2. Electrostatic Potential And Capacitance

Potential turns an electric-force problem into an energy problem. Capacitance then connects charge storage to geometry, materials and voltage.

The potential description grew from the study of conservative forces. Early charge-storage devices led to the capacitor; later field theory located its stored energy in the electric field between conductors.

Moving a charge can store or release energy even when its final position is unchanged in height.

Electric potential gives an energy account for electrostatics. It often avoids the component calculations needed for force. A potential is a scalar assigned to a position; potential energy belongs to a charge or a configuration of charges. Keep those two ideas separate throughout this chapter.

Electrostatic forces are conservative. The work they do between two positions does not depend on the path. Therefore a potential difference can be defined. A battery or a time-varying magnetic field needs extra care; the present chapter concerns electrostatic fields.

An external agent moves a charge slowly against an electric force and does 0.018 J of work. How does potential energy change?

Slow motion means no net change in kinetic energy. External work is stored as electrostatic potential energy.

Result: It increases by 0.018 J.

1. Does conservative work depend on path length?

Answers

1. No; it depends on the endpoints.

Potential difference is work per unit positive charge.

Define $V_B-V_A=(U_B-U_A)/q=-\int_A^B\mathbf E\cdot d\mathbf l$. The field’s work is $W_{\rm field}=-\Delta U=q(V_A-V_B)$. A slow external transfer requires $W_{\rm ext}=\Delta U$. One volt is one joule per coulomb.

A reference fixes the otherwise arbitrary zero of potential. For a finite, isolated charge distribution it is often convenient to take V=0 at infinity. That choice is not available for every idealised infinite distribution. Potential differences remain physically meaningful even if the reference changes.

A +4 μC charge moves from 90 V to 30 V. Find its change in potential energy and the field’s work.

Use $\Delta U=q(V_B-V_A)$.

Result: $\Delta U=-0.24\ \mathrm{mJ}$; the field does +0.24 mJ of work.

A positive charge released from rest tends to move to lower potential. A negative charge tends to move to higher potential. In both cases its potential energy decreases if only the electric field speeds it up.

1. What is the potential difference if 0.30 J is needed to move +0.020 C slowly?

Answers

1. 15 V, final minus initial.

The potential of a point source falls as inverse distance.

With zero at infinity, a stationary point source gives $V(r)=kq/r$. Integrating the radial field from infinity produces the $1/r$ dependence. The sign of V follows the source charge. Unlike electric field, no direction belongs to this scalar value.

For two points, $V_B-V_A=kq(1/r_B-1/r_A)$. Do not use the distance between A and B in place of each point’s distance from the source. At an ideal point charge the expression diverges; this is a limitation of the point-source model.

Find the potential 0.45 m from a −5 nC source using k≈9 × 10⁹ SI units.

Use the signed source charge in $V=kq/r$.

Result: -100 V.

1. When distance from a point charge doubles, what happens to V relative to infinity?

Answers

1. It halves.

A neutral dipole can have nonzero potential away from its equatorial plane.

For charges +q and −q, the exact potential is $V=kq(1/r_+-1/r_-)$. The distances run from the observation point to each charge. At a point on the equatorial plane they are equal, so V=0 with the usual reference.

Far from a dipole of moment p, expand the two inverse distances to first order in separation. The result is $V\approx kp\cos\theta/r^2=k\mathbf p\cdot\hat{\mathbf r}/r^2$. Here θ is measured from p and the distance is much larger than the charge separation. The angular sign is as important as the magnitude.

A short dipole has p=6 nC m. Estimate V at r=3 m and θ=60°.

Use the far-field formula only because the separation is assumed much smaller than r.

Result: 3 V.

The equatorial plane is a useful counterexample: V is zero everywhere in that plane, while the field generally is not zero there. The field depends on spatial changes of V in all directions, not just changes within one plane.

1. What sign is the far dipole potential on the negative-axis side?

Answers

1. Negative, since cos θ is negative there.

Potentials from different sources add algebraically.

For point sources, $V(\mathbf r)=\sum_i kq_i/|\mathbf r-\mathbf r_i|$. First find each source-to-point distance, then add signed scalar contributions. A continuous distribution gives $V=\int k\,dq/r$. No vector components are needed for the addition itself.

A point is 0.30 m from +4 nC and 0.60 m from −6 nC. Find its potential.

The two terms use their own distances. Their signs are opposite.

Result: 30 V.

This shortcut does not imply that electric fields add as scalars. Two charges can give zero potential and a large resultant field. A system’s potential at a source charge itself excludes that charge’s singular self-potential when calculating interaction energy.

1. Equal and opposite charges are equally distant from a point. What is their total V there?

Answers

1. Zero, with zero at infinity.

An equipotential surface joins points with the same electric potential.

No electrostatic work is needed to move a charge along an equipotential because ΔV=0. The electric field is perpendicular to such a surface wherever E is nonzero. If it had a tangential component, moving along that tangent would change V, contradicting the definition.

The component relation is $E_x=-dV/dx$ in a one-dimensional variation. More generally E is the negative spatial gradient of V. Closely spaced equipotentials with equal voltage steps indicate a stronger field. Around an isolated point charge they are spheres; in a uniform field they are parallel planes.

Potential falls from 120 V to 75 V over 0.15 m along +x in a uniform field.

Use $E_x=-\Delta V/\Delta x$.

Result: 300 V/m along +x.

1. Can two equipotential surfaces with different voltages intersect?

Answers

1. No; one point cannot have two potentials.

Assembling charges requires interaction energy.

Bring q₁ from infinity first. With no other source present, its interaction-energy cost is zero. Bringing q₂ next requires $kq_1q_2/r_{12}$. Thus a pair has $U=kq_1q_2/r_{12}$ relative to infinite separation. Positive U for like signs means work must be supplied to bring them together; negative U for unlike signs means energy is released.

For several charges, include each pair once: $U=\sum_{i<j}kq_iq_j/r_{ij}$. Equivalently, $U=\tfrac12\sum_iq_iV_{\rm others}(\mathbf r_i)$. The half prevents double counting. The expression excludes point-charge self-energies.

Three +2 μC charges occupy an equilateral triangle of side 0.80 m. Find the interaction energy.

There are three distinct pairs, each with the same separation and product.

Result: 0.135 J.

1. What sign is the interaction energy of an isolated positive-negative pair?

Answers

1. Negative when zero is defined at infinite separation.

External fields contribute energy without changing the source charges in the model.

A small charge q in a prescribed external potential has $U_{\rm ext}=qV_{\rm ext}$. For two added charges include both external terms and their mutual term: $U=q_1V_{\rm ext}(\mathbf r_1)+q_2V_{\rm ext}(\mathbf r_2)+kq_1q_2/r_{12}$. This assumes the added charges do not significantly rearrange the external sources.

For a dipole in a uniform external field, $U=-\mathbf p\cdot\mathbf E=-pE\cos\theta$, choosing zero at θ=90°. Integrating the torque gives the energy change between orientations. Parallel alignment has minimum energy; antiparallel alignment has maximum energy.

A dipole with p=10 nC m is rotated slowly from parallel to antiparallel in E=300 N/C. Find external work.

The change is $pE-(-pE)=2pE$.

Result: 6 μJ.

1. What work does the field do when a dipole moves from perpendicular to parallel?

Answers

1. It does +pE of work, reducing U by pE.

In electrostatic equilibrium, mobile charges have stopped redistributing.

• The field inside the conducting material is zero; otherwise free carriers would continue to move.
• The conductor is an equipotential, including its surface.
• Any excess free charge resides on the surface of an isolated conductor.
• The field just outside a smooth surface is normal and has magnitude |σ|/ε₀ in vacuum.
• Tangential surface field would cause charge flow, so it must vanish.

The zero-interior-field statement applies to the conducting material, not automatically to every cavity. An empty closed cavity with no enclosed charges is shielded from an external electrostatic field. If a charge is placed inside a cavity, induced charges on the inner surface create a field in the cavity while the surrounding metal remains field-free.

A hollow metal enclosure can therefore shield instruments from external static fields. Openings, time-varying fields and imperfect conductors introduce practical limits. The local surface charge density need not be uniform; sharp curvature can create a large nearby field.

An isolated conducting shell has net charge +3 nC. A +5 nC point charge is placed in an internal cavity. Find total induced charges on the inner and outer surfaces.

A Gaussian surface in the metal has zero flux, so the cavity charge plus inner-surface charge sums to zero. The conductor’s own net charge stays +3 nC.

Result: Inner surface: −5 nC; outer surface: 8 nC.

1. Can different points of a single equilibrium conductor have different V?

Answers

1. No.

A dielectric responds by shifting bound charges rather than letting them flow freely across the sample.

In nonpolar molecules an applied field slightly separates the centres of positive and negative charge, inducing dipoles. Polar molecules already have dipole moments; a field tends to align them against thermal agitation. Polarisation is electric dipole moment per unit volume.

Bound charges appear on the dielectric surfaces. In a uniformly filled parallel-plate capacitor their field opposes the original field. The net field is reduced for the same free charge on the plates. The material’s relative permittivity K compares its permittivity with that of vacuum for the stated conditions.

A linear dielectric is an approximation in which polarisation is proportional to field. Real materials can be nonlinear, frequency dependent or break down when the field becomes too large.

A dielectric reduces the field of an isolated, uniformly filled capacitor from 800 V/m to 200 V/m. Find K.

For the same free plate charge, E=E₀/K.

Result: 4.

1. Do polar and nonpolar dielectrics both polarise in an applied field?

Answers

1. Yes; their microscopic mechanisms differ.

Capacitance measures charge stored per potential difference.

For two conductors with equal and opposite charges ±Q, $C=Q/(V_+-V_-)$. Use the magnitude Q on one plate, not the sum of both magnitudes. The farad, F, is C/V. In the ideal linear model, geometry and medium determine C; doubling Q doubles V without changing C.

A capacitor stores separated charge and electric-field energy. Its rating specifies limits such as maximum voltage; it is not a source of unlimited charge. Stray capacitance also exists between nearby conductors even when no component is deliberately installed.

A capacitor stores 18 μC at 6 V. Find its capacitance.

Use $C=Q/V$.

Result: 3 μF.

1. What happens to Q if the voltage across an ideal 5 μF capacitor doubles?

Answers

1. Q doubles.

A parallel-plate capacitor is a controlled example of field, voltage and geometry.

Take large plates of area A separated by d, with $d\ll\sqrt A$, and neglect edge effects. Their two fields add between the plates: $E=\sigma/\varepsilon_0=Q/(\varepsilon_0A)$. The potential difference is $V=Ed$. Therefore $C=\varepsilon_0A/d$.

Larger area allows more charge at a given voltage. Greater separation requires more voltage for the same charge and reduces C. Fringing fields near the edges make the real result depart from the simple uniform-field model.

Vacuum plates have A=0.060 m² and d=0.30 mm. Use ε₀≈8.85 × 10⁻¹² SI units. Find C.

Convert millimetres to metres. Report the small capacitance in nanofarads.

Result: 1.77 nF.

1. If area doubles and separation triples, how does C change?

Answers

1. It becomes two thirds of its original value.

A dielectric increases capacitance, but the charge and voltage response depends on the connection.

If the entire gap is filled by a linear dielectric, $C=K\varepsilon_0A/d=KC_0$. An isolated capacitor retains Q, so $V=V_0/K$ and $E=E_0/K$. A battery-connected capacitor retains V, so Q increases by K and the field V/d stays fixed.

For a dielectric slab of thickness t filling the plate area but not the whole gap, treat the vacuum and dielectric layers in series. With negligible fringing, $C=\varepsilon_0A/[d-t+t/K]$. Each layer has the same normal electric displacement when no free charge lies at the interface.

A 2 μF capacitor charged to 12 V is disconnected, then fully filled with K=3 dielectric.

Charge stays fixed. Capacitance triples and voltage falls by the same factor.

Result: Q=24 μC; C=6 μF; V=4 V.

1. For the same insertion with the battery still connected, what happens to Q?

Answers

1. It triples while V remains 12 V.

Connections determine whether capacitors share voltage or charge magnitude.

All capacitors connect between the same two nodes, so each has the same V. Their stored charges add: $Q=\sum_iC_iV$, giving $C_{\rm eq}=\sum_iC_i$.

Initially uncharged floating junctions cannot acquire net charge from outside. Adjacent plates therefore acquire equal and opposite charges, so every series capacitor carries the same Q magnitude. Voltages add: $V=Q\sum_i1/C_i$, giving $1/C_{\rm eq}=\sum_i1/C_i$.

Capacitors of 4 μF and 12 μF are connected in series across 16 V. Find equivalent capacitance, charge and each voltage.

First use $C_{\rm eq}=C_1C_2/(C_1+C_2)$, then Q=Ceq V and $V_i=Q/C_i$.

Result: Ceq=3 μF; Q=48 μC; voltages are 12 V and 4 V.

For parallel capacitors, the equivalent exceeds any individual value. For series capacitors, it is smaller than the smallest individual value. These bounds are useful error checks.

1. What is the parallel equivalent of 4 μF and 12 μF?

Answers

1. 16 μF.

Charging a capacitor requires progressively more work as its voltage rises.

When the capacitor already holds charge q, its voltage is q/C. Adding dq slowly costs $dW=(q/C)dq$. Integrating from zero to Q gives $U=Q^2/(2C)=\tfrac12QV=\tfrac12CV^2$. The factor one half comes from the rising voltage during charging.

For a vacuum parallel-plate capacitor, substitute C and V=Ed to obtain $U=\tfrac12\varepsilon_0E^2Ad$. Thus the field-energy density is $u=\tfrac12\varepsilon_0E^2$. For a linear dielectric under suitable electrostatic conditions, replace ε₀ by ε.

Find the energy stored by a 5 μF capacitor at 20 V.

Use $U=CV^2/2$ and convert microfarads to farads.

Result: 1 mJ.

If two capacitors at unequal voltages are joined, charge redistributes. The final stored electrostatic energy can be less than the initial value. The difference becomes heat, radiation or other energy during the transient; total energy is still conserved.

1. At fixed capacitance, what happens to stored energy when voltage doubles?

Answers

1. It becomes four times as large.

Potential is scalar: $V=\sum kq_i/r_i$. Field is its negative spatial gradient. Work by the field is $-\Delta U$.

For a linear capacitor, $C=Q/V$ and $U=CV^2/2$. Parallel capacitors share V; simple uncharged series junctions enforce equal charge magnitudes. Always identify what the battery or isolation holds fixed.

• A zero of potential is a choice; a potential difference is measurable.
• Shielding concerns an empty closed cavity and electrostatic conditions.
• A dielectric changes both field response and energy accounting; specify the circuit connection.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

2.1 A +6 μC charge moves slowly through a potential rise of 25 V. Find external work.

2.2 An electron moves to a higher potential. Does its electric potential energy rise?

2.3 Find V at 0.80 m from +12 nC.

2.4 Two sources +3 nC and −5 nC are each 0.50 m from a point. Find V.

2.5 Can V vanish where E is nonzero? Give a configuration.

2.6 A short dipole has p=4 nC m. Find its far axial potential at r=2 m.

2.7 Potential changes by −84 V over +0.28 m in a uniform field. Find Ex.

2.8 How much work does the field do moving a charge along an equipotential?

2.9 Charges +3 μC and −4 μC are 0.60 m apart. Find interaction energy.

2.10 Why is a factor 1/2 needed in one half of the sum qiVi?

2.11 A dipole p=7 nC m turns from 90° to 0° in E=200 N/C. Find ΔU.

2.12 State the electric field in the material of a conductor at electrostatic equilibrium.

2.13 A conducting shell with net charge −2 nC encloses +4 nC in a cavity. Find inner and outer surface totals.

2.14 Why is a closed metal enclosure useful for electrostatic shielding?

2.15 A capacitor has Q=42 μC and V=14 V. Find C.

2.16 For vacuum plates A=0.080 m² and d=0.40 mm, find C using ε₀≈8.85 × 10⁻¹² SI units.

2.17 What change in C follows from halving plate separation?

2.18 An isolated charged capacitor is filled with K=5 dielectric. State changes in C, Q and V.

2.19 Repeat the preceding question with its ideal battery connected.

2.20 Find the series equivalent of 6 μF and 3 μF.

2.21 Find their parallel equivalent.

2.22 The series pair is placed across 18 V. Find charge magnitude and individual voltages.

2.23 A 4 μF capacitor is at 15 V. Find its stored energy.

2.24 At fixed Q, doubling C changes energy by what factor?

2.25 Explain why a nonzero electric field may remain outside a conductor while E is zero within its material.

2.26 Why is potential zero at infinity inappropriate for an ideal infinite charged sheet?

Answers

2.1 0.15 mJ.

2.2 No; q is negative, so U=qV decreases.

2.3 135 V.

2.4 -36 V.

2.5 Yes: a point on the equatorial plane of a nonzero dipole.

2.6 9 V on its positive side.

2.7 300 V/m.

2.8 Zero.

2.9 -0.18 J.

2.10 Each pair otherwise appears twice; Vi must exclude the self-potential.

2.11 -1.4 μJ.

2.12 Zero.

2.13 Inner −4 nC; outer +2 nC.

2.14 Its mobile charges rearrange so an empty closed cavity is at constant potential with zero electrostatic field.

2.15 3 μF.

2.16 1.77 nF.

2.17 C doubles, within the parallel-plate approximation.

2.18 C multiplies by five; Q stays fixed; V divides by five.

2.19 C and Q multiply by five; V stays fixed.

2.20 2 μF.

2.21 9 μF.

2.22 Q=36 μC; 6 μF has 6 V and 3 μF has 12 V.

2.23 0.45 mJ.

2.24 One half, from U=Q²/(2C).

2.25 Surface charge rearranges to cancel the internal field; its external field generally remains.

2.26 The potential difference to infinity does not converge; choose a finite reference instead.