D. Fields

State the three laws

1. A planet follows an elliptical orbit with the star at one focus.
2. The line from the star to the planet sweeps out equal areas in equal time intervals.
3. For bodies orbiting the same central star, the square of the orbital period is proportional to the cube of the semi-major axis: $T^2\propto a^3$.

Interpret the geometry

The semi-major axis $a$ is half the longest diameter of the ellipse. In an elliptical orbit the planet is closer to the star at one focus-side end and farther away at the other. Equal swept areas mean the planet moves faster when it is closer to the star and slower when it is farther away.

Compare orbital systems

For two planets around the same star,
$$\frac{T_Y}{T_X}=\left(\frac{a_Y}{a_X}\right)^{3/2}$$
Use the semi-major axes, not automatically the instantaneous distance from the star. The third law comparison assumes the same central mass.

Common trap

The star is at a focus, not generally at the centre of the ellipse. Also, the second law refers to equal swept areas, not equal arc lengths or equal distances travelled.

Core idea

Any two point masses attract along the line joining them. In the equation below, $m_1$ and $m_2$ are the masses, $r$ is their centre-to-centre separation, and $G=6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$.

Worked example — Earth and a book

For $m=1.0\,\mathrm{kg}$, $M=6.0\times10^{24}\,\mathrm{kg}$ and $r=6.4\times10^6\,\mathrm{m}$, $F=(6.67\times10^{-11})(1.0)(6.0\times10^{24})/(6.4\times10^6)^2=9.8\,\mathrm{N}$. This is the book’s weight; the book attracts Earth with the same force magnitude.

Read the scaling

Doubling either mass doubles the force. Doubling the separation reduces the force to one quarter. The two masses exert equal-magnitude forces on each other in opposite directions; the equation gives the interaction force, not two independent forces on the same object.

Choose the model

Use the point-mass equation when the bodies can be treated as point masses, or when a spherically symmetric body is outside its surface and the centre-to-centre separation is used. For extended irregular bodies, the simple centre-to-centre model may not be valid.

Common trap

Do not use the radius of one body as $r$ unless the other mass is effectively at its centre. Convert kilometres to metres before substituting, and remember that gravitational force is attractive rather than repulsive.

Core idea

The point-mass model replaces an extended body by a single mass at a representative point, usually its centre of mass. It is suitable when the body’s size is negligible compared with the separation involved, or when the body is spherically symmetric and the point of interest is outside it.

Use spherical symmetry

A satellite orbiting a spherical planet of uniform density can be modelled as if the planet’s entire mass were concentrated at its centre. The gravitational force then depends on the centre-to-centre distance. The same approximation can be used for two spherically symmetric bodies when their separation is measured between centres.

Check the limit

The approximation is not automatically valid for nearby irregular bodies, for points inside an extended body, or when the object’s size is comparable with the separation. In those cases different parts of the mass are at significantly different distances and their contributions cannot be represented by one point without further justification.

Common trap

Do not justify the model only by saying that the planet is “large”. The relevant reasons are small satellite-to-planet size ratio and/or spherical symmetry with an external point of interest.

Define the field

Gravitational field strength is force per unit test mass. For a point or spherical source of mass $M$, it depends on distance $r$ from the source centre. It is a vector directed toward the source, with unit $\mathrm{N\,kg^{-1}}$, numerically equivalent to $\mathrm{m\,s^{-2}}$.

Worked example — surface field

For $M=4.87\times10^{24}\,\mathrm{kg}$ and $r=6.05\times10^6\,\mathrm{m}$, $g=(6.67\times10^{-11})(4.87\times10^{24})/(6.05\times10^6)^2=8.87\,\mathrm{N\,kg^{-1}}$. The result is the force per kilogram at the surface, directed inward.

Read the scaling

At a fixed distance, $g$ is proportional to $M$. At a fixed source mass, doubling $r$ reduces $g$ to one quarter. The field direction is toward the source mass; the scalar expression gives the magnitude. Near a surface, the weight of a mass $m$ is $W=mg$.

Common trap

Do not use the object’s own mass in $g=GM/r^2$ as $M$, and do not use altitude alone for $r$: use distance from the source centre.

Interpret a field line

A gravitational field line is a drawn line whose tangent gives the direction of the gravitational field at each point. Arrows point toward the mass creating the field because gravity is attractive. Field lines are a representation of the vector field, not physical paths followed by test masses.

Read the pattern

Around an isolated point mass or spherical mass, field lines are radial and point inward. Where lines are closer together, the field is stronger; where they are farther apart, it is weaker. This matches the inverse-square decrease of field strength with distance.

Combine sources

For more than one source mass, the net field is the vector sum of the individual fields. At a point between two stars, draw each contribution along the line joining that star to the point and point each arrow toward its source; the resultant can then be found by vector addition.

Common trap

Do not draw field lines as zigzags, make them cross, or point them away from a positive-looking “source” label: gravitational field arrows always point toward mass. Line density indicates relative strength, not a separate force on each line.

Use the charge signs

There are two types of electric charge. Like charges repel: positive–positive and negative–negative. Unlike charges attract: positive–negative. The force on each charge acts along the line joining the two charges, with equal magnitude and opposite direction.

Draw the interaction

For two like point charges, draw arrows away from each other. For two unlike point charges, draw arrows toward each other. The direction is determined by the sign combination; the force magnitude also depends on charge magnitudes and separation, which Coulomb’s law quantifies.

Extend to a third charge

If a third charge is present, find the force from each other charge separately and add the force vectors. Do not decide the net direction by charge sign alone: compare the individual vectors and their magnitudes.

Common trap

Do not say that a negative charge always repels or that a positive charge always attracts. Attraction and repulsion depend on the pair of charges, and Newton’s third-law pair acts on different charges.

Use the inverse-square law

For two point charges, $r$ is their centre-to-centre separation. In a medium of permittivity $\varepsilon$, $k=1/(4\pi\varepsilon)$; in vacuum, $k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}$. Calculate magnitude, then use charge signs to state attraction or repulsion.

Worked example — unlike charges in air

For $q_1=4.5\times10^{-8}\,\mathrm{C}$, $q_2=-1.3\times10^{-7}\,\mathrm{C}$ and $r=3.2\times10^{-2}\,\mathrm{m}$, $F=(8.99\times10^9)|q_1q_2|/r^2=5.1\times10^{-2}\,\mathrm{N}$. The force is attractive because the charges have opposite signs.

Read the scaling

Doubling either charge doubles the force. Doubling the separation reduces the force to one quarter. If the medium has permittivity $\varepsilon$ rather than $\varepsilon_0$, use $k=1/(4\pi\varepsilon)$; greater permittivity reduces the force for the same charges and separation.

Choose the point-charge model

Spherical charged bodies can be treated as point charges at their centres when the geometry permits. Use centre-to-centre separation and convert charge units, such as microcoulombs, before substitution. For several charges, calculate each force vector and add them.

Common trap

Do not use diameter or a single radius as r, and do not forget that a change in separation is squared. Keep the force magnitude positive in the calculation, then state attraction or repulsion separately.

Core idea

Electric charge is conserved: in an isolated system, the total charge before an interaction equals the total charge after it. Charge can move between objects, but it is not created or destroyed in the transfer.

Use it at a junction

In a steady circuit, charge does not accumulate at a junction. The current entering equals the current leaving, for example $I_1=I_2+I_3$. This is a consequence of charge conservation, not a separate rule that overrides it.

Track the system boundary

When charge appears to change on one object, include the other object, the conductor or the ground in the system. Electrons may move across the chosen boundary, so the object’s charge changes while the total charge of the larger isolated system remains constant.

Common trap

Do not answer “Kirchhoff’s law” alone when asked for the fundamental law behind current balance. State conservation of electric charge.

Set the force balance

Millikan observed charged oil drops between parallel plates. By adjusting the potential difference, the electric force on a drop can balance its weight so the drop is stationary. With $E=V/d$, the balance is
$$qE=mg\quad\Rightarrow\quad q=\frac{mg}{E}=\frac{mgd}{V}$$
for the simplified model in which buoyancy is neglected.

Read the evidence

Repeating the measurement for many drops gives charges that are integer multiples of a smallest value, the elementary charge $e$: $q=ne$, where $n$ is an integer. This pattern is evidence that electric charge is quantized rather than continuously variable.

Explain the method

The experiment varies the electric field until a drop is held stationary, then uses the known mass and field to infer its charge. It is the repeated integer-multiple pattern—not one isolated drop—that supports the quantization conclusion.

Common trap

Do not say that Millikan directly measured a continuous range of charge or that the drop is uncharged when it is stationary. Stationary means the electric and gravitational forces balance; the charge is non-zero and can be calculated from the balance.

Transfer by friction

Rubbing two insulating materials can move electrons from one surface to the other. One object becomes negatively charged and the other positively charged; the total charge of the pair is conserved. The material that loses electrons is positive, and the material that gains electrons is negative.

Transfer by induction

Bring a charged object near a conductor without touching it. Charges in the conductor separate by repulsion and attraction. If the conductor is connected to ground while the charged object remains nearby, electrons can flow to or from Earth. Disconnect the ground first, then remove the external charged object, leaving the conductor with a net charge.

Transfer by contact and grounding

Touching a charged conductor to another conductor allows charge to redistribute between them. Grounding connects an object to a very large charge reservoir: electrons can leave an object or enter it, depending on the nearby charge and the object’s potential.

Common trap

Induction does not require contact with the charged rod. In a grounding sequence, remove the ground before removing the inducing charge; reversing the order can leave the conductor neutral.

Define the field

Electric field strength is force per unit positive test charge. For a point source $Q$, it is directed away from positive $Q$ and toward negative $Q$. Its SI unit is $\mathrm{N\,C^{-1}}$.

Worked example — point-charge field

At $r=1.0\,\mathrm{m}$ from $Q=+2.9\times10^{-8}\,\mathrm{C}$, $E=(8.99\times10^9)(2.9\times10^{-8})/(1.0)^2=2.6\times10^2\,\mathrm{N\,C^{-1}}$. Because the source is positive, the field points radially outward.

Add fields as vectors

For more than one source, calculate each electric-field vector at the point and add them. Do not add magnitudes unless all field vectors point in the same direction. The force on a particular charge is then $F=qE$, with its direction reversed from the field if the charge itself is negative.

Common trap

The field direction is defined using a positive test charge, not the sign of the test charge in the question. Also distinguish field strength $E$ from force $F$: changing the test charge changes F but not the source field E.

Interpret a field line

Electric field lines show the direction of the force on a small positive test charge. Their arrows point away from positive charges and toward negative charges. The tangent to a line gives the local field direction.

Read qualitative strength

Where field lines are closer together, the field is stronger; where they spread out, it is weaker. This is a qualitative representation unless the diagram specifies equal field-line intervals or a scale.

Common trap

Do not point electric field lines from negative to positive, make them cross, or draw them tangent to equipotential surfaces. Field lines follow the positive-test-charge convention.

Core idea

In a field-line diagram, greater line density represents a stronger electric field. Compare density over equal areas or equal widths of the same diagram; the visual spacing is a qualitative encoding of $|E|$, not a new physical force.

Connect density to distance

Around an isolated point charge, field lines spread as distance increases, so the field becomes weaker. A denser pattern near the charge is consistent with the inverse-square dependence of field strength. In a uniform field, equal spacing indicates constant field strength.

Check the representation

Density comparisons are meaningful only when the diagram uses the same line convention and potential/field intervals. Do not infer exact numerical values from arbitrary artwork; use labels or a scale if a calculation is required.

Common trap

Do not count field lines as individual objects or compare the total number of lines in two drawings with different scales. It is the local density that represents relative field strength.

Use the uniform-field model

Between two large opposite parallel plates, away from the edges, field strength equals potential difference $V$ divided by perpendicular plate separation $d$. The field points from the positive plate to the negative plate; edge regions are not uniform.

Worked example — required potential difference

For $E=1.0\times10^6\,\mathrm{V\,m^{-1}}$ and $d=0.50\,\mathrm{cm}=5.0\times10^{-3}\,\mathrm{m}$, $V=Ed=(1.0\times10^6)(5.0\times10^{-3})=5.0\times10^3\,\mathrm{V}$. The result applies to the uniform central region.

Read the direction

Electric field lines point from the positive plate to the negative plate, so a positive charge accelerates in that direction and a negative charge accelerates oppositely. The field strength can be expressed in $\mathrm{N\,C^{-1}}$ or equivalently $\mathrm{V\,m^{-1}}$.

Check the boundary

The formula assumes a uniform region and neglects edge effects. Use the perpendicular plate separation in metres; do not use the diagonal distance or the plate length.

Common trap

Do not reverse the field direction because the test charge is negative. Field direction is defined by a positive test charge; the force on a negative charge is opposite.

Interpret a magnetic field line

Magnetic field lines show the local direction of the magnetic field; a compass north pole or a suitable test direction follows the arrow convention. Unlike isolated electric field lines, magnetic field lines form continuous closed loops.

Use the right-hand rule

Around a long straight current-carrying wire, the field lines are concentric circles centred on the wire. Point the right thumb in the conventional current direction; curled fingers give the magnetic-field direction. If electrons move into the page, conventional current is out of the page, so reverse the electron-motion direction before applying the rule.

Common trap

Do not use electron motion as though it were conventional current, and do not draw magnetic field lines starting or ending on an isolated magnetic pole.

Start with the force

A charge in a uniform electric field experiences $F=qE$. A positive charge accelerates in the field direction; a negative charge accelerates opposite to it. In vacuum, if the field is uniform, the acceleration is constant: $a=qE/m$.

Read the trajectory

A particle initially moving perpendicular to a uniform electric field has constant velocity in the direction perpendicular to the field and constant acceleration along the field, so its path is parabolic. A particle initially at rest accelerates along a field line.

Solve the motion

Find the force and acceleration direction first, then use constant-acceleration equations. The time in the field comes from motion along the entry direction; the transverse displacement comes from the electric acceleration.

Common trap

Do not make an electron accelerate in the electric-field direction. The field direction is defined for a positive charge; an electron accelerates toward the positive plate.

Identify the magnetic force

A moving charge in a magnetic field experiences a force perpendicular to both its velocity and the field. A stationary charge, or a charge moving parallel to the field, has zero magnetic force.

Explain and measure the circular path

For velocity perpendicular to a uniform magnetic field, the force is always perpendicular to the velocity and supplies the centripetal force. The direction changes continuously while speed and kinetic energy remain constant.

Worked example — charge-to-mass ratio

A particle beam with $v=2.5\times10^7\,\mathrm{m\,s^{-1}}$ follows a circle of radius $r=7.8\,\mathrm{cm}=0.078\,\mathrm{m}$ in $B=1.8\,\mathrm{mT}=1.8\times10^{-3}\,\mathrm{T}$. Then $|q|/m=v/(Br)=1.8\times10^{11}\,\mathrm{C\,kg^{-1}}$. The path direction is needed separately to determine the sign of the charge.

Use the full motion picture

A velocity component parallel to the field is unchanged, while the perpendicular component produces circular motion. Together they can form a helical path. Magnetic force does no work because it is perpendicular to instantaneous velocity, so kinetic energy stays constant.

Common trap

Do not say a magnetic field speeds up a charge or changes its kinetic energy. It changes direction, not speed, when no electric field is present.

Separate the two forces

With perpendicular uniform electric and magnetic fields, a charged particle can experience an electric force $F_E=qE$ parallel to the electric field and a magnetic force $F_B=qvB$ perpendicular to both velocity and magnetic field. For the correct geometry, these forces can point in opposite directions.

Find the undeflected speed

For the geometry in which the two forces oppose, a particle travels straight when their magnitudes are equal. Charge magnitude and mass do not determine this selected speed.

Worked example — crossed-field selector

For $v=5.9\times10^6\,\mathrm{m\,s^{-1}}$ and $B=42\,\mathrm{mT}=0.042\,\mathrm{T}$, the balancing field is $E=vB=(5.9\times10^6)(0.042)=2.5\times10^5\,\mathrm{V\,m^{-1}}$. Across plates $0.10\,\mathrm{m}$ apart, $V=Ed=2.5\times10^4\,\mathrm{V}$.

Check the geometry

The cancellation condition depends on velocity being perpendicular to both fields and on the electric and magnetic force directions being opposite. If the particle is deflected, compare the vector forces rather than applying E/B blindly.

Common trap

Do not use the electric-field direction alone to predict the path, and do not insert the particle’s mass into v=E/B. This is a force-balance condition.

Use the magnitude equation

The angle $\theta$ is measured between the particle velocity and the magnetic field. Use charge magnitude for the force magnitude; determine direction separately with the right-hand rule and reverse it for a negative charge.

Worked example — oblique proton motion

For $|q|=1.60\times10^{-19}\,\mathrm{C}$, $v=3.4\times10^5\,\mathrm{m\,s^{-1}}$, $B=5.3\times10^{-3}\,\mathrm{T}$ and $\theta=32^\circ$, $F=|q|vB\sin\theta=1.5\times10^{-16}\,\mathrm{N}$. Only the velocity component perpendicular to the field contributes.

Find the direction

The magnetic force is perpendicular to both velocity and field. Use the right-hand rule for a positive charge; reverse the result for a negative charge. This force bends the path but does no work.

Connect to circular motion

For perpendicular motion, set $F=|q|vB$ equal to $mv^2/r$ to obtain $r=mv/(|q|B)$. Increasing $|q|$ or $B$ reduces the radius; increasing $m$ or $v$ increases it.

Common trap

Do not use the right-hand rule without reversing for an electron, and do not use $\theta$ as the angle between the field and the force. It is the angle between velocity and field.

Use the conductor equation

Here $L$ is only the conductor length inside the uniform field and $\theta$ is the angle between conventional current and the field. Direction follows the force rule for conventional current.

Worked example — measuring a field

A perpendicular wire carries $I=1.64\,\mathrm{A}$ through $L=8.13\,\mathrm{cm}=0.0813\,\mathrm{m}$ and experiences $F=4.12\times10^{-4}\,\mathrm{N}$. Thus $B=F/(IL)=(4.12\times10^{-4})/(1.64\times0.0813)=3.09\times10^{-3}\,\mathrm{T}$.

Read the angle limits

The force is zero when the wire is parallel to the field and maximum when it is perpendicular. Use conventional current direction for the force rule, not the electron drift direction.

Connect the models

The conductor formula is the combined magnetic force on moving charge carriers: $I=q/t$ and $L=vt$ lead from $F=qvB\sin\theta$ to $F=BIL\sin\theta$.

Common trap

Do not use electron motion as the current direction, and do not include wire length outside the region where the magnetic field exists.

Use the force-per-length equation

For two long, straight, parallel wires, $r$ is their perpendicular separation and $\mu_0$ is the permeability of free space. The force acts along the line joining the wires.

Worked example — opposite currents

For $I_1=3.7\,\mathrm{A}$, $I_2=1.6\,\mathrm{A}$ and $r=12\,\mathrm{cm}=0.12\,\mathrm{m}$, $F/L=\mu_0I_1I_2/(2\pi r)=9.9\times10^{-6}\,\mathrm{N\,m^{-1}}$. Opposite current directions mean the wires repel; each force has the same magnitude and opposite direction.

Read attraction and repulsion

Parallel currents in the same direction attract; currents in opposite directions repel. Each wire experiences an equal-magnitude force in the opposite direction. The result follows because each wire lies in the magnetic field produced by the other.

Check the scaling

The force per unit length increases with either current and decreases inversely with separation. Keep $F/L$ units as $\mathrm{N\,m^{-1}}$, equivalently $\mathrm{kg\,s^{-2}}$.

Common trap

Do not reverse same-direction current behaviour, and do not confuse force on a finite length with force per unit length.