Electric potential | PhysicsRank

A charge sees an energy landscape that we can describe without following every force arrow. But changing the charge sign changes which way is downhill.

We connect potential differences to work, equipotential geometry, particle motion and the energy needed to assemble a charge arrangement.

What you will learn to do

Distinguish potential from the energy of a particular charge.
Predict work signs for positive and negative probes.
Use scalar superposition without confusing zero potential with zero field.
Interpret equipotentials and conductor interiors.
Check particle and assembly energy with explicit reference conventions.

Useful before this chapterElectric forces, fields and charge distributions →Work, stored energy and power →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

Does moving to a higher potential always increase a particle’s potential energy?

Compare your reasoning

No. The change is qΔV, so the sign of the moving charge matters. A negative charge loses electric potential energy when potential increases.

Revisit the explanation →

01 / Electric potential and energy

Voltage tells a charge what an energy change would cost

A voltmeter does not count how much energy is stored in an object. It compares two places. If a charge moves between them, the potential difference tells us the change in electric potential energy per unit charge. That distinction is why a small charge and a large charge can cross the same voltage but exchange different amounts of energy.

Name both the source charges and the moving charge. In this chapter the sources are fixed, their fields are electrostatic, and the moving charge is small enough not to rearrange them. Choose a zero of potential once and use it consistently. Adding the same constant to every potential changes no potential difference and no electric force.

Positive and negative charges read the same potential map but have opposite potential-energy changes. A positive charge moving to lower potential loses electric potential energy. A negative charge making the same trip gains it. “Downhill” in potential therefore means downhill in energy only after you specify the charge sign.

\Delta U=q(V_B-V_A),\qquad W_{\rm field}=-\Delta U

Potential difference is measured in J/C = V. The field’s work is the energy lost by the electric interaction, with the fixed sources included in that interaction.

Explain this equation

Read it in words

A charge gains electric potential energy equal to its charge multiplied by the change in electric potential. The electric field does the opposite amount of work.

Symbols and units

$\Delta U$: Final electric potential energy minus initial electric potential energy.Joules (J).
$q$: The signed charge being moved; retain a negative sign for an electron.Coulombs (C).
$V_A,\ V_B$: Electric potentials at the starting point A and finishing point B, using the same reference.Volts (V); 1 V = 1 J/C.
$W_{\rm field}$: Work done by the electric field on the moving charge.Joules (J).

Signs and operations

Δ means final minus initial. The minus sign in field work says that energy given to motion by the field comes from electric potential energy. It does not mean that field work is always negative.

When it applies

Use a fixed electrostatic source configuration and a test charge that does not significantly rearrange it. If a changing magnetic field induces a nonconservative electric field, a scalar potential difference alone does not determine all electric work.

Make it concrete

For q = −2 μC and a potential rise of +30 V, ΔU = −60 μJ and the field does +60 μJ of work. A negative charge can lose potential energy while moving to higher potential.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \Delta U=q(V_B-V_A),\qquad W_{\rm field}=-\Delta U
Read it as: A charge gains electric potential energy equal to its charge multiplied by the change in electric potential. The electric field does the opposite amount of work.
Symbols:
\Delta U: Final electric potential energy minus initial electric potential energy.; Joules (J).
q: The signed charge being moved; retain a negative sign for an electron.; Coulombs (C).
V_A,\ V_B: Electric potentials at the starting point A and finishing point B, using the same reference.; Volts (V); 1 V = 1 J/C.
W_{\rm field}: Work done by the electric field on the moving charge.; Joules (J).
Signs and operations: Δ means final minus initial. The minus sign in field work says that energy given to motion by the field comes from electric potential energy. It does not mean that field work is always negative.
Conditions: Use a fixed electrostatic source configuration and a test charge that does not significantly rearrange it. If a changing magnetic field induces a nonconservative electric field, a scalar potential difference alone does not determine all electric work.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

A negative charge climbs the energy hill

A charge of −4.0 nC moves from A at 180 V to B at 90 V in a fixed electrostatic field. Find its potential-energy change and the field work.

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STOP & THINK

Add 50 V to the assigned potential at every point. Which changes?

The electric field
The voltage between two fixed points
Neither the field nor that voltage

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Focus: Voltage tells a charge what an energy change would cost
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.
Equations in this section (LaTeX):
\Delta U=q(V_B-V_A),\qquad W_{\rm field}=-\Delta U
Meaning: Potential difference is measured in J/C = V. The field’s work is the energy lost by the electric interaction, with the fixed sources included in that interaction.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

02 / Electric potential and energy

A longer route does not cost more electrostatic work

Imagine carrying a test charge between two marked points around a fixed source. One route goes straight across; another wanders sideways first. The electrostatic force can vary throughout either journey, but its total work depends only on the endpoints. Extra distance alone does not determine work: only the component of force along each small displacement contributes.

The mathematical statement is a line integral. The minus sign makes potential decrease in the direction of the electric field. In a uniform field, only the displacement along that field enters the potential change; sideways segments contribute zero. A complete closed electrostatic trip returns to the same potential.

This statement has a boundary. A changing magnetic flux can produce an induced electric field whose closed-loop circulation is nonzero. Use the induction chapter for that case. Here, path independence belongs to the specified stationary electrostatic model.

V_B-V_A=-\int_A^B\mathbf E\cdot d\mathbf l

For a uniform field this becomes −E·(rB−rA). A curved route does not alter the endpoints.

Explain this equation

Read it in words

To find the potential change from A to B, add the electric field component along each tiny step of the path, then reverse the sign.

Symbols and units

$V_B-V_A$: Potential at the end minus potential at the start.Volts (V).
$\mathbf E$: Electric field vector at each point on the chosen route.Newtons per coulomb (N/C), equivalently V/m.
$d\mathbf l$: A tiny displacement vector pointing along the route from A toward B.Metres (m).
$\int_A^B$: An integral: add the tiny contributions along the full route.An operation; E · dl contributes volts.
$\cdot$: Dot product: multiply the field component along a step by that step’s length.An operation; perpendicular contributions are zero.

Signs and operations

The minus sign makes potential decrease when a step follows the electric field. Reversing the route reverses the result. Bold symbols carry directions; potential is a scalar.

When it applies

This path-independent potential relation applies to an electrostatic field. In a time-varying magnetic environment, the full electric field generally cannot be represented by a scalar potential gradient alone.

Make it concrete

In a uniform field of 40 V/m toward the right, moving 0.30 m right gives ΔV = −12 V. A further vertical step gives no additional change because it is perpendicular to the field.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): V_B-V_A=-\int_A^B\mathbf E\cdot d\mathbf l
Read it as: To find the potential change from A to B, add the electric field component along each tiny step of the path, then reverse the sign.
Symbols:
V_B-V_A: Potential at the end minus potential at the start.; Volts (V).
\mathbf E: Electric field vector at each point on the chosen route.; Newtons per coulomb (N/C), equivalently V/m.
d\mathbf l: A tiny displacement vector pointing along the route from A toward B.; Metres (m).
\int_A^B: An integral: add the tiny contributions along the full route.; An operation; E · dl contributes volts.
\cdot: Dot product: multiply the field component along a step by that step’s length.; An operation; perpendicular contributions are zero.
Signs and operations: The minus sign makes potential decrease when a step follows the electric field. Reversing the route reverses the result. Bold symbols carry directions; potential is a scalar.
Conditions: This path-independent potential relation applies to an electrostatic field. In a time-varying magnetic environment, the full electric field generally cannot be represented by a scalar potential gradient alone.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

An original cross-section of a positive point charge’s equipotentials. A and C share radius r and potential; B is at 2r. The curved route A→C→B and the radial route A→B have the same electrostatic work.

WORKED EXAMPLE

Separate along-field motion from sideways motion

A uniform electric field is 150 N/C in the +x direction. A test charge moves from (0, 0) to (0.40 m, 0.30 m). Find the potential change.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Focus: A longer route does not cost more electrostatic work
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.
Equations in this section (LaTeX):
V_B-V_A=-\int_A^B\mathbf E\cdot d\mathbf l
Meaning: For a uniform field this becomes −E·(rB−rA). A curved route does not alter the endpoints.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

03 / Electric potential and energy

Add potentials as signed scalars, then ask separately about fields

For an isolated point charge with zero potential at infinity, integrating its inverse-square field gives a potential proportional to inverse distance. A negative source contributes negative potential. Distance is always positive; the source sign belongs in the numerator.

For several fixed point charges, add their potentials as ordinary signed numbers. You do not resolve potential into x and y components. Electric field is a vector, however, and its contributions must still be added with directions. A location with zero potential can have a large field; a location with zero field can have nonzero potential.

The zero-at-infinity convention works for finite localized charge arrangements. Some ideal infinite distributions need a different reference because the integral to infinity does not converge. State a finite reference when the model requires one rather than silently using a point-charge convention everywhere.

V(P)=\sum_i\frac{kQ_i}{r_i},\qquad k=\frac1{4\pi\varepsilon_0}

Every ri is the distance from source i to the observation point. Do not include the test charge in the source sum used to find its external potential.

WORKED EXAMPLE

Zero potential can hide a strong field

A +3.0 nC source is 0.30 m to the left of P and a −2.0 nC source is 0.20 m to its right. Take k = 9.0×10⁹ N m²/C². Find potential and field at P.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Focus: Add potentials as signed scalars, then ask separately about fields
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.
Equations in this section (LaTeX):
V(P)=\sum_i\frac{kQ_i}{r_i},\qquad k=\frac1{4\pi\varepsilon_0}
Meaning: Every ri is the distance from source i to the observation point. Do not include the test charge in the source sum used to find its external potential.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

04 / Electric potential and energy

An equipotential is a contour, not a region with no field

On a map of height, walking along a contour keeps your altitude fixed. An equipotential plays a similar geometric role: every point on it has the same V. Moving a test charge along it changes no electric potential energy, even if a nonzero electric force acts perpendicular to the route.

Where the field is nonzero, it is perpendicular to a smooth equipotential surface. Any tangential field component would do work along that surface and contradict constant potential. Closely spaced contours indicate a large field only when the labelled voltage intervals are equal; crowding lines with arbitrary labels is not a quantitative measurement.

In electrostatic equilibrium, an ideal conductor has zero field within its material and is at a single potential. That potential need not be zero. An empty closed cavity with no enclosed charge also has zero electrostatic field. A cavity containing charge requires a separate argument; do not extend the empty-cavity result without checking the contents.

WORKED EXAMPLE

Zero field inside does not mean zero potential

An isolated spherical conducting shell of radius 0.20 m carries +6.0 nC. Its cavity is empty and external fields are absent. Take zero potential at infinity and k = 9.0×10⁹. Find potential at its centre.

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STOP & THINK

A positive test charge moves along one equipotential. Which conclusion follows?

The field is zero at every point
The electrostatic work is zero
The test charge has zero potential energy

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Focus: An equipotential is a contour, not a region with no field
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

05 / Electric potential and energy

Choose energy for speed, force for the path

A potential map often gives a speed more directly than a trajectory. When only the electrostatic force does work, kinetic plus electric potential energy is conserved. You can find the final kinetic energy from the endpoint voltage without knowing how long the trip takes.

This shortcut does not prove that every mathematically possible destination is dynamically reachable. Geometry, the direction of the force and any angular momentum can restrict the actual path. It also does not provide the direction of final velocity: energy is a scalar account.

For a particle released from rest, require the predicted kinetic energy to be nonnegative. An electron naturally gains kinetic energy toward higher electric potential, because its charge is negative. At sufficiently high energies use relativistic kinetic energy; the voltage-to-energy change qΔV remains the starting account.

K_B-K_A=-q(V_B-V_A)

This relation assumes no other work or energy loss. If an external force acts, include its work explicitly.

WORKED EXAMPLE

An electron moves toward higher potential

An electron starts with 30 eV of kinetic energy at A and reaches B, whose potential is 70 V higher. No other force does work. Find its final kinetic energy.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Focus: Choose energy for speed, force for the path
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.
Equations in this section (LaTeX):
K_B-K_A=-q(V_B-V_A)
Meaning: This relation assumes no other work or energy loss. If an external force acts, include its work explicitly.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

06 / Electric potential and energy

Count each interacting pair once

The expression qV describes the interaction of one chosen charge with a specified external source field. To assemble a whole set of charges from infinity, add charges one at a time. The first has no pair interaction yet; each later charge interacts with those already placed.

This process counts every distinct pair exactly once. Summing qiVi over all particles without a factor of one half counts each pair twice, because each charge appears once as a source and once as a probe. Ignore the divergent self-energy of ideal point charges; the assembly formula below describes mutual interaction energy.

Negative assembly energy means an external agent can extract energy during a sufficiently slow assembly from infinity. It does not mean every force vanishes or that the configuration is automatically stable. Supports may be required to keep the final charges fixed.

U=\sum_{i<j}\frac{kq_iq_j}{r_{ij}}=\frac12\sum_i q_iV_{\rm others}(i)

The zero is infinitely separated particles, and each Vi excludes the charge’s own field.

WORKED EXAMPLE

A zero total assembled energy

Three fixed charges +2 nC, +2 nC and −1 nC occupy the corners of an equilateral triangle of side 0.30 m. Use k = 9×10⁹. Find their mutual interaction energy.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Focus: Count each interacting pair once
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.
Equations in this section (LaTeX):
U=\sum_{i<j}\frac{kq_iq_j}{r_{ij}}=\frac12\sum_i q_iV_{\rm others}(i)
Meaning: The zero is infinitely separated particles, and each Vi excludes the charge’s own field.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Electric potential and energy

Use the quantity that answers the question

Begin by naming the requested output. A force or acceleration needs the vector field and the test charge. A voltage needs a scalar source sum or a field integral. Work and speed often need only a potential difference. Assembly energy needs all distinct pairs. These are related tools, not interchangeable formulas.

For a continuous source, replace the scalar sum by an integral over small charge elements. On the axis of a uniformly charged ring every element has the same distance from the observation point, so the potential integral is especially simple. Differentiating that potential recovers the axial field and exposes why the field is zero at the centre while potential is not.

Finish by testing a limiting case, a reference shift or an independent energy balance. A plausible-looking number is not enough: the sign should agree with the charge’s physical response and the units should match the requested quantity.

V(x)=\frac{kQ}{\sqrt{R^2+x^2}},\qquad E_x=-\frac{dV}{dx}=\frac{kQx}{(R^2+x^2)^{3/2}}

Uniform thin ring of fixed radius R in vacuum, with zero potential at infinity. The derivative is with respect to the axial coordinate.

WORKED EXAMPLE

Potential remains when symmetric fields cancel

A thin ring carries +4.0 nC uniformly around its circumference and has radius 0.12 m. In vacuum, find potential and axial field at its centre using k = 9.0×10⁹ and zero potential at infinity.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Electric potential and energy
Focus: Use the quantity that answers the question
Learning goals:
- Distinguish potential from the energy of a particular charge.
- Predict work signs for positive and negative probes.
- Use scalar superposition without confusing zero potential with zero field.
- Interpret equipotentials and conductor interiors.
- Check particle and assembly energy with explicit reference conventions.
Equations in this section (LaTeX):
V(x)=\frac{kQ}{\sqrt{R^2+x^2}},\qquad E_x=-\frac{dV}{dx}=\frac{kQx}{(R^2+x^2)^{3/2}}
Meaning: Uniform thin ring of fixed radius R in vacuum, with zero potential at infinity. The derivative is with respect to the axial coordinate.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

State the source configuration, test charge and potential reference.
Electrostatic work depends on endpoints; a longer path alone does not mean more work.
Equal potential does not imply zero field.
Count each pair once when assembling a charge system.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

Two routes join the same endpoints in a stationary electrostatic field. Which field work is larger?

The longer route
The route with more bends
They are equal

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SELF CHECK 2

A +3 nC probe moves from 20 V to 80 V. What work does the field do, in nJ?

Answer in nJ.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 3

Why is a factor 1/2 needed in ΣqiVothers(i)?

Each pair was counted twice
Potential is a vector
Every charge is only half assembled

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

Can potential vanish at a location where the electric field is nonzero?

Compare your reasoning

Yes. Signed scalar potentials may cancel while directed field contributions add. Potential value and local potential slope answer different questions.

Revisit the explanation →

A few days later: change the situation

You know qV for one particle. Is adding that expression for every particle enough to get total mutual interaction energy?

Compare your reasoning

Only after correcting for double counting and excluding self-potentials. Each pair appears in both particles’ terms, so the full sum needs a factor of one half.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

Back to chapter contents ↑