D. Fields

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.

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.

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.

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 $W=mg$.

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.

Interpret the sign

Moving a mass farther from an attracting body increases gravitational potential energy toward zero and requires positive external work if done slowly. Moving it inward makes the potential energy more negative; the gravitational field can do positive work and transfer potential energy into kinetic energy.

Use the field model

Gravitational force is conservative, so the work between two fixed positions depends only on the endpoints, not the path. Near Earth’s surface, where $g$ is approximately constant, changes can be approximated by $\Delta E_p=mg\Delta h$; for large distances use the field-based potential model rather than a constant-g approximation.

Worked example — 1.0 kg at Earth’s surface

Using $M=6.0\times10^{24}\,\mathrm{kg}$, $m=1.0\,\mathrm{kg}$ and $r=6.4\times10^6\,\mathrm{m}$, $E_p=-(6.67\times10^{-11})(6.0\times10^{24})(1.0)/(6.4\times10^6)=-6.3\times10^7\,\mathrm{J}$. The negative result means energy must be supplied to separate the pair to infinity.

Interpret the negative sign

At every finite separation, $E_p<0$ because the masses form a bound configuration relative to infinity. Increasing $r$ makes $E_p$ less negative; decreasing $r$ makes it more negative. The change in potential energy is what matters when comparing two positions.

Connect to a circular orbit

For a satellite in a circular orbit, the gravitational potential energy is still $-GMm/r$. If the orbital relation gives $K=GMm/(2r)$, then the total mechanical energy is $E_T=K+E_p=-GMm/(2r)$. This orbit result is a consequence of the circular-orbit model, not a replacement for the general two-body potential-energy equation.

Worked example — orbital potential

At $r=7.9\times10^6\,\mathrm{m}$ from Earth’s centre, with $M=6.0\times10^{24}\,\mathrm{kg}$, $V_g=-(6.67\times10^{-11})(6.0\times10^{24})/(7.9\times10^6)=-5.1\times10^7\,\mathrm{J\,kg^{-1}}$. The negative value is potential energy per kilogram relative to zero at infinity.

Interpret the sign

At finite distance the potential is negative because the field does work as an attracting mass is brought inward from infinity. Moving outward increases $V_g$ toward zero; moving inward makes it more negative. The potential difference between two points is what determines work: $W=m\Delta V_g$.

Worked example — graph gradient

If a tangent changes by $3.8\times10^8\,\mathrm{J\,kg^{-1}}$ over $4.2\times10^7\,\mathrm{m}$, then $|g|=(3.8\times10^8)/(4.2\times10^7)=9.0\,\mathrm{J\,kg^{-1}m^{-1}}=9.0\,\mathrm{N\,kg^{-1}}$. Direction comes from the negative gradient.

Read a potential–distance graph

The gradient is $\Delta V_g/\Delta r$, with units $\mathrm{J\,kg^{-1}m^{-1}}=\mathrm{N\,kg^{-1}}$. A negative slope gives a positive outward radial magnitude only after the vector direction and sign convention are interpreted. Near a source, the potential changes more rapidly with distance, so the field is stronger.

Connect to work

For a mass $m$ moved between two points, $W=m\Delta V_g$ is the work done on the mass by the external agent under the stated sign convention. The field strength relation is local; potential difference and work compare endpoints.

Worked example — changing orbit

For $m=850\,\mathrm{kg}$, $V_{g,1}=-5.07\times10^7\,\mathrm{J\,kg^{-1}}$ and $V_{g,2}=-5.40\times10^7\,\mathrm{J\,kg^{-1}}$, $W_{\mathrm{on}}=850[(-5.40)-(-5.07)]\times10^7=-2.8\times10^9\,\mathrm{J}$. The negative result means the satellite must lose energy to enter the lower orbit.

Distinguish the work agent

The work done by the gravitational field is the negative of the work done on the mass by an external agent when the motion is quasistatic:
$$W_{\text{field}}=-m\Delta V_g=m(V_{g,1}-V_{g,2})$$
Moving outward raises potential toward zero, so the field does negative work; moving inward lowers potential, so the field does positive work.

Read a graph

If a potential–distance graph gives $V_{g,1}$ and $V_{g,2}$, use their difference, not the area under the graph. An area under a force–distance graph can represent work, but a potential graph already gives work per unit mass through its vertical difference.

Read the geometry

Around an isolated spherical mass, equipotential surfaces are concentric spheres; in a two-dimensional diagram they appear as concentric circles. For multiple masses, the shape is distorted by the scalar sum of the individual potentials. The numerical spacing of drawn surfaces is a choice, so use labelled potential values rather than assuming equal physical spacing means equal potential difference.

Compare movements

The external work needed to move a mass slowly between surfaces is $W_{\text{on}}=m\Delta V_g$. The greatest work for a fixed mass occurs for the largest potential difference, not automatically for the longest geometric path. Equipotentials help identify where the potential changes and where the field is strong.

Apply it to a radial field

Around an isolated spherical mass, equipotential surfaces are concentric spheres and field lines are radial. In a two-dimensional sketch, draw concentric equipotential circles and radial field arrows crossing them normally toward the mass.

Read field strength

If equal potential intervals are drawn, closer equipotential lines mean a larger potential gradient and therefore a stronger field. Farther spacing indicates a weaker field. The line/surface geometry gives direction and relative strength; the potential labels give the quantitative difference.

Derive the model

At the limiting escape condition, initial kinetic energy supplies the increase in potential energy from $-GMm/r$ to zero at infinity. The escaping object’s mass cancels.

Worked example — Earth

With $M=6.0\times10^{24}\,\mathrm{kg}$ and $r=6.4\times10^6\,\mathrm{m}$, $v_{\mathrm{esc}}=\sqrt{2(6.67\times10^{-11})(6.0\times10^{24})/(6.4\times10^6)}=1.1\times10^4\,\mathrm{m\,s^{-1}}$, about $11\,\mathrm{km\,s^{-1}}$. This is the minimum no-drag speed for zero speed at infinity.

Read the scaling

Escape speed increases with the square root of source mass and decreases with the square root of distance from its centre. For bodies with the same density, $M\propto R^3$, so at the surface $v_{\rm esc}\propto R$. Always use the source centre-to-point distance $r$.

Worked example — lunar orbit

At $100\,\mathrm{km}$ above the Moon, $r=1.737\times10^6+0.100\times10^6=1.837\times10^6\,\mathrm{m}$. With $M=7.35\times10^{22}\,\mathrm{kg}$, $v=\sqrt{(6.67\times10^{-11})(7.35\times10^{22})/(1.837\times10^6)}=1.63\times10^3\,\mathrm{m\,s^{-1}}$.

Compare orbits

At the same central mass, orbital speed decreases as $r^{-1/2}$. A satellite at a smaller circular-orbit radius moves faster. The satellite mass does not affect the required speed in this ideal model.

Explain the speed increase

For a circular orbit, $v=\sqrt{GM/r}$. As drag lowers $r$, the new circular orbital speed is larger, so the satellite speeds up as it spirals inward even though drag is an opposing force at every instant.

Follow the feedback

At lower altitude the atmosphere is generally denser, and the higher orbital speed can increase the drag effect. Continued energy loss can therefore make the orbit decay further, eventually leading to re-entry, burning or impact depending on the body and conditions.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

Interpret energy transfer

Work done by an external agent changes the electric potential energy. If a positive charge moves toward a negative source, the electric field can do positive work while the potential energy decreases. If an electron is accelerated through a potential difference, the lost electric potential energy can become kinetic energy.

Track the system

For several charges, electric potential energy belongs to the whole charge configuration and is built from pair interactions. State the reference and the system before assigning a sign; do not confuse the energy of a charge configuration with electric potential at one point.

Worked example — two negative conductors

Radii $2.5\,\mathrm{cm}$ and $1.5\,\mathrm{cm}$, separated by a $1.7\,\mathrm{cm}$ surface gap, give $r=5.7\times10^{-2}\,\mathrm{m}$. For charges $-4.7\times10^{-8}\,\mathrm{C}$ and $-6.3\times10^{-8}\,\mathrm{C}$, $E_p=(8.99\times10^9)q_1q_2/r=+4.7\times10^{-4}\,\mathrm{J}$. Positive energy matches repulsion between like charges.

Interpret the sign

Like charges give positive potential energy because work is required to bring them together. Unlike charges give negative potential energy because the electric field releases energy as they approach. Increasing separation moves the energy toward zero.

Use changes in energy

When a charge configuration changes, calculate the final minus initial potential energy. The external work and work done by the electric field have opposite signs under a quasistatic convention. Use the actual separation between charge centres, not the physical radius of either object.

Add potentials algebraically

Electric potential is a scalar, so contributions from several point charges add with their signs:
$$V_e=\sum_i k\frac{Q_i}{r_i}$$
There is no vector-angle calculation when combining potential values.

Interpret the result

A positive source contributes positive potential and a negative source contributes negative potential. A point can have zero net potential because positive and negative contributions cancel, even though the electric field there is not necessarily zero.

Worked example — negative source

For $Q=-1.00\times10^{-8}\,\mathrm{C}$ at $r=1.00\,\mathrm{m}$, $V_e=(8.99\times10^9)(-1.00\times10^{-8})/(1.00)=-89.9\,\mathrm{V}$. At $2.00\,\mathrm{m}$, it is $-45.0\,\mathrm{V}$: farther away, the negative potential increases toward zero.

Combine sources

For several point charges, calculate each $kQ_i/r_i$ and add the scalar values. Use centre-to-point distance and convert all distances and charges before substitution. The potential does not depend on the test charge used to define it.

Check conducting spheres

Inside a charged conducting sphere in electrostatic equilibrium, the electric field is zero and the potential is constant throughout the interior. The potential need not be zero; it equals the surface potential for the ideal spherical case.

Worked example — tangent gradient

If a tangent changes from $-26\,\mathrm{kV}$ to $0\,\mathrm{V}$ over $8.0\,\mathrm{cm}=8.0\times10^{-2}\,\mathrm{m}$, $E=-[0-(-26\times10^3)]/(8.0\times10^{-2})=-3.3\times10^5\,\mathrm{V\,m^{-1}}$. The negative sign gives the field direction in the chosen coordinate.

Interpret the sign

The negative sign means the electric field points toward decreasing potential. A negative slope of $V_e$ against position corresponds to a positive field component in that coordinate direction; state whether the question wants a signed component or a magnitude.

Connect field to motion

A negative charge experiences force opposite to the electric field. Therefore its acceleration direction is opposite to the direction of decreasing potential, even though the field itself is always defined using a positive test charge.

Worked example — moving between equipotentials

A $+2.0\,\mathrm{C}$ charge moves from $40\,\mathrm{V}$ to $20\,\mathrm{V}$. Then $W_{\mathrm{on}}=(2.0)(20-40)=-40\,\mathrm{J}$. Its electric potential energy falls by $40\,\mathrm{J}$; if free, that energy can become kinetic energy.

Connect to kinetic energy

If only the electric field does work, $W_{\rm field}=-q\Delta V_e$, and the change in kinetic energy equals this work. A positive charge moving to lower potential can gain kinetic energy; a negative charge may gain kinetic energy moving to higher potential.

Use endpoints

Because electrostatic fields are conservative, the work between two points does not depend on the path. Motion along an equipotential has $\Delta V_e=0$ and therefore zero work by the field.

Read spacing carefully

If equal potential intervals are drawn, closer equipotential lines indicate a larger potential gradient and stronger electric field. Use labels and the drawing convention; arbitrary visual spacing alone does not provide a numerical field.

Apply it between plates

For oppositely charged parallel plates, field lines are approximately straight and perpendicular to the plates; equipotential surfaces are parallel to the plates. This is why the potential changes across the separation but remains constant along a plate.

Predict motion

A positive charge accelerates along the electric field, toward lower potential. A negative charge accelerates opposite to the field, toward higher potential. The field direction and particle-force direction must be kept separate.

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.

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.

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.

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.

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 $F=BIL\sin\theta$.

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 $\mathrm{kg\,s^{-2}}$.

Worked example — square loop

For side $6.2\,\mathrm{cm}=6.2\times10^{-2}\,\mathrm{m}$, $A=(6.2\times10^{-2})^2\,\mathrm{m^2}$. In $B=4.3\times10^{-4}\,\mathrm{T}$ with $\theta=45^\circ$, $\Phi=BA\cos\theta=1.2\times10^{-6}\,\mathrm{Wb}$. For $N$ identical turns, flux linkage is $N\Phi$.