1. Electric Charges And Fields

Electric charge can stay on a surface while its influence reaches across empty space. This chapter develops a consistent language for that influence: force, field and flux.

Coulomb studied electrical forces with a torsion balance in the eighteenth century. Gauss later provided a flux relation that links electric fields to enclosed charge and makes symmetric problems much simpler.

A small spark after walking across a dry surface is evidence of separated electric charge.

Electrostatics studies charges that remain at rest on the time scale of interest. Start by separating the objects, the charges they carry and the forces between them. A charged object can attract a neutral object because the neutral object can redistribute its charges.

Rubbing transfers some electrons between materials; it does not manufacture charge. An electroscope detects charge through the separation of its conducting leaves. Attraction alone cannot identify the sign of an unknown object; repulsion from a known charged object is a more useful sign test.

Two initially neutral objects exchange electrons. Object A gains electrons. What happens to B?

The pair is isolated, so their total charge stays zero. Electrons carry negative charge.

Result: A becomes negative and B becomes equally positive.

1. Can a charged comb attract uncharged paper?

Answers

1. Yes. Charge redistribution produces a closer region of opposite sign and a net attraction.

Charge is the property that determines an object’s electric interaction.

There are two signs of charge. Like signs repel and unlike signs attract. The SI unit is the coulomb, C. A proton carries +e and an electron carries −e. In ordinary charging, electrons move between objects while nuclei remain bound in the material.

For calculations, $e=1.602176634\times10^{-19}\ \mathrm C$ exactly. A convenient rounded value in school problems is $1.60\times10^{-19}\ \mathrm C$. State which approximation you use.

An object has an excess of 25 million electrons. Find its charge using the rounded elementary charge.

Use $q=-ne$; “excess electrons” fixes the negative sign.

Result: $q=-4\ \mathrm{pC}$.

1. What sign remains when a neutral object loses electrons?

Answers

1. Positive.

Materials differ in how easily their charges can move.

A conductor contains charge carriers that can move over macroscopic distances. In a metal these carriers are electrons. In an electrolyte, positive and negative ions can both move. An insulator has charges bound much more tightly to local positions. Real materials lie on a range; the classification depends on conditions.

To charge a conductor by contact, let charge flow from another charged object. To charge it by induction, bring a charged object near without contact, connect the conductor to Earth, disconnect Earth first, then remove the inducing object. The conductor is left with charge opposite in sign to the inducing object.

Earthing supplies or removes electrons through a conducting path. The order of operations matters: removing the nearby charge while Earth is still connected usually allows the conductor to become neutral again.

A negative rod is brought near a neutral metal sphere on an insulating stand. The sphere is earthed, then disconnected, then the rod is removed.

The rod repels some electrons into Earth. Disconnecting traps the remaining imbalance.

Result: The sphere ends positively charged.

1. Why hold a conductor on an insulating stand?

Answers

1. To reduce charge leakage through the support to Earth.

Three accounting rules make charge problems manageable.

The total charge is the algebraic sum: $Q=\sum_i q_i$. Positive and negative contributions can cancel even though both kinds of charged particles remain present.

The total charge of an isolated system stays constant. Charge may move across a boundary; then the chosen subsystem is not isolated. Pair creation can produce charged particles, but the net charge of the pair is zero.

For an ordinary isolated body, $q=ne$, where $n$ is an integer. A large body contains so many carriers that a continuous-charge approximation is usually useful. That approximation does not remove the underlying quantisation.

Charges +7 nC, −12 nC and +9 nC are placed on identical metal spheres and the spheres are connected together.

Add the initial charges algebraically. Equal isolated spheres at the same potential share the charge equally after separation in a symmetric arrangement.

Result: Total charge is 4 nC; each sphere receives 1.3333 nC.

1. Can an ordinary isolated object have charge 2.5e?

Answers

1. No. Its excess or deficit of elementary charges is an integer.

The electric force depends strongly on separation.

For two stationary point charges in vacuum, the magnitude is $F=k|q_1q_2|/r^2$, with $k=1/(4\pi\varepsilon_0)\approx9.0\times10^9\ \mathrm{N\,m^2\,C^{-2}}$. The force lies along the line joining them. Determine attraction or repulsion from the signs before adding vectors.

If $\hat{\mathbf r}_{21}$ points from charge 1 to charge 2, the force on charge 2 is $\mathbf F_{21}=kq_1q_2\hat{\mathbf r}_{21}/r_{21}^{2}$. The reverse force is equal in magnitude and opposite in direction. For extended objects, the point-charge formula needs appropriate size separation or spherical symmetry.

Two charges +4 μC and −3 μC are 0.60 m apart in vacuum. Find their mutual force.

Convert microcoulombs to coulombs before using SI units. The unlike signs imply attraction.

Result: The magnitude is 0.3 N, directed towards the other charge.

A torsion balance compares the electric turning effect with the restoring twist of a thin fibre. Coulomb’s experiments helped establish the inverse-square relation; modern treatments express the force using vectors and SI units.

1. If separation triples while charges stay fixed, what happens to force?

Answers

1. It becomes one ninth of its original magnitude.

With several charges, add the individual forces as vectors.

Superposition means that each source contributes the force it would exert by itself. Calculate each contribution at the charge of interest, resolve components on the same axes, then add the components. Adding magnitudes works only when every contribution points in the same direction.

A +1 μC test charge at the origin feels a 0.18 N force along +x and a 0.24 N force along +y from two other sources. Find the resultant.

The components are already on perpendicular axes. Use $F=\sqrt{F_x^2+F_y^2}$ and $\tan\theta=F_y/F_x$.

Result: $F=0.3\ \mathrm N$ at $53.1301^\circ$ above +x.

Symmetry can eliminate components before arithmetic. For equal source charges placed symmetrically above and below an axis, the transverse components at a point on that axis cancel. Draw their directions first so that cancellation is justified.

1. Two equal positive charges lie at x=−a and x=+a. What force acts on any test charge at the origin?

Answers

1. Zero: the two electric fields cancel there.

A field describes the electric influence at each position before a test charge is added.

Define $\mathbf E=\lim_{q_0\to0}\mathbf F/q_0$ using a positive test charge. The limiting idea ensures that the probe does not substantially rearrange the sources. Units are $\mathrm{N\,C^{-1}}$, equivalently $\mathrm{V\,m^{-1}}$. The force on any charge is $\mathbf F=q\mathbf E$.

A point charge produces $\mathbf E=kq\hat{\mathbf r}/r^2$. A positive source gives an outward field; a negative source gives an inward field. For multiple sources, $\mathbf E=\sum_i\mathbf E_i$. The field exists whether a test charge is present or not.

The field at a point is 850 N/C towards +x. What force acts on a −6 μC charge there?

Multiply the field vector by the signed charge. A negative multiplier reverses direction.

Result: 5.1 mN towards −x.

1. If the test charge doubles without changing the sources, does E double?

Answers

1. No. Its force doubles, but the source field is unchanged.

Field lines translate a vector field into a drawing.

A tangent to a field line gives the local direction of E. Closer lines indicate a stronger field in a consistent drawing. Lines begin on positive charges and end on negative charges or at infinity. They cannot cross: a crossing would assign two field directions to one point. Electrostatic field lines do not form closed loops.

Field lines are a representation, not material threads or particle tracks. A charged particle with an initial velocity need not follow a field line. At a conductor in electrostatic equilibrium, the external field is normal to its surface.

1. Can two electrostatic field lines intersect?

Answers

1. No; the field has a unique direction wherever it is nonzero.

Flux measures how much field crosses an oriented surface.

For a uniform field through a flat area, $\Phi_E=\mathbf E\cdot\mathbf A=EA\cos\theta$. The area vector is perpendicular to the surface, and $\theta$ is the angle between E and that normal. Flux has units $\mathrm{N\,m^2\,C^{-1}}$. For a curved surface, add small pieces: $\Phi_E=\int\mathbf E\cdot d\mathbf A$.

A closed surface uses outward normals. Outward crossing is positive and inward crossing is negative. A field parallel to a surface contributes zero flux, even when the field is strong. Flux is a scalar with a sign; it is not the field magnitude.

A field of 240 N/C crosses a 0.030 m² flat surface at 60° to its normal.

Use the angle to the normal, not the angle to the plane.

Result: $\Phi_E=3.6\ \mathrm{N\,m^2\,C^{-1}}$.

1. What is the flux if the same plane is parallel to the field?

Answers

1. Zero.

A dipole has zero net charge but still produces an electric field.

An idealised electric dipole consists of −q and +q separated by $2a$. Its moment is $\mathbf p=q(2a)\hat{\mathbf p}$, directed from negative to positive. Its unit is C m. Neutral molecules can have permanent dipole moments when their positive and negative charge centres do not coincide.

On the axis, at a distance $r>a$ from the centre on the positive-charge side, subtract the opposing fields: $E_{\rm ax}=kq[(r-a)^{-2}-(r+a)^{-2}]=2kpr/(r^2-a^2)^2$. Far away, $r\gg a$, this becomes $E_{\rm ax}\approx2kp/r^3$, along p.

On the equatorial plane, transverse components cancel and the axial components add opposite p: $\mathbf E_{\rm eq}=-k\mathbf p/(r^2+a^2)^{3/2}\approx-k\mathbf p/r^3$. Both far fields fall as $r^{-3}$, faster than a monopole field. These approximations fail close to either charge.

A small dipole has moment 8 nC m. Estimate its axial field 2 m from its centre, assuming its separation is much smaller than 2 m.

Use the far-field axial expression $E=2kp/r^3$.

Result: 18 N/C along its moment.

1. At the same large distance, what is the ratio of axial to equatorial field magnitudes?

Answers

1. 2:1.

A uniform field can rotate a dipole without translating its centre.

The two charges feel equal and opposite forces. Their net force is zero, but the pair forms a couple unless the dipole is parallel or antiparallel to the field. Its torque is $\boldsymbol\tau=\mathbf p\times\mathbf E$, with magnitude $pE\sin\theta$. The torque tends to align p with E.

Parallel alignment is stable: a small displacement produces a restoring torque. Antiparallel alignment is unstable: a small displacement moves the dipole farther away. A nonuniform field can also exert a net force because the forces at the two ends need not balance.

A dipole of moment 12 nC m makes 30° with a uniform field of 500 N/C. Find its torque magnitude.

Use $\tau=pE\sin\theta$.

Result: 3 μN m.

1. At what angle is dipole torque greatest?

Answers

1. 90°.

Large collections of charges can be modelled by charge per length, area or volume.

For a thin line, $\lambda=dq/dl$ in C/m. For a sheet, $\sigma=dq/dA$ in C/m². For a volume, $\rho=dq/dV$ in C/m³. Uniform distributions give $Q=\lambda L$, $Q=\sigma A$ or $Q=\rho V$.

To find a field, choose a small source element $dq$, write $d\mathbf E=k\,dq\,\hat{\mathbf r}/r^2$, resolve its components and integrate over the source. The vector $\mathbf r$ runs from each source element to the observation point; it generally changes during integration.

A uniformly charged 0.75 m wire has line density −8 nC/m. Find its total charge.

Multiply the signed density by the wire length.

Result: -6 nC.

1. What density is most useful for a very thin charged foil?

Answers

1. Surface charge density σ, provided thickness can be neglected.

Gauss’s law connects total outward flux with enclosed charge.

For any closed surface, $\displaystyle\oint\mathbf E\cdot d\mathbf A=Q_{\rm enc}/\varepsilon_0$. All charges contribute to E on the surface, but only enclosed charge determines the net flux. External charges can produce equal inward and outward contributions.

For a charge at the centre of a sphere, E is radial and constant on the sphere. Thus flux is E times its area. Moving the charge away from the centre changes the distribution of field over the surface, but the total flux remains q/ε₀ as long as it stays inside. A zero enclosed charge gives zero net flux; it does not force E to be zero everywhere.

A closed surface encloses +6 nC and −2 nC. Use ε₀ ≈ 8.85 × 10⁻¹² C²/(N m²) to find its outward flux.

Only the enclosed algebraic sum enters Gauss’s law.

Result: $\Phi_E=451.9774\ \mathrm{N\,m^2\,C^{-1}}$.

1. A charge lies outside an empty closed surface. What net flux does it contribute?

Answers

1. Zero, even though it generally produces nonzero field on that surface.

Choose a Gaussian surface that matches the source symmetry.

Use a coaxial cylinder of radius r and length L, away from the ends of a sufficiently long wire. The curved surface has constant radial E. The flat ends have zero flux because their normals are axial. Therefore $E(2\pi rL)=\lambda L/\varepsilon_0$, so $E=\lambda/(2\pi\varepsilon_0r)$.

Use a pillbox crossing a uniformly charged sheet. Each flat face contributes EA and the curved side contributes zero. Hence $2EA=\sigma A/\varepsilon_0$ and $E=\sigma/(2\varepsilon_0)$ on either side. The field points away for positive σ. A finite sheet approximates this only near its centre and close relative to its width.

Outside a shell of radius R, a concentric sphere gives $E=kQ/r^2$. Inside, the enclosed charge is zero and spherical symmetry makes E constant on the Gaussian sphere; therefore E=0. At the charged surface the ideal field has different inner and outer limits. Do not apply the outside formula inside the shell.

An ideal long line has λ=3 nC/m. Find the field at r=0.40 m using k=9 × 10⁹ SI units.

Since $1/(2\pi\varepsilon_0)=2k$, use $E=2k\lambda/r$.

Result: 135 N/C, radially outward.

1. What is E at a point inside a uniformly charged spherical shell?

Answers

1. Zero, by spherical symmetry and Gauss’s law.

Use $\mathbf F=q\mathbf E$, superposition and charge conservation before choosing an equation. Coulomb’s law is a point-source law; Gauss’s law is a closed-surface law.

A dipole has moment from − to +. Far fields scale as $r^{-3}$. A long line has $E\propto r^{-1}$; an ideal sheet has constant E; a spherical shell has zero interior E.

• Neutral does not mean empty of charge.
• Zero flux does not imply zero field.
• State symmetry and approximation limits before using a compact field formula.

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

1.1 An object loses 40 million electrons. Find its charge using e=1.60 × 10⁻¹⁹ C.

1.2 Charges −15 nC and +21 nC are combined on an isolated conductor. Find the total.

1.3 Explain why a charged rod attracts a neutral conductor.

1.4 Why must an induction experiment disconnect Earth before removing the inducing rod?

1.5 Charges 2 μC and 5 μC are 0.50 m apart. Find the force magnitude.

1.6 A force is 0.72 N at one separation. What is it at twice that separation?

1.7 Two perpendicular force components are 0.5 N and 1.2 N. Find the resultant magnitude.

1.8 Equal positive charges lie symmetrically about an origin. Explain E there.

1.9 A charge of −8 μC experiences 0.024 N along +x. Find E.

1.10 A +3 μC charge is placed in a field of 700 N/C. Find the force.

1.11 Why do electrostatic field lines never cross?

1.12 Why is a field line not generally the trajectory of an electron?

1.13 A 0.08 m² area is normal to a 120 N/C field. Find flux.

1.14 The same area is tilted so its normal is at 60° to E. Find flux.

1.15 A closed surface encloses no charge. Must its field be zero?

1.16 Charges ±4 nC are 6 mm apart. Find their dipole moment magnitude.

1.17 For a short dipole, compare the axial fields at r and 3r.

1.18 A 5 nC m dipole is perpendicular to a 400 N/C field. Find torque.

1.19 What distinguishes stable and unstable dipole alignment?

1.20 A sheet of area 0.12 m² carries uniform density 25 nC/m². Find charge.

1.21 A sphere encloses +9 nC. If its radius doubles, how does total flux change?

1.22 For a long charged line, how does E change when r doubles?

1.23 For an ideal infinite sheet, how does E change when distance doubles?

1.24 A charged spherical shell has radius 0.20 m and charge 8 nC. Find E at 0.40 m.

1.25 For the same shell, find E at 0.10 m.

1.26 What assumption permits E to be taken outside a Gaussian flux integral?

Answers

1.1 +6.4 pC.

1.2 6 nC.

1.3 Mobile charges separate; the closer opposite charge gives the stronger attraction.

1.4 Otherwise charge can flow back and remove the imbalance.

1.5 0.36 N.

1.6 0.18 N.

1.7 1.3 N.

1.8 Their equal opposite fields cancel; E=0.

1.9 3000 N/C along −x.

1.10 2.1 mN along E.

1.11 A nonzero field cannot have two directions at one point.

1.12 Motion depends on initial velocity and acceleration; the electron’s acceleration is opposite E.

1.13 9.6 N m²/C.

1.14 4.8 N m²/C.

1.15 No. Only the net outward flux must be zero.

1.16 24 pC m.

1.17 The second is one twenty-seventh of the first.

1.18 2 μN m.

1.19 Parallel alignment is stable; antiparallel alignment is unstable.

1.20 3 nC.

1.21 It is unchanged; enclosed charge is unchanged.

1.22 It halves.

1.23 It is unchanged.

1.24 450 N/C outward.

1.25 Zero.

1.26 Its relevant normal component must be constant on the chosen part of the surface, justified by symmetry.