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.

Define the reference

Gravitational potential energy is defined as the work done to assemble the masses of a system from infinite separation. Set the potential energy at infinite separation to zero. Because gravity is attractive, bringing masses together releases energy, so the gravitational potential energy of a bound system is negative.

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.

Common trap

Do not make gravitational potential energy positive simply because the mass is high above a planet. With zero at infinity, every finite point in the isolated attractive field has negative potential energy.

Use the two-body expression

For two point masses, or spherical bodies represented at their centres, use centre-to-centre separation $r$. The zero reference is infinite separation, so a finite bound pair has negative potential energy.

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.

Common trap

Do not omit the minus sign or use $r^2$ in the potential-energy expression. $r^2$ belongs to force and field-strength laws; potential energy varies as $1/r$.

Define potential at a point

Gravitational potential $V_g$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point. Set $V_g=0$ at infinity. Its SI unit is $\mathrm{J\,kg^{-1}}$, and it is a scalar quantity.

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: $W=m\Delta V_g$.

Common trap

Do not confuse potential $V_g$ in J kg⁻¹ with potential energy $E_p$ in joules, and do not use $r^2$: potential follows $1/r$, whereas field strength follows $1/r^2$.

Use the gradient relationship

Gravitational field strength is the negative spatial gradient of gravitational potential. For a graph, use the tangent gradient at the required point; the negative sign makes the field point toward decreasing potential.

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.

Common trap

Do not use the average slope over a wide curved section as the field at one point unless the question’s graph is effectively linear there. Do not drop the negative sign without stating whether you are reporting a vector component or a magnitude.

Use the potential difference

For a mass $m$ moving from point 1 to point 2, external work in the stated convention equals the change in gravitational potential energy. Potential is scalar, so only the endpoints matter.

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.

Common trap

Check whether the question asks for work by the field or work done on the mass. Reversing the order of the potential values changes the sign.

Define an equipotential

An equipotential surface is a surface on which every point has the same gravitational potential $V_g$. Moving a mass along one equipotential gives $\Delta V_g=0$, so no work is done by the field and no external work is required for quasistatic motion along the surface.

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.

Common trap

Do not claim that every move between nearby-looking surfaces requires equal work. Read the potential labels and the starting and ending surfaces; motion along one surface has zero potential difference.

Use the perpendicular relationship

Gravitational field lines cross equipotential surfaces at right angles. The field points in the direction of decreasing gravitational potential, so the field-line arrow is normal to the equipotential and toward lower $V_g$.

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.

Common trap

Do not draw field lines along equipotentials. Moving along an equipotential has zero potential difference, while the gravitational field points across it, toward lower potential.

Define escape speed

Escape speed is the minimum speed an object must have at a point—usually the surface of a planet—to reach infinity with zero remaining speed, assuming no air resistance and no other significant gravitational fields. It is not the speed needed to enter a circular orbit.

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 $r$.

Common trap

Do not use the circular-orbit speed $\sqrt{GM/r}$ or say that escape means “overcoming gravity” at a finite boundary. The limiting condition is reaching infinity with zero final speed.

Set the circular-orbit model

For a small mass $m$ in a circular orbit of radius $r$ around a much larger mass $M$, gravitational force supplies the centripetal force. The satellite mass cancels.

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.

Common trap

Do not use escape speed for a bound circular orbit: $v_{\rm esc}=\sqrt2\,v_{\rm orbital}$ at the same radius. Also add the planet’s radius to altitude before using $r$.

Start with energy loss

Atmospheric drag opposes the satellite’s motion and removes mechanical energy from the orbit. The total orbital energy becomes more negative, so the satellite moves to a lower orbit. It does not simply slow while remaining at the same radius.

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.

Common trap

Do not conclude that drag causes the final orbital speed to decrease merely because drag opposes motion. Distinguish the instantaneous force from the speed of the new lower circular orbit.

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.

Set the reference

Electric potential energy of a system is the work done to assemble its charges from infinite separation to their present positions. Define the energy as zero at infinite separation. The sign depends on the charge combination: bringing opposite charges together lowers the energy, while bringing like charges together raises it.

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.

Common trap

Electric potential energy is not always positive. Unlike charges have negative pair energy relative to infinity; like charges have positive pair energy. The sign is determined by the interaction and reference, not by whether a charge is an electron.

Use the pair-energy expression

For two point charges or spherical conductors represented at their centres, keep the signs of $q_1$ and $q_2$ and use centre separation $r$. The zero reference is infinite separation.

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.

Common trap

Do not use absolute values for q1q2 before deciding the sign of the energy, and do not put r² in the potential-energy formula; r² belongs to Coulomb force.

Define electric potential

Electric potential $V_e$ at a point is the work done per unit positive test charge in bringing it from infinity to that point. Its unit is $\mathrm{J\,C^{-1}}$, equivalent to volts. Set $V_e=0$ at infinity.

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.

Common trap

Do not add electric field magnitudes as though they were scalar, and do not infer zero electric field from zero potential. Potential and field are different quantities.

Use point-charge potential

Electric potential is signed and has zero at infinity. For source charge $Q$, use distance $r$ from the charge centre. Positive $Q$ gives positive potential; negative $Q$ gives negative potential.

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.

Common trap

Do not use $kQ/r^2$ for potential, and do not assume zero field means zero potential inside a conductor.

Use the gradient relationship

Electric field strength is the negative spatial gradient of electric potential. On a potential–distance graph, use the tangent gradient at the required point and state whether the answer is a signed component or a magnitude.

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.

Common trap

Do not use the graph’s potential value instead of its local gradient, and do not reverse the particle’s force direction without checking the particle’s charge sign.

Use potential difference

External work on a charge equals its change in electric potential energy. Keep the sign of $q$ and calculate final potential minus initial potential. Work by the field has the opposite sign.

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.

Common trap

Check which agent’s work the question asks for and keep the charge sign. Do not assume that moving a negative charge toward lower potential necessarily increases its kinetic energy.

Define an equipotential

An electric equipotential surface joins points with the same electric potential. Moving a charge along one surface gives $\Delta V_e=0$, so the electric field does no work.

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.

Common trap

Do not confuse equipotential lines with field lines. A charge can move along an equipotential without field work, even though the electric field may be non-zero perpendicular to the path.

Use the perpendicular relationship

Electric field lines cross equipotential surfaces at 90°. The field points toward decreasing electric potential, so the field-line arrow is normal to the equipotential and in the direction of the negative potential gradient.

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.

Common trap

Do not draw field lines parallel to equipotentials or assume a negative charge accelerates in the field direction. The field is defined using a positive test charge.

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.