Capacitance | PhysicsRank

01 / Potential, capacitors and dipoles

Potential is energy per unit charge, not another name for field

Electrostatic forces are conservative. Moving a charge q between two positions changes its electric potential energy by q times the potential difference: ΔU = qΔV. The electric field does work W_field = −qΔV. Potential V is a scalar, while electric field E is a vector.

A positive charge released from rest under electric force alone tends to accelerate toward lower potential, losing potential energy. A negative charge accelerates toward higher potential while also losing potential energy, because q is negative. The direction of decreasing U depends on the test-charge sign.

Only potential differences affect this work. Adding the same constant to V everywhere changes no field or force. A chosen zero, such as infinity for an isolated finite charge distribution, is a reference convention rather than a location with a universal physical privilege.

For a point source Q in vacuum with V = 0 at infinity, V = kQ/r. Potentials from several sources add algebraically. Opposite signs can cancel a potential at a point even when their vector fields do not cancel; conversely a field can vanish where potential is nonzero.

\Delta U=q\Delta V,\qquad W_{\mathrm{field}}=-q\Delta V,\qquad V_{\mathrm{point}}=\frac{kQ}{r}

The point-source expression uses zero potential at infinity. Potential has units J/C = V.

WORKED EXAMPLE

Follow energy for a negative charge

A charge −2.0 μC moves from a point at 30 V to one at 80 V. Find its potential-energy change and the work done by the electrostatic field.

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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: Potential, capacitors and dipoles
Focus: Potential is energy per unit charge, not another name for field
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
\Delta U=q\Delta V,\qquad W_{\mathrm{field}}=-q\Delta V,\qquad V_{\mathrm{point}}=\frac{kQ}{r}
Meaning: The point-source expression uses zero potential at infinity. Potential has units J/C = V.

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.

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02 / Potential, capacitors and dipoles

The electric field points down the potential gradient

For a small displacement dr, the field’s work on q is qE·dr, while the potential-energy change is q dV. Their conservative relation gives dV = −E·dr. Integrating along a path determines a potential difference from the field.

In one dimension E_x = −dV/dx. A steep decrease of potential with x gives a large positive E_x; a flat potential gives zero component along that direction. The potential value itself is not the field: a constant 100 V and a constant 0 V both have zero gradient.

In three dimensions, E = −∇V. Differentiate with respect to each coordinate while holding the others fixed. A zero derivative along one direction does not establish that every field component is zero.

An equipotential surface has no potential change along any tangent displacement, so the electrostatic field is perpendicular to it where the field is nonzero. Closer-spaced equal-potential increments indicate a stronger field. A charge moved along an equipotential has zero electrostatic work, even though a force may still act normally to the surface.

\Delta V=-\int_i^f\mathbf E\cdot d\mathbf r,\qquad\mathbf E=-\nabla V

For electrostatic fields, the potential difference is path independent. Time-varying magnetic induction requires a more general field description.

Read field direction from a potential graph

Inspect a plotted point: x (m) = 0

Potential: 8

This idealized one-dimensional potential is piecewise linear. Use the slope inside each segment, not at the corners where the derivative is not defined by a single value. Lines connect the plotted samples.

WORKED EXAMPLE

Extract three fields from the same potential plot

For the plotted potential, find E_x in 0 < x < 2 m, 2 < x < 4 m and 4 < x < 6 m.

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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: Potential, capacitors and dipoles
Focus: The electric field points down the potential gradient
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
\Delta V=-\int_i^f\mathbf E\cdot d\mathbf r,\qquad\mathbf E=-\nabla V
Meaning: For electrostatic fields, the potential difference is path independent. Time-varying magnetic induction requires a more general field description.

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.

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03 / Potential, capacitors and dipoles

Differentiate a scalar result and verify it against a field calculation

For some symmetric charge distributions, calculating potential is easier than adding vector field components because each element contributes the scalar k dq/r. Once V is known in a neighbourhood, differentiation recovers the field.

On the axis of a uniform ring, every source element is at the same distance √(R² + z²), so V(z) = kQ/√(R² + z²) with zero at infinity. Differentiating gives the axial field found by direct vector integration.

Knowing V only along one line gives the derivative along that line. Inferring the full three-dimensional field from that information requires symmetry or additional spatial data. On a ring’s axis, rotational symmetry eliminates transverse components, making the axial derivative sufficient.

At the centre of a positively charged ring, the field is zero but potential is kQ/R. Likewise, a charged conductor in electrostatic equilibrium has constant potential throughout its material, but that constant need not be zero. Zero field means no spatial variation of potential within a connected region.

WORKED EXAMPLE

Recover the ring field from potential

A uniform ring has axial potential V(z) = kQ(R² + z²)^(−1/2). Derive E_z and evaluate V and E_z at the centre.

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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: Potential, capacitors and dipoles
Focus: Differentiate a scalar result and verify it against a field calculation
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
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 ↑

04 / Potential, capacitors and dipoles

Capacitance describes how much charge separation a voltage supports

An ideal two-conductor capacitor carries equal and opposite charges +Q and −Q on its conductors. Its charge magnitude Q is related to their potential difference ΔV by Q = CΔV. The net charge of the two-conductor pair can be zero while its separated charge and stored energy are nonzero.

For a linear capacitor with fixed geometry and dielectric properties, C is independent of Q and ΔV. Doubling the charge doubles the voltage; it does not double capacitance. Capacitance changes when geometry or the material response changes.

The unit is the farad, C/V. Microfarads and nanofarads are common convenient units. A value of Q/C has volts only when charge and capacitance prefixes are handled consistently; μC divided by μF already gives V.

To derive a capacitance, assume charge Q, find its field, integrate to get the conductor potential difference and divide Q by that difference. This connects capacitance to electrostatics rather than treating it as a separate collection of geometry formulas.

C=\frac{Q}{\Delta V}

Q is the magnitude on either conductor for the ideal equal-and-opposite-charge capacitor; ΔV is the positive potential difference from negative to positive conductor.

STOP & THINK

A linear capacitor with fixed geometry has its charge doubled. What happens to its capacitance and voltage?

Both double
Capacitance stays fixed and voltage doubles
Capacitance doubles and voltage stays fixed

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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: Potential, capacitors and dipoles
Focus: Capacitance describes how much charge separation a voltage supports
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
C=\frac{Q}{\Delta V}
Meaning: Q is the magnitude on either conductor for the ideal equal-and-opposite-charge capacitor; ΔV is the positive potential difference from negative to positive conductor.

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 / Potential, capacitors and dipoles

Derive the parallel-plate result from field and potential

For broad parallel plates carrying surface charge densities ±σ in vacuum, neglecting edge effects, the field between them is σ/ε₀. Since Q = σA and the field is uniform across separation d, the potential difference magnitude is Ed = Qd/(ε₀A). Thus C = ε₀A/d.

Larger area accommodates more charge for the same voltage, while larger separation requires more voltage for the same charge. The approximation requires separation small compared with the plate dimensions and an observation region where fringing is negligible.

A homogeneous linear dielectric fully filling the gap replaces ε₀ by ε = κ ε₀ in this ideal model, giving C = κ ε₀A/d. Bound polarization charges alter the relationship between free plate charge and field. Partial filling may behave as a series or parallel arrangement depending on the geometry.

For example, side-by-side filled and unfilled regions between the same plates share the same voltage and add capacitances in parallel if edge effects between regions are neglected. Layers stacked through the gap share a normal free-displacement constraint and their voltage drops add, producing a series relation. The orientation of the dielectric matters.

C_{\mathrm{plates}}=\frac{\varepsilon_0A}{d},\qquad C_{\mathrm{full\ dielectric}}=\kappa\frac{\varepsilon_0A}{d}

Uniform ideal parallel plates with negligible edge effects; the dielectric is linear and completely fills the stated gap.

WORKED EXAMPLE

Treat a partly filled gap as two voltage drops

A parallel-plate capacitor has area A and total separation d. A dielectric of relative permittivity κ fills half the gap thickness, while the other half is vacuum. Neglect fringing and interface free charge. Find C in terms of the empty capacitance C₀ = ε₀A/d.

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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: Potential, capacitors and dipoles
Focus: Derive the parallel-plate result from field and potential
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
C_{\mathrm{plates}}=\frac{\varepsilon_0A}{d},\qquad C_{\mathrm{full\ dielectric}}=\kappa\frac{\varepsilon_0A}{d}
Meaning: Uniform ideal parallel plates with negligible edge effects; the dielectric is linear and completely fills the stated gap.

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 / Potential, capacitors and dipoles

Parallel and series follow from node potentials and charge conservation

Capacitors are in parallel when each connects to the same pair of electrical nodes. Ideal wires make each node equipotential, so the capacitors share a voltage. The total supplied charge is the sum C_iV, giving C_parallel = ΣC_i.

For a simple series chain with initially neutral isolated intermediate conductors, equal and opposite charges build on the facing plates at each internal node. The series charge magnitudes are equal, while voltages add. Hence 1/C_series = Σ1/C_i.

The equal-charge rule has conditions. If an isolated intermediate node starts with a specified net charge, its plate charges need not have equal magnitudes. In that case write charge conservation for the node and voltage relations directly rather than applying the neutral-node shortcut.

A drawing can be bent or rearranged without changing its nodes. Decide series or parallel from connectivity, not from whether symbols look side by side on the page. Reduce a mixed network only where the necessary shared-node or unbranched neutral-node conditions hold.

C_{\parallel}=\sum_iC_i,\qquad\frac1{C_{\mathrm{series}}}=\sum_i\frac1{C_i}

The ordinary series rule assumes the intermediate isolated nodes carry no net initial charge and the ideal capacitor model applies.

WORKED EXAMPLE

Find individual voltages after reducing a series pair

Initially uncharged capacitors of 3 μF and 6 μF are connected in series across 12 V, with a neutral isolated intermediate node. Find equivalent capacitance, series charge and each voltage.

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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: Potential, capacitors and dipoles
Focus: Parallel and series follow from node potentials and charge conservation
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
C_{\parallel}=\sum_iC_i,\qquad\frac1{C_{\mathrm{series}}}=\sum_i\frac1{C_i}
Meaning: The ordinary series rule assumes the intermediate isolated nodes carry no net initial charge and the ideal capacitor model applies.

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 / Potential, capacitors and dipoles

Undo a reduction to recover the internal quantities

A reduced equivalent tells you how a whole network responds at its terminals. It does not automatically give the charge or voltage of every internal capacitor. After finding the terminal response, reverse the reduction steps and use the relevant shared voltage or shared charge at each stage.

Keep charge labels attached to a capacitor or node. In a parallel group, its total charge divides among branches in proportion to capacitance at the shared voltage. In a series pair, the same charge magnitude appears on each capacitor under the neutral-node conditions.

Capacitance alone does not determine which component reaches a large voltage. A small series capacitor can carry the largest voltage even though it stores less charge capacity per volt. In a physical design, voltage ratings and dielectric limits would therefore require checking each component, not only the terminal voltage.

WORKED EXAMPLE

Reduce and then reconstruct a three-capacitor network

A 4 μF capacitor is in parallel with a series pair of 3 μF and 6 μF. The network is connected across 10 V, with initially uncharged capacitors. Find total source charge and the voltage across the 3 μF capacitor.

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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: Potential, capacitors and dipoles
Focus: Undo a reduction to recover the internal quantities
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
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 ↑

08 / Potential, capacitors and dipoles

Charging work builds up as the capacitor voltage rises

As a linear capacitor is charged quasistatically from zero, its instantaneous voltage is q/C. Moving an additional charge dq between conductors requires work (q/C)dq. Integrating from zero to Q gives stored energy U = Q²/(2C).

Using Q = CV gives the equivalent expressions ½CV² and ½QV. They describe the same state, but different forms make different constraints easy to see. At fixed Q, energy decreases when C increases. At fixed V, energy increases when C increases.

For a vacuum parallel-plate capacitor, substituting C = ε₀A/d and V = Ed gives U = ½ε₀E²(Ad). The factor Ad is the field volume, leading to energy density u = ½ε₀E². This connects stored capacitor energy to the electric field itself.

The charging source can transfer more energy than the final field stores if energy is dissipated during charging. The capacitor energy formula accounts for the field state; a complete circuit energy balance must also include the source, resistance, radiation or other pathways.

U=\frac{Q^2}{2C}=\frac12CV^2=\frac12QV,\qquad u_{\mathrm{vacuum}}=\frac12\varepsilon_0E^2

The capacitance is linear and fixed for the charging integral. A changing geometry requires mechanical energy accounting as well.

Explain this equation

Read it in words

The three expressions give the same stored energy of a linear capacitor. The last expression gives electric-field energy per unit volume in vacuum.

Symbols and units

$U$: Energy stored in the capacitor.Joules (J).
$Q$: Magnitude of charge on either plate of the two-terminal capacitor.Coulombs (C).
$C$: Capacitance, assumed constant during charging.Farads (F) = C/V.
$V$: Potential difference magnitude across the capacitor.Volts (V).
$u_{\mathrm{vacuum}}$: Local electric-field energy density in vacuum. Lowercase u differs from total U.Joules per cubic metre (J/m³).
$\varepsilon_0$: Vacuum permittivity.Farads per metre (F/m).
$E$: Magnitude of the local electric field.Volts per metre (V/m).

Signs and operations

The square makes stored energy nonnegative. The factor one-half appears because the capacitor voltage rises from zero during charging; it is not at its final voltage throughout that process.

When it applies

The capacitor expressions assume a linear, fixed-geometry capacitor with Q = CV. The vacuum energy-density expression must not be used unchanged in arbitrary dielectric materials. When geometry changes, whether Q or V stays fixed depends on the circuit.

Make it concrete

A 4 μF capacitor at 10 V carries 40 μC and stores 200 μJ. If voltage doubles at fixed capacitance, energy becomes four times larger, not twice as large.

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): U=\frac{Q^2}{2C}=\frac12CV^2=\frac12QV,\qquad u_{\mathrm{vacuum}}=\frac12\varepsilon_0E^2
Read it as: The three expressions give the same stored energy of a linear capacitor. The last expression gives electric-field energy per unit volume in vacuum.
Symbols:
U: Energy stored in the capacitor.; Joules (J).
Q: Magnitude of charge on either plate of the two-terminal capacitor.; Coulombs (C).
C: Capacitance, assumed constant during charging.; Farads (F) = C/V.
V: Potential difference magnitude across the capacitor.; Volts (V).
u_{\mathrm{vacuum}}: Local electric-field energy density in vacuum. Lowercase u differs from total U.; Joules per cubic metre (J/m³).
\varepsilon_0: Vacuum permittivity.; Farads per metre (F/m).
E: Magnitude of the local electric field.; Volts per metre (V/m).
Signs and operations: The square makes stored energy nonnegative. The factor one-half appears because the capacitor voltage rises from zero during charging; it is not at its final voltage throughout that process.
Conditions: The capacitor expressions assume a linear, fixed-geometry capacitor with Q = CV. The vacuum energy-density expression must not be used unchanged in arbitrary dielectric materials. When geometry changes, whether Q or V stays fixed depends on the circuit.

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

Check equivalent-energy and component-energy calculations

The 3 μF and 6 μF series pair from the 12 V example stores charge magnitude 24 μC on each capacitor. Calculate total stored energy two ways.

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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: Potential, capacitors and dipoles
Focus: Charging work builds up as the capacitor voltage rises
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
U=\frac{Q^2}{2C}=\frac12CV^2=\frac12QV,\qquad u_{\mathrm{vacuum}}=\frac12\varepsilon_0E^2
Meaning: The capacitance is linear and fixed for the charging integral. A changing geometry requires mechanical energy accounting as well.

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 ↑

09 / Potential, capacitors and dipoles

An isolated capacitor and a battery-connected capacitor obey different constraints

If a charged capacitor is disconnected from any conducting path, its free plate charge remains fixed in the ideal leakage-free model. Changing its geometry or inserting a dielectric changes capacitance, so voltage must adjust through V = Q/C.

If it remains connected to an ideal voltage source, the source fixes V while charge can flow to or from the plates. The same increase in capacitance then increases Q. Ask which quantity is physically constrained before choosing an energy formula.

For a dielectric inserted quasistatically into a fixed-charge capacitor, increasing C lowers field energy. That energy can become mechanical work or another transfer. For fixed voltage, field energy increases while the battery supplies additional energy; the battery must be included before inferring the mechanical force from an energy change.

In the fixed-voltage case, the relevant combined energy accounting differs from simply minimizing ½CV² for the capacitor alone. Ignoring the source can predict the wrong force direction. Dielectric attraction is compatible with an increase in capacitor field energy because the source contributes energy.

A capacitance increase by a factor κ
Constraint	Charge	Voltage	Stored field energy
Isolated, fixed Q	Unchanged	Divided by κ	Divided by κ
Ideal source, fixed V	Multiplied by κ	Unchanged	Multiplied by κ

WORKED EXAMPLE

Compare the same material change under two connections

An initially 2 μF capacitor is charged to 10 V. A dielectric doubles its capacitance. Compare final charge, voltage and energy if it is first isolated versus if it remains attached to an ideal 10 V source.

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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: Potential, capacitors and dipoles
Focus: An isolated capacitor and a battery-connected capacitor obey different constraints
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
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 ↑

10 / Potential, capacitors and dipoles

Charge conservation does not imply conservation of capacitor field energy

When two isolated capacitors are connected in parallel with specified polarity, charge redistributes until their voltages agree. Conserve the net charge on each connected conductor group, then use the common final voltage. Do not separately conserve each capacitor’s original charge after the connection is made.

For a charged capacitor C₁ at V₀ connected with matching polarity to an initially uncharged C₂, the final voltage is C₁V₀/(C₁ + C₂). The initial field energy generally exceeds the final field energy. The difference can appear in resistance, radiation or other transient electromagnetic and mechanical effects.

An ideal circuit with exactly zero resistance and nonzero inductance may oscillate rather than settle immediately to the static final state. The common final-voltage calculation assumes a relaxation mechanism eventually establishes equilibrium. A zero-resistance shortcut does not make the missing energy disappear.

This example is a useful reminder to conserve the correct quantity at the correct boundary. Charge remains conserved, total energy remains conserved, but the portion stored in capacitor fields can decrease.

WORKED EXAMPLE

A 4 μF capacitor at 12 V is disconnected from its source and connected with matching polarity to an initially uncharged 2 μF capacitor. Assume transients eventually decay. Find the final voltage and the decrease in stored field energy.

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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: Potential, capacitors and dipoles
Focus: Charge conservation does not imply conservation of capacitor field energy
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
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 ↑

11 / Potential, capacitors and dipoles

A neutral pair can experience a turning effect in a uniform field

An electric dipole consists of equal and opposite charges separated by a displacement vector from the negative charge to the positive charge. Its dipole moment is p = qd, directed from negative to positive, with units C·m.

In a uniform electric field, the forces on the two charges are equal and opposite, so the net force is zero. They generally act along different lines and form a couple. The torque is τ = p × E, with magnitude pE sin θ.

The torque tends to align p with E. At θ = 0 and θ = π, torque vanishes, but the two orientations have different stability. A small angular displacement from alignment produces a restoring torque; a small displacement from anti-alignment drives the dipole farther away.

A nonuniform field can produce a net force as well as torque because the two charges experience different field values. The zero-force statement therefore requires uniform E and the ideal fixed dipole model.

\mathbf p=q\mathbf d,\qquad\boldsymbol\tau=\mathbf p\times\mathbf E

The separation vector d points from −q to +q. The uniform-field torque result is exact for the ideal two-charge dipole.

WORKED EXAMPLE

Calculate a torque while checking the net force

A rigid dipole has moment magnitude 4.0 × 10⁻⁸ C·m in a uniform field of 5000 N/C. Its moment makes 30° with the field. Find torque magnitude and net electric force.

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: Potential, capacitors and dipoles
Focus: A neutral pair can experience a turning effect in a uniform field
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
\mathbf p=q\mathbf d,\qquad\boldsymbol\tau=\mathbf p\times\mathbf E
Meaning: The separation vector d points from −q to +q. The uniform-field torque result is exact for the ideal two-charge dipole.

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 ↑

12 / Potential, capacitors and dipoles

Potential energy distinguishes stable and unstable zero-torque orientations

Integrating the angular work of the uniform-field torque gives dipole energy U = −p·E = −pE cos θ, with zero chosen at θ = 90°. The aligned orientation has the lowest energy −pE; anti-alignment has the highest +pE.

The quasistatic external work to rotate the dipole between orientations equals ΔU when the dipole begins and ends without a kinetic-energy change and other losses are neglected. The work done by the electric field is the negative of that change.

For a freely rotating rigid dipole with moment of inertia I, small angular oscillations near alignment satisfy Iθ̈ ≈ −pEθ, so ω = √(pE/I). This connects dipole dynamics to the same restoring-torque structure used for pendulums.

The field is treated as externally maintained and the dipole moment magnitude fixed. Induced dipoles, deformable molecules or changing external-source arrangements require additional energy accounting and should not inherit every permanent-dipole formula automatically.

U=-pE\cos\theta,\qquad W_{\mathrm{ext,quasistatic}}=\Delta U

Uniform externally specified field and a fixed permanent dipole; the energy reference is chosen at θ = 90°.

WORKED EXAMPLE

Use energy rather than averaging torque over a large angle

A fixed dipole in a uniform field has pE = 0.20 mJ. It is rotated quasistatically from alignment to 120°. Find the required external work, neglecting losses.

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

STOP & THINK

A permanent dipole is exactly anti-aligned with a uniform field. Which statement is correct?

Its torque is zero and the orientation is stable
Its torque is zero but the orientation is unstable
Its torque has maximum magnitude

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: Potential, capacitors and dipoles
Focus: Potential energy distinguishes stable and unstable zero-torque orientations
Learning goals:
- Relate potential difference to signed charge energy and field work.
- Recover field components from slopes and gradients.
- Derive parallel-plate and layered-dielectric capacitance.
- Use node connections and charge conservation in capacitor networks.
- Compute stored energy and account for sources during changes.
- Distinguish conserved charge from capacitor field energy during sharing.
- Derive uniform-field dipole torque, orientation energy and stability.
Equations in this section (LaTeX):
U=-pE\cos\theta,\qquad W_{\mathrm{ext,quasistatic}}=\Delta U
Meaning: Uniform externally specified field and a fixed permanent dipole; the energy reference is chosen at θ = 90°.

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

Potential is scalar energy per charge; electric field is minus its spatial gradient.
Zero field does not require zero potential.
Capacitance belongs to a specified geometry and linear material response.
Parallel capacitors share voltage; the ordinary series equal-charge rule needs neutral intermediate nodes.
Reduce a network, then reverse the steps to recover internal voltages and charges.
Choose fixed Q or fixed V from the physical connection before predicting energy changes.
Capacitor field energy can transfer into other forms during charge sharing.
A uniform field can exert a dipole torque with zero net force.
Dipole alignment is stable; anti-alignment has zero torque but is unstable.

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

Capacitors of 2 μF and 5 μF share the same two nodes. What is their equivalent capacitance?

Answer in μF.

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

SELF CHECK 2

An isolated ideal capacitor’s capacitance triples while its free charge is conserved. What is the new-to-old stored-energy ratio? Give at least three significant figures, or an exact fraction.

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

SELF CHECK 3

A one-dimensional potential decreases by 12 V over 3 m at a constant rate toward positive x. What is E_x?

Answer in V/m.

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

SELF CHECK 4

Adding a constant 20 V to the potential at every point does what to the electric field?

Adds 20 V/m everywhere
Leaves the field unchanged
Reverses its direction

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

SELF CHECK 5

Does a charged isolated intermediate node always allow the usual equal-charge series-capacitor shortcut?

Yes, all series drawings imply equal charges
No; write the node’s charge conservation explicitly
Only if capacitances are equal

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

SELF CHECK 6

A capacitor stays connected to an ideal voltage source while a dielectric increases its capacitance. Which quantity is fixed?

Free plate charge
Potential difference
Stored field energy

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

An isolated charged parallel-plate capacitor has its plate separation increased. What stays fixed, and what happens to voltage?

Compare your reasoning

Charge stays fixed if leakage is negligible. Capacitance decreases, so voltage rises. The external agent supplies the increased field energy by working against the attraction.

Revisit the explanation →

A few days later: change the situation

Two capacitors at different voltages are connected through a resistor and settle to a common voltage. Why need their initial field energy exceed the final field energy?

Compare your reasoning

Charge can be conserved while field energy is transferred into resistor heating during redistribution. Conserved charge does not imply conserved capacitor energy. Include the transfer in the full energy account.

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 ↑

What you will learn to do

Bring back one idea

Potential is energy per unit charge, not another name for field

Follow energy for a negative charge

The electric field points down the potential gradient

Read field direction from a potential graph

Extract three fields from the same potential plot

Differentiate a scalar result and verify it against a field calculation

Recover the ring field from potential

Capacitance describes how much charge separation a voltage supports

A linear capacitor with fixed geometry has its charge doubled. What happens to its capacitance and voltage?

Derive the parallel-plate result from field and potential

Treat a partly filled gap as two voltage drops

Parallel and series follow from node potentials and charge conservation

Find individual voltages after reducing a series pair

Undo a reduction to recover the internal quantities

Reduce and then reconstruct a three-capacitor network

Charging work builds up as the capacitor voltage rises

Read it in words

Symbols and units

Signs and operations

When it applies

Make it concrete

Check equivalent-energy and component-energy calculations

An isolated capacitor and a battery-connected capacitor obey different constraints

Compare the same material change under two connections

Charge conservation does not imply conservation of capacitor field energy

Track both charge and energy in a redistribution

A neutral pair can experience a turning effect in a uniform field

Calculate a torque while checking the net force

Potential energy distinguishes stable and unstable zero-torque orientations

Use energy rather than averaging torque over a large angle

A permanent dipole is exactly anti-aligned with a uniform field. Which statement is correct?

What should stay with you

Try it without the examples

Capacitors of 2 μF and 5 μF share the same two nodes. What is their equivalent capacitance?

An isolated ideal capacitor’s capacitance triples while its free charge is conserved. What is the new-to-old stored-energy ratio? Give at least three significant figures, or an exact fraction.

A one-dimensional potential decreases by 12 V over 3 m at a constant rate toward positive x. What is E_x?

Adding a constant 20 V to the potential at every point does what to the electric field?

Does a charged isolated intermediate node always allow the usual equal-charge series-capacitor shortcut?

A capacitor stays connected to an ideal voltage source while a dielectric increases its capacitance. Which quantity is fixed?

Take the ideas into a problem

What can you explain later?

Next study session

A few days later: change the situation