droplet, Millikan found charges of \( 1.6\times10^{-19} \), \( 3.2\times10^{-19} \), \( 4.8\times10^{-19} \) C — and nothing in between. Charge is quantized: it comes in whole-number multiples of the elementary charge \( e = 1.60\times10^{-19} \) C, which is the magnitude of the charge on a single electron.

✏️Worked example 3 — how many electrons?

An oil drop of weight \( 3.0\times10^{-14} \) N is held stationary between horizontal plates 12 mm apart with 750 V across them. Find the charge on the drop, and how many excess electrons it carries.

Field between the plates. \( E = V/d = 750/0.012 = 6.25\times10^{4} \) V m\(^{-1}\).

Charge, from the balance of forces.

\[ q = \frac{mg}{E} = \frac{3.0\times10^{-14}}{6.25\times10^{4}} = 4.8\times10^{-19}\ \text{C} \]
\[ \frac{q}{e} = \frac{4.8\times10^{-19}}{1.60\times10^{-19}} = 3.0 \]

So the drop carries 3 excess electrons.

The whole number is the point. Getting exactly 3.0 rather than 3.4 is what makes the result mean something — it is evidence that charge is quantized, not just a measurement of one droplet. If your answer is not close to an integer, check the arithmetic before concluding anything about physics. Note too that the drop being stationary is what removes the drag force from the problem; a moving drop would need it.

🧲Magnetic fields are made by moving charge

There are no magnetic monopoles — no isolated N or S pole has ever been found — and one consequence shows up in every diagram you will draw: magnetic field lines always form closed loops. They have no starting point and no end, which is the clearest single difference from electric field lines.

Three magnetic field patterns. First, a bar magnet with field lines emerging from the north pole, curving round the outside of the magnet and entering the south pole, then continuing through the inside of the magnet back to the north pole so that every line is a closed loop. Second, a straight current-carrying wire seen end-on, with concentric circular field lines around it and a right hand gripping the wire, thumb along the conventional current and fingers curling in the direction of the field. Third, a solenoid, with the field inside drawn uniform and parallel along the axis and the outside field looking exactly like that of a bar magnet, with one end behaving as a north pole.
Three patterns to know cold. The solenoid is the one that ties the page together: it is a bar magnet you can switch off, and it is made of nothing but a moving charge going round in circles.

The direction of the field round a current is given by the right-hand grip rule, and it is worth learning as a physical gesture rather than a sentence — it is the rule you will reach for most often in this theme.

The right-hand grip rule drawn as a hand gripping a vertical wire, with the thumb extended along the direction of the conventional current and the four fingers curling around the wire in the direction of the magnetic field. Beside it, the same rule applied twice more: to a flat circular coil, where curling the fingers along the current makes the thumb point along the field through the middle of the coil, and to a solenoid, where curling the fingers along the current in the windings makes the thumb point to the end that behaves as the north pole. A note stresses that this is the RIGHT hand and that it uses conventional current, the direction positive charge moves.
One gesture, three uses. For a straight wire the thumb is the current and the fingers are the field; for a coil or a solenoid it works the other way round — the fingers follow the current and the thumb gives the field, and so the north pole.

Two more patterns follow from it. A flat circular coil has the circular field of a straight wire wrapped into a ring, so the contributions reinforce through the middle and the field there is strong and nearly straight. A solenoid is many such coils in a row: inside it the field is uniform and parallel, and from outside it is indistinguishable from a bar magnet — one whose poles swap when you reverse the current, and vanish when you switch it off.

A comparison of electric and magnetic fields set out in rows. Symbol: E against B. Caused by: charges against magnets or electric currents. Affects: charges against magnets or currents. Two types of: positive and negative charge, against north and south poles. Simple force rule: like charges repel and unlike attract, against like poles repel and unlike attract. Then three differences: a magnet feels no force in an electric field; a stationary charge feels no force in a magnetic field; and isolated charges exist whereas isolated poles do not, so magnetic field lines are always closed loops while electric ones start and end on charges.
The two field types are strikingly parallel until the last three rows — and those three are where nearly all the exam marks are, because they are the ones that cannot be guessed from the pattern.

One further pattern is worth knowing because it is where the naming comes from: the Earth has a magnetic field much like a bar magnet’s, and a compass needle lines up along it. The needle’s north pole points towards the geographic North Pole — which means that, magnetically, there is a south pole up there.

➡️What these fields DO to charges and currents

A current-carrying wire placed across a magnetic field feels a force — the motor effect — and so does a single moving charge. Both of those, together with Fleming’s left-hand rule for working out which way that force points, belong to D.3 Motion in electromagnetic fields, and are treated in full there.

The division is worth keeping straight: this page is about the fields themselves — what makes them, what they look like, how strong they are. D.3 is about what happens to something placed in them.

HLElectric potential and potential energy

Electric potential\( V_e = \dfrac{kQ}{r} \)
Electric potential energy\( E_p = \dfrac{kQ_1Q_2}{r} \)

Both look exactly like their gravitational counterparts, but with one crucial difference: they are not always negative. The sign now follows the charges.

A graph of electric potential against distance with two curves. The curve for a positive charge is entirely above the axis, very large close in and falling towards zero at large distance. The curve for a negative charge is its mirror image below the axis, rising towards zero from below. Both approach zero at infinity, which is where the zero is defined. Beside the graph, two cases of potential energy are set out: two like charges have positive potential energy, and would fly apart if released, so they are unbound; a positive and a negative charge have negative potential energy, are bound together, and work must be done to separate them.
Gravitational potential has only the lower branch, because mass has only one sign. Electric potential has both, and reading the sign correctly is usually the whole difficulty in these questions.

Potential is a scalar, so the potential at a point due to several charges is just the arithmetic sum of the individual contributions — signs included, no angles, no components. Field strength is a vector and must be added as one. This is the same split you met in D.1, and it has the same consequence: a point can have zero field and a large potential, or zero potential and a large field.

Two stacked graphs against distance from the centre of a charged conducting sphere of radius a. The upper graph is electric field strength: it is zero everywhere inside the sphere, jumps to its maximum value at the surface, and then falls away outside as one over r squared. The lower graph is potential: it is constant everywhere inside the sphere at the value it has at the surface, and outside falls away as one over r. A note explains that the charge sits entirely on the outside surface, that there is no field inside so no work is needed to move within the sphere, and that the potential is therefore the same at every interior point.
Inside a charged conductor the field is zero and the potential is constant — not zero. No field means no work to move around inside, which is exactly what constant potential means. Outside, both behave as if all the charge sat at the centre.

Negative potential energy still means bound, exactly as it did for orbits in D.1 — work must be supplied to separate the pair and reach zero. Positive potential energy means the opposite: released, they fly apart, converting that energy into kinetic energy.

✏️HLWorked example 4 — the work to assemble a pair

How much work must be done to bring two \( +5.0\ \mu\text{C} \) charges from very far apart to a separation of 0.20 m? What happens if they are then released?

At infinity the potential energy is zero, so the work done equals the final potential energy:

\[ E_p = \frac{kQ_1Q_2}{r} = \frac{8.99\times10^{9} \times (5.0\times10^{-6})^{2}}{0.20} = +1.1\ \text{J} \]

So 1.1 J of work must be supplied. Released, the charges repel, and that 1.1 J reappears as kinetic energy — shared between them, so 0.56 J each if their masses are equal.

The positive sign is the answer to “is it bound?” It is not: positive potential energy means the pair will separate on their own. Contrast the gravitational case, where \( E_p \) is negative for every pair of masses without exception, so nothing ever flies apart of its own accord.

HLThe potential gradient, and the electronvolt

Exactly as in gravitation, the field is minus the rate at which the potential changes with distance:

\[ E = -\frac{\Delta V_e}{\Delta r} \]

This is why the two units for electric field strength, N C\(^{-1}\) and V m\(^{-1}\), are the same thing. Between parallel plates the potential falls steadily across the gap, so the gradient is constant at \( V/d \) — which is where \( E = V/d \) came from earlier on this page.

Two parallel plates with 70 volts across them, drawn with a series of evenly spaced horizontal equipotential lines between them labelled 70, 60, 50, 40, 30, 20, 10 and 0 volts from the positive plate down to the earthed negative plate. Field lines run perpendicular to the equipotentials, from the positive plate to the negative one. The equipotentials are equally spaced because the field is uniform, and the field strength is marked as the potential difference divided by the plate separation. A note gives the general relation, that field equals minus the potential gradient, and points out that this is why volts per metre and newtons per coulomb are the same unit.
Equally spaced equipotentials mean a uniform field, and vice versa. The field lines cross them at 90°, as they must — the same rule as for gravitational equipotentials in D.1.

At the scale of a single electron the joule is hopelessly large, so physicists use the electronvolt: the energy gained by a charge of \(e\) moving through a potential difference of 1 V.

\[ 1\ \text{eV} = 1\ \text{V} \times 1.60\times10^{-19}\ \text{C} = 1.60\times10^{-19}\ \text{J} \]

The usual prefixes apply — keV, MeV, GeV — and in particle physics they are used far more often than joules.

✏️HLWorked example 5 — accelerating an electron

An electron starts from rest and is accelerated in a vacuum through a potential difference of 1000 V. Find its energy in eV and in joules, and hence its final speed. (\( m_e = 9.11\times10^{-31} \) kg.)

Energy. A charge of \(e\) through 1000 V gains 1000 eV = 1.00 keV, and \( 1000 \times 1.60\times10^{-19} = 1.60\times10^{-16} \) J.

Speed, from \( \tfrac{1}{2}mv^{2} = qV \):

\[ v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 1.60\times10^{-16}}{9.11\times10^{-31}}} = 1.9\times10^{7}\ \text{m s}^{-1} \]
Is that allowed? \( 1.9\times10^{7} \) m s\(^{-1}\) is about 6% of the speed of light, so treating the kinetic energy as \( \tfrac{1}{2}mv^{2} \) is still safe. Push the accelerating voltage into the hundreds of kilovolts and it stops being safe — that is the point at which A.5’s relativistic treatment takes over. Note also how little algebra the electronvolt saved you here: the energy in eV is just the voltage, with no constants at all.

HLGravitational and electric fields, side by side

Examiners ask for this comparison directly, so it is worth being able to produce it from memory.

A two-column comparison of gravitational and electric fields. Force: G M m over r squared, against k Q q over r squared. Field strength: G M over r squared, against k Q over r squared. Potential: minus G M over r, against k Q over r. Constant: G is six point six seven times ten to the minus eleven, against k which is eight point nine nine times ten to the nine, a difference of about twenty orders of magnitude. Source: mass, against charge. Sign: mass has one sign only, whereas charge has two. Direction: always attractive, against attractive or repulsive. Shielding: impossible for gravity, whereas an electric field can be screened out by a conductor. Relative strength: vastly weaker, against vastly stronger.
The top three rows are the same algebra twice. Every row below them is a consequence of the one fact that charge has two signs and mass has one.
Why gravity wins at large scales despite losing by \( 10^{36} \). Because charge cancels. A planet contains a staggering quantity of positive and negative charge in almost exactly equal amounts, so its net electric field is essentially zero, while every gram of its mass adds to its gravitational field with nothing to subtract. The weaker force wins because it is the one that only ever adds up.

🔭See it happen

PhET, Charges and Fields. Drop charges on the canvas and the field vectors appear live; place two like charges and hunt for the null point between them with the field sensor. Then switch on the equipotential tool and confirm that every surface it draws meets the field lines at 90°, which is the D.1 result reappearing unchanged.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Two charges of \( +4.0\ \mu\text{C} \) and \( +4.0\ \mu\text{C} \) are 0.10 m apart. Find the force on each, and state its direction.
\( F = \dfrac{kQ_1Q_2}{r^{2}} = \dfrac{8.99\times10^{9} \times (4.0\times10^{-6})^{2}}{(0.10)^{2}} = \dfrac{0.1438}{0.010} = 14 \) N. The charges are alike, so the force is repulsive: each is pushed directly away from the other, and the two forces are equal in magnitude.
2. Find the electric field strength 0.30 m from a point charge of \( +4.0 \) nC.
\( E = \dfrac{kQ}{r^{2}} = \dfrac{8.99\times10^{9} \times 4.0\times10^{-9}}{(0.30)^{2}} = \dfrac{35.96}{0.090} = 4.0\times10^{2} \) N C\(^{-1}\), directed radially away from the charge because it is positive.
3. Two parallel plates 2.0 cm apart have a potential difference of 600 V. Find the field between them, and the force on a charge of \( +3.0 \) nC placed anywhere in the gap.
\( E = V/d = 600/0.020 = 3.0\times10^{4} \) V m\(^{-1}\). \( F = qE = 3.0\times10^{-9} \times 3.0\times10^{4} = 9.0\times10^{-5} \) N. The word anywhere is the point of the question: the field is uniform, so the answer does not depend on where in the gap the charge sits.
4. In a Millikan-type experiment an oil drop of weight \( 1.6\times10^{-14} \) N hangs stationary between plates 10 mm apart with 500 V across them. Find the charge on the drop and the number of excess electrons.
\( E = V/d = 500/0.010 = 5.0\times10^{4} \) V m\(^{-1}\). For a stationary drop the electric force balances the weight, and the drag is zero because nothing is moving, so \[ q = \frac{mg}{E} = \frac{1.6\times10^{-14}}{5.0\times10^{4}} = 3.2\times10^{-19}\ \text{C} \] Dividing by the elementary charge, \( 3.2\times10^{-19} / 1.60\times10^{-19} = 2.0 \), so the drop carries 2 excess electrons. The whole-number answer is the evidence that charge is quantized — Millikan never found a fraction of \(e\), and nor has anyone since.
5. A vertical wire carries a conventional current upwards. Describe the magnetic field around it, and state the direction of the field at a point due north of the wire.
Concentric horizontal circles centred on the wire, closer together nearer it. By the right-hand grip rule — thumb up, along the conventional current — the fingers curl anticlockwise seen from above. At a point due north of the wire the field therefore points west. (Check with a second point: due east of the wire it points north, which is the same anticlockwise sense.)
6. Sketch the field pattern for two equal positive charges, and state what is special about the midpoint.
Lines radiate outwards from each charge and curve away from the other, so that no line passes through the region directly between them. At the midpoint the two fields are equal and opposite, so the resultant field is zero — a null point, exactly as between two masses in D.1. For two opposite charges the pattern is completely different: lines run from the positive to the negative and are most crowded between them, and there is no null point anywhere.
7. HLFind the electric potential 0.20 m from a point charge of \( -6.0 \) nC.
\( V_e = \dfrac{kQ}{r} = \dfrac{8.99\times10^{9} \times (-6.0\times10^{-9})}{0.20} = -2.7\times10^{2} \) V. The sign here comes straight from the charge, and is negative because the charge is negative — not, as in gravitation, because of the definition. Around a positive charge the potential would be \( +2.7\times10^{2} \) V at the same distance.
8. HLA charge of \( +3.0 \) nC and one of \( -4.0 \) nC are 0.15 m apart. Find their potential energy, and state whether the pair is bound.
\( E_p = \dfrac{kQ_1Q_2}{r} = \dfrac{8.99\times10^{9} \times 3.0\times10^{-9} \times (-4.0\times10^{-9})}{0.15} = -7.2\times10^{-7} \) J. It is negative, so the pair is bound: \( 7.2\times10^{-7} \) J of work must be supplied to separate them to infinity. Here the negative sign is genuine information about the physics, because the two charges have opposite signs.
9. HLState three ways in which gravitational and electric fields are the same, and three ways in which they differ.
Same: both obey an inverse-square law for force and for field strength; both have a potential that goes as \(1/r\) and is defined as zero at infinity; in both, the force on each of the two bodies is equal in magnitude and opposite in direction. Different: gravity is always attractive whereas the electric force can attract or repel; gravitational potential is always negative whereas electric potential takes the sign of the charge; and an electric field can be shielded by a conductor whereas nothing screens gravity. A fourth difference worth a mark: their constants differ by about twenty orders of magnitude.
10. The electric repulsion between two protons is about \( 10^{36} \) times their gravitational attraction. Explain why gravity nonetheless determines the structure of the Solar System.
Because charge cancels and mass does not. Bulk matter is electrically neutral to an extraordinary precision, so the vast positive and negative charges in a planet produce almost exactly no net electric field outside it. Mass has only one sign, so every particle’s gravitational contribution adds with nothing to subtract. Over astronomical amounts of matter the force that only adds up wins, despite being the far weaker one between any individual pair.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Classroom — Static Electricity and Magnetism
  • HyperPhysics — electric field, potential and magnetic field concepts