Electrostatic Potential, Potential Energy and Capacitance | ISC Class 12 Physics Notes

This note covers electrostatic potential and potential difference, point charges and dipoles, equipotential surfaces, electrostatic potential energy, conductors and dielectrics, capacitance, parallel plate capacitors, capacitor combinations, stored energy and energy density.

What do electrostatic potential and potential difference measure?

Electrostatics concerns charges at rest. The electrostatic force is conservative: its work between two positions is independent of the path. This allows us to associate a change in potential energy with a change in the arrangement of charges.

A test charge is a charge small enough not to disturb the source charges. Let q₀ denote its charge, W the work done by an external agency, U its potential energy and V the electric potential. Subscripts identify positions or individual charges.

Definition: Potential difference is the external work done per unit positive test charge in moving it between two points without acceleration. Electric potential at a point uses infinity as the reference when the potential there is chosen as zero.

For motion from position B to position A, let WBA be the external work and ΔU the final potential energy minus its initial value. The symbol Δ denotes a change. Then VA − VB = WBA/q₀ and ΔU = q₀(VA − VB).

How are the units and dimensions expressed?

SI means the International System of Units. In dimensional expressions, M, L, T and I represent mass, length, time and electric current, the rate of flow of charge. A scalar has magnitude and sign without a spatial direction; a vector also has direction.

The SI unit of charge is the coulomb. The SI unit of work is the joule. The SI unit of potential is the volt: 1 V = 1 J C⁻¹. The SI unit of electric field is the newton per coulomb, equivalent to the volt per metre.

The electric force does work opposite in sign to W when the charge moves without a change in kinetic energy, its energy of motion. Only potential differences have physical significance; changing the common reference does not change them.

How is the potential of a point charge obtained?

A point charge has negligible size compared with the distances considered. Let q be a fixed source charge and r its distance from the observation point. Distance is measured in metres, symbol m, with dimensions [L]. Take the potential at infinity as zero.

Write k = 1/(4πε₀), where π is the circle constant and ε₀ is the permittivity of free space, the vacuum constant in Coulomb's law. Its unit is C² N⁻¹ m⁻² and dimensions are [M⁻¹ L⁻³ T⁴ I²].

The constant k has unit N m² C⁻², dimensions [M L³ T⁻⁴ I⁻²], and approximate value 9 × 10⁹ N m² C⁻². In vacuum, the signed radial field component is Eᵣ = kq/r²; positive means radially outwards.

Derivation: potential and potential difference of a point charge

1. Let rA and rB be the distances of A and B from q. Choose a radial path because electrostatic work is independent of the path.
2. For a small outward displacement dr, the external work per unit test charge is −Eᵣ dr. The symbol d denotes an infinitesimal change; ∫ denotes integration, or continuous summation.
3. Integrating gives VA − VB = −∫ from rB to rA of kq/r² dr = kq(1/rA − 1/rB).
4. Place B at infinity and set VB = 0. The term 1/rB vanishes, giving the potential at distance r.

V = kq/r. Positive q gives positive potential; negative q gives negative potential. The result applies away from the location of the point charge.

The horizontal axis is r. The curves labelled V and E decrease towards zero, showing the 1/r and 1/r² dependences on separately scaled vertical units. Their plotted intersection does not mean equal physical quantities.

See Fig. 2.4 in your NCERT textbook

Worked example 1. A charge q = 4 × 10⁻⁷ C is 9 cm from P. Find the potential and external work to bring q₀ = 2 × 10⁻⁹ C from infinity to P without acceleration. Use k = 9 × 10⁹ N m² C⁻² and zero potential at infinity.

Formula: V = kq/r; W = q₀V. Substitute: r = 0.09 m; V = (9 × 10⁹)(4 × 10⁻⁷)/0.09.

Answer: V = 40000 V = 4 × 10⁴ V; W = (2 × 10⁻⁹)(4 × 10⁴) = 8 × 10⁻⁵ J. The work is independent of the path.

How are potentials due to several charges combined?

The superposition principle for potential requires algebraic addition. For charges q₁, q₂ and q₃ at distances r₁, r₂ and r₃ from an observation point, V = k(q₁/r₁ + q₂/r₂ + q₃/r₃). Each distance is positive; each charge keeps its sign.

Potential is a scalar, so there is no need to resolve contributions into directions. Electric field is a vector and must be added with directions included. Consequently, a point of zero potential need not have zero electric field.

How should a zero-potential position be found?

First specify the zero reference, usually infinity for a finite collection of charges. Divide the line into regions separated by the charge positions. In each region, express the physical distances correctly before setting the algebraic sum of the potentials equal to zero.

Worked example 2. Charges +3 × 10⁻⁸ C and −2 × 10⁻⁸ C are separated by 15 cm. Find the finite zero-potential points on their joining line, taking zero potential at infinity. Put the positive charge at coordinate x = 0 and the negative charge at x = 15 cm; x measures position along that line.

Formula: V = k(q₁/r₁ + q₂/r₂) = 0. Between the charges, substitute distances x and 15 − x, both in centimetres: 3/x = 2/(15 − x).

Answer: x = 9 cm. Beyond the negative charge, 3/x = 2/(x − 15), giving x = 45 cm. To the left of the positive charge its larger charge magnitude and smaller distance prevent cancellation.

At the point between these opposite charges, both electric field contributions point from the positive charge towards the negative one. Thus their potentials cancel while their fields reinforce. This illustrates why potential and field must be calculated separately.

For a continuous charge distribution, divide the charge into small elements, find each element's potential and integrate their contributions. This is the same superposition procedure with a continuous sum in place of a finite one.

What is the potential due to an electric dipole?

An electric dipole consists of equal and opposite charges +q and −q separated by a small distance 2l, where l is half their separation. Its dipole moment has magnitude p = 2ql and points from the negative charge towards the positive charge.

The SI unit of dipole moment is C m and its dimensions are [I T L]. Let r be the distance from the dipole centre to P, and θ the angle between the dipole moment and the line from the centre to P. Angles are dimensionless.

What are the exact axial and equatorial results?

If r₊ and r₋ are P's distances from +q and −q respectively, V = kq(1/r₊ − 1/r₋). On the axis beyond the positive charge, r₊ = r − l and r₋ = r + l, for r greater than l.

Subtracting the reciprocals gives Vaxial = kp/(r² − l²) on that side. Here Vaxial denotes axial potential. At an equal distance beyond the negative charge, the potential has the same magnitude and opposite sign.

On the equatorial plane, the plane through the centre perpendicular to the dipole axis, both charges are equally distant. Their potentials cancel exactly: Vequatorial = 0. This statement does not require a large-distance approximation.

What changes for a short dipole?

For r much greater than 2l, the dipole is short compared with the observation distance. To first order, 1/r₊ is approximately (1 + l cos θ/r)/r and 1/r₋ is approximately (1 − l cos θ/r)/r.

Subtracting gives V ≈ kp cos θ/r². The symbol ≈ means approximately equal. Axial values are approximately +kp/r² and −kp/r²; the equatorial value is zero. The general expression is exact for an ideal point dipole.

The dipole potential depends on distance and angle. At large distances it decreases as 1/r², whereas a single point charge's potential decreases as 1/r. Do not use the short-dipole approximation when the observation distance is comparable to the separation.

What do equipotential surfaces show about the electric field?

An equipotential surface has the same potential at every point. Moving a test charge along it gives no change in potential energy, so the electrostatic field does no work on that motion. A nonzero field can still exist perpendicular to the surface.

If the field had a component along the surface, motion against that component would require work. That would contradict the constant potential. Therefore, wherever the field is nonzero, it is normal, meaning perpendicular, to the equipotential surface.

Part (a) shows concentric circular sections of spherical equipotential surfaces around +q. Part (b) shows straight radial field arrows directed outwards from the positive central charge.

See Fig. 2.9 in your NCERT textbook

Parallel planes cut across parallel horizontal arrows labelled E. The planes represent equipotential surfaces, and the arrows represent a field perpendicular to them.

See Fig. 2.10 in your NCERT textbook

How is field related to the potential gradient?

Let s be distance measured in the direction of the electric field. The potential gradient, dV/ds, is potential change per unit distance, with unit V m⁻¹ and the same dimensions as electric field. Then E = −dV/ds.

The minus sign means that potential decreases along the field. For a uniform field, the magnitude of the potential difference across perpendicular separation d is Ed. Closely spaced surfaces representing equal potential increments indicate a stronger field.

For a uniformly charged thin spherical shell of radius R and total charge q, the outside potential is kq/r. On the surface and throughout the interior it is kq/R. Here R is a length measured in metres.

The shell potential is continuous at the surface. Inside, the field is zero because the potential is constant, not necessarily because it is zero. Outside, the field and potential behave as those of a point charge at the centre.

How is the potential energy of a charge system calculated?

In a prescribed external potential V, the potential energy of a charge q is U = qV. External means that V is produced by other charges. The charge's own potential must not be included in this expression.

The source charges must remain fixed or be negligibly affected by q. When q moves from B to A without acceleration, the external work is q(VA − VB). The sign of q matters: negative charges have lower potential energy at higher external potential.

How are two or three charges assembled?

1. Bring the first charge q₁ from infinity. With no other charges or external field present, no interaction work is required.
2. Bring q₂ to a distance r₁₂ from q₁, holding q₁ fixed. The external work is kq₁q₂/r₁₂.
3. For a third charge q₃, let r₁₃ and r₂₃ be its distances from q₁ and q₂. Its additional assembly work is kq₁q₃/r₁₃ + kq₂q₃/r₂₃.
4. Add the pair energies, counting each distinct pair once. The result depends on the final arrangement, not the assembly order.

U₁₂ = kq₁q₂/r₁₂ is the two-charge interaction energy. For three charges, U₁₂₃ = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃). The subscripts identify the charges included in each energy.

With zero interaction energy at infinite separation, like charges have positive pair energy. Unlike charges have negative pair energy. Separating unlike charges to infinity requires positive external work, since their interaction energy must increase to zero.

Worked example 3. Charges +7 μC and −2 μC are 18 cm apart in vacuum, without an external field. Find their interaction energy and the work required to separate them to infinity without acceleration. The prefix μ means 10⁻⁶. Use k = 9 × 10⁹ N m² C⁻².

Formula: U = kq₁q₂/r₁₂; W = 0 − U. Substitute: U = (9 × 10⁹)(7 × 10⁻⁶)(−2 × 10⁻⁶)/0.18.

Answer: U = −0.7 J and W = +0.7 J. The negative interaction energy reflects the attractive pair relative to infinite separation.

Why does a dipole have orientation-dependent potential energy?

A dipole in a uniform electric field experiences equal and opposite forces on its charges. Their resultant force is zero. They can nevertheless produce a torque, a turning effect, unless the dipole moment is parallel or antiparallel to the field.

Let τ denote torque magnitude. Its SI unit is N m and its dimensions are [M L² T⁻²]. If θ is the angle between dipole moment and field, τ = pE sin θ. The torque tends to reduce the angle.

Derivation: potential energy of a dipole in a uniform field

1. Rotate the dipole very slowly using an external torque that balances the electric torque, so there is no change in rotational kinetic energy.
2. For a small increase dθ, the external work is pE sin θ dθ. Integrate from initial angle θ₀ to final angle θ.
3. The work is pE(cos θ₀ − cos θ), which equals the change in potential energy.
4. Choose the energy to be zero at θ₀ = 90°. Since cos 90° = 0, the orientation-dependent energy is −pE cos θ.

U = −pE cos θ. Equivalently, this is minus the scalar product of the dipole-moment vector and electric-field vector. It describes interaction with the external field; the fixed internal energy of the dipole adds only a constant.

Note: A torque tending to align the dipole does not guarantee that it settles along the field. Without energy dissipation it can oscillate; damping is required for eventual alignment.

How do conductors and dielectrics respond to an electric field?

A conductor contains mobile charge carriers. In a metal, free electrons can move within the material, while positive ions and bound electrons remain associated with fixed lattice positions. Bound charges cannot migrate through the material as free carriers do.

Electrostatic equilibrium is the static condition after charge redistribution, with no current. The field within the conducting material is zero; otherwise free charges would move. Excess charge resides on its surface, and potential is constant throughout and on that surface.

The field immediately outside a charged conductor is normal to its surface. Its signed outward component is σ/ε₀, where surface charge density σ is charge per unit area. Its unit is C m⁻² and dimensions are [I T L⁻²].

Electrostatic shielding means that a charge-free cavity enclosed by a conductor is protected from external electrostatic fields. The field inside such a cavity is zero. The condition that no charge is placed in the cavity must be retained.

What is polarisation?

A dielectric is a non-conducting substance with no, or a negligible number of, free charge carriers. Its charges cannot redistribute freely as in a conductor. An applied field instead stretches or reorients molecular charge distributions, producing a net dipole moment.

In non-polar molecules, positive and negative charge centres coincide without an applied field. Oxygen and hydrogen molecules are examples. A field can displace these centres relative to each other and induce a dipole moment.

In polar molecules, the centres are already separated, giving a permanent dipole moment. Water is an example. Thermal agitation randomises orientations without an external field. Applied fields make the dipoles tend to align, while thermal motion opposes alignment.

There may also be an induced-dipole contribution in polar molecules, but generally the alignment effect is more important. Polarisation, denoted P, is dipole moment per unit volume. Its SI unit is C m⁻² and dimensions are [I T L⁻²].

For the dielectric slab considered here, polarisation produces bound surface charges whose field opposes the original field. It reduces that field without cancelling it exactly. The extent of the reduction depends on the dielectric material. In a linear isotropic dielectric, polarisation is proportional to the field and points in its direction.

What determines the capacitance of a parallel plate capacitor?

A capacitor consists of two conductors separated by an insulating region. Usually, in practice, its conductors carry equal and opposite charges +Q and −Q. Here Q is the magnitude on either conductor; V denotes their potential difference.

Capacitance is the charge magnitude per unit potential difference: C = Q/V. For fixed geometry and dielectric properties in the linear regime, Q and V change proportionally. Their ratio depends on conductor shape, size, separation and the intervening medium.

The SI unit of capacitance is the farad: 1 F = 1 C V⁻¹. From C = Q/V, its dimensions are [M⁻¹ L⁻² T⁴ I²]. A microfarad is 10⁻⁶ F, a nanofarad is 10⁻⁹ F and a picofarad is 10⁻¹² F, written μF, nF and pF.

Derivation: capacitance with vacuum between parallel plates

Let A be each plate's area, measured in m² with dimensions [L²], and d their separation in metres. Assume d² is much smaller than A and consider the region sufficiently far from the edges.

1. The facing plates have approximately uniform surface charge densities +σ and −σ, with σ = Q/A.
2. The field of each ideal plane sheet has magnitude σ/(2ε₀). Between the plates the contributions add, giving E = σ/ε₀ = Q/(ε₀A).
3. Outside the ideal plates the contributions cancel. Between them, the uniform field gives potential difference V = Ed = Qd/(ε₀A).
4. Divide the plate charge by the potential difference: C = Q/V = ε₀A/d.

C = ε₀A/d. A larger plate area increases capacitance; a larger separation decreases it, provided the other quantities remain constant. An isolated conductor likewise has capacitance Q/V, using its potential relative to the chosen reference at infinity.

Plate 1 lies above plate 2. The labels show plate area A, separation d, surface charge densities σ and −σ, and downward electric-field arrows E between the positive and negative plates.

See Fig. 2.25 in your NCERT textbook

Fringing is the outward bending of field lines near plate edges. Finite plates do not have a perfectly uniform field everywhere. The ideal calculation neglects this edge effect under the stated small-separation condition.

How does a dielectric change capacitance under different conditions?

The dielectric constant K is the dimensionless relative permittivity: K = ε/ε₀, where ε is the permittivity of the material, with the same units and dimensions as ε₀. Relative permittivity is also written εᵣ. For vacuum, K = 1.

Let C₀ be the capacitance before inserting a dielectric and C′ the capacitance afterwards. A prime labels the changed state. If a linear dielectric completely fills the gap, C′ = KC₀ = Kε₀A/d, so K = C′/C₀.

Is the charge or the potential difference held constant?

For a disconnected capacitor with no leakage, charge remains Q. Polarisation weakens the field from E₀ to E′ = E₀/K. The potential difference falls from V₀ to V′ = V₀/K. The subscript 0 identifies the original vacuum state.

For a capacitor kept connected to a source maintaining the same potential difference, V′ = V₀. The source supplies additional charge, so Q′ = KQ. With separation unchanged, E′ = V₀/d equals the original field between the plates.

What if the slab fills only part of the separation?

Let t be the slab thickness, a length in metres, and assume the slab covers the full plate area A. The remaining vacuum gap has thickness d − t. Neglect fringing and use a uniform linear dielectric of constant K.

The potential drops across the two regions add: V = (Q/ε₀A)(d − t + t/K). Hence C′ = ε₀A/(d − t + t/K). The limits t = 0 and t = d recover the vacuum and completely filled results.

For the slab thickness t = 3d/4, substitution gives C′ = 4KC₀/(K + 3). With charge fixed, the potential becomes V′ = V₀(K + 3)/(4K). This thickness arrangement differs from a dielectric covering only part of the plate area.

How are capacitors combined in series and parallel?

Equivalent capacitance, written Cₑ, is the capacitance of a single capacitor that would take the same terminal charge at the same applied potential difference as the combination. Determine which conductors share a potential before selecting a formula.

How does a series combination work?

In a series combination, capacitors are joined end to end. For initially uncharged capacitors with neutral isolated intermediate connections, each capacitor acquires the same charge magnitude Q. Their individual potential differences add to the applied total.

For capacitances C₁, C₂ and C₃, write V = V₁ + V₂ + V₃ = Q/C₁ + Q/C₂ + Q/C₃. Since V = Q/Cₑ, division by Q gives 1/Cₑ = 1/C₁ + 1/C₂ + 1/C₃.

Worked example 4. Three initially uncharged 9 pF capacitors are connected in series across 120 V. Find their equivalent capacitance and individual potential differences, with neutral intermediate connections. Use 1 pF = 10⁻¹² F.

Formula: 1/Cₑ = 3/C; Q = CₑV; V₁ = Q/C. Substitute: Cₑ = 9/3 pF = 3 pF; Q = (3 × 10⁻¹²)(120) C.

Answer: Cₑ = 3 pF; Q = 3.6 × 10⁻¹⁰ C. Each potential difference is (3.6 × 10⁻¹⁰)/(9 × 10⁻¹²) = 40 V, and the three drops add to 120 V.

How does a parallel combination work?

In a parallel combination, corresponding terminals connect to the same two conductors, so each capacitor has the same V. The charges add: Q = C₁V + C₂V + C₃V. Therefore Cₑ = C₁ + C₂ + C₃.

Worked example 5. Capacitors of 2 pF, 3 pF and 4 pF are connected in parallel across 100 V. Find the equivalent capacitance and each charge magnitude. Use 1 pF = 10⁻¹² F.

Formula: Cₑ = C₁ + C₂ + C₃; Q₁ = C₁V, with corresponding expressions for Q₂ and Q₃. Substitute: Cₑ = 2 + 3 + 4 pF; Q₁ = (2 × 10⁻¹²)(100) C.

Answer: Cₑ = 9 pF; Q₁ = 2 × 10⁻¹⁰ C, Q₂ = 3 × 10⁻¹⁰ C and Q₃ = 4 × 10⁻¹⁰ C. Every capacitor has the full 100 V across it.

How much energy does a capacitor store?

A charged capacitor stores electrostatic energy, released when it discharges. For capacitance C, charge magnitude Q and potential difference V, the equivalent formulae are U = Q²/(2C), U = CV²/2 and U = QV/2.

Choose the expression matching the quantities known or held fixed. At fixed charge, increasing capacitance reduces stored energy. At fixed potential difference, increasing capacitance increases it. In the connected case, energy exchange with the source must be included in a complete energy account.

What is electric energy density?

Energy density u is energy per unit volume. Its SI unit is J m⁻³ and dimensions are [M L⁻¹ T⁻²]. For an electric field in vacuum, u = ε₀E²/2. In the ideal parallel plate region, volume is Ad and U = uAd.

Worked example 6. A 900 pF capacitor is charged by a 100 V battery. Find its stored energy, using 1 pF = 10⁻¹² F.

Formula: Q = CV; U = QV/2. Substitute: Q = (900 × 10⁻¹²)(100) = 9 × 10⁻⁸ C.

Answer: U = (9 × 10⁻⁸)(100)/2 = 0.0000045 J = 4.5 × 10⁻⁶ J. The formula CV²/2 gives the same result.

What happens when charge is shared?

Disconnect that charged capacitor and connect it in parallel to an identical uncharged 900 pF capacitor. The total charge is conserved. Both settle at the same potential difference, so each receives half the original charge and the common potential difference is 50 V.

The combined energy becomes 2.25 × 10⁻⁶ J, half the initial value. During redistribution, a transient current, meaning a temporary flow of charge, transfers energy into heat and electromagnetic radiation. Conserved total charge does not imply conserved stored electrostatic energy.

Note: The capacitor's charge Q means the magnitude on one plate, not the sum of the two plate charges. The plates carry +Q and −Q, while their electrostatic energy is positive.

Glossary

• Electrostatic potential — External work per unit positive test charge brought without acceleration from the chosen zero-potential reference to a point.
• Potential difference — Difference between two potentials, equal to external work per unit test charge moved without acceleration between those positions.
• Conservative force — A force whose work between two positions is independent of the path taken.
• Equipotential surface — A surface on which every point has the same electric potential.
• Electric dipole — A pair of equal and opposite charges separated by a small distance.
• Dipole moment — Charge magnitude multiplied by charge separation, directed from the negative charge towards the positive charge.
• Electrostatic shielding — Protection of a charge-free cavity within a conductor from external electrostatic fields.
• Dielectric — A non-conducting substance with no, or a negligible number of, free charge carriers.
• Polarisation — The electric dipole moment per unit volume of a dielectric material.
• Capacitance — Charge magnitude on one capacitor conductor divided by the potential difference between its conductors.
• Relative permittivity — The dimensionless ratio of a material's permittivity to the permittivity of free space.
• Energy density — The energy stored in a region divided by the volume of that region.

Common errors and misconceptions

• Misconception: Zero potential at a point means that its electric field is zero. Correct: Potential contributions can cancel while vector field contributions reinforce, as between the opposite charges in the zero-potential example.
• Misconception: External work and work done by the electric field have the same sign. Correct: For motion without acceleration they are opposite; external work equals the potential-energy increase.
• Misconception: The short-dipole potential applies at every distance. Correct: The finite-dipole approximation requires an observation distance much larger than the charge separation.
• Misconception: A conductor with zero internal field must have zero potential. Correct: Its potential is constant throughout, and can be nonzero relative to infinity.
• Misconception: Inserting a dielectric reduces voltage even with a fixed-voltage source connected. Correct: Connection holds voltage constant while additional charge flows onto the plates.
• Misconception: Series capacitors have equal potential differences even when their capacitances differ. Correct: Under the usual neutral-intermediate-connection condition, they have equal charge magnitudes; voltage varies inversely with capacitance.
• Misconception: Capacitor energy is QV. Correct: The stored energy is QV/2; using the final voltage for every increment of transferred charge overestimates the energy.

Exam-style questions with model answers

Q1. Define electrostatic potential relative to infinity and state its SI unit. [2 marks]

1. Electrostatic potential is the external work per unit positive test charge in bringing it from infinity to the point without acceleration, taking the potential at infinity as zero.
2. Its SI unit is the volt: one volt equals one joule per coulomb.

Q2. A fixed charge of +4 × 10⁻⁷ C is 9 cm from P in vacuum. Find the potential at P and external work to bring +2 × 10⁻⁹ C from infinity to P without acceleration. Take zero potential at infinity and k = 9 × 10⁹ N m² C⁻². [3 marks]

1. The distance must be in metres: r = 0.09 m. For a point charge q, the potential relative to infinity is V = kq/r.
2. Substitution gives V = (9 × 10⁹)(4 × 10⁻⁷)/0.09 = 4 × 10⁴ V at P.
3. For the arriving charge q₀, external work W = q₀V = (2 × 10⁻⁹)(4 × 10⁴) = 8 × 10⁻⁵ J. Positive work is required against repulsion.

Q3. A rigid dipole of moment magnitude p makes angle θ with a uniform electric field of magnitude E. Starting from torque magnitude pE sin θ, derive its orientation-dependent potential energy, choosing zero energy at 90°, and identify the stable orientation. [5 marks]

1. The electric torque tends to align the dipole with the field. Rotate it slowly with an external torque that balances the electric torque, preventing angular acceleration.
2. For a small angular increase dθ, the external work is dW = pE sin θ dθ. This work is stored as potential energy.
3. Between initial angle θ₀ and final angle θ, integration gives W = pE(cos θ₀ − cos θ).
4. Choose θ₀ = 90° and U(90°) = 0. The result is U(θ) = −pE cos θ, since cos 90° is zero.
5. At θ = 0°, U = −pE is a minimum. This parallel orientation is stable because a small angular displacement produces a restoring torque.

Q4. Three point charges q₁, q₂ and q₃ are held in vacuum at pair separations r₁₂, r₁₃ and r₂₃. With no external field and zero energy when infinitely separated, obtain their total assembly energy. Use k = 1/(4πε₀), where ε₀ is vacuum permittivity. [3 marks]

1. Bring q₁ first from infinity. No other charge is present, so no interaction work is needed. Bringing q₂ next requires work kq₁q₂/r₁₂.
2. The potential at the third position due to the first two charges is k(q₁/r₁₃ + q₂/r₂₃). Bringing q₃ there requires q₃ times this potential.
3. Adding gives U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃). Each distinct pair contributes once, and every charge retains its algebraic sign.

Q5. Derive the vacuum capacitance of parallel plates of area A, separation d and charges +Q and −Q. Assume d² is much smaller than A and neglect fringing. Use the field magnitude σ/(2ε₀) for one infinite charged sheet, where σ is charge per unit area and ε₀ is vacuum permittivity. [5 marks]

1. The large, closely spaced plates have approximately uniform facing surface charge densities +σ and −σ, where σ = Q/A.
2. Between the plates the two sheet fields point in the same direction. Their magnitudes add to E = σ/(2ε₀) + σ/(2ε₀) = σ/ε₀.
3. Substitute σ = Q/A to obtain E = Q/(ε₀A). In the ideal approximation the contributions outside the plates cancel.
4. The internal field is uniform, so the potential difference is V = Ed = Qd/(ε₀A), where V is the positive plate-to-plate voltage magnitude.
5. By definition C = Q/V. Substitution gives C = ε₀A/d, valid away from edges under the stated large-plate approximation.

Q6. A linear dielectric of constant K completely fills an initially vacuum-filled parallel plate capacitor without changing its plate area or separation d. Neglect fringing. Initially its capacitance, charge magnitude and voltage are C₀, Q and V₀. Compare the final capacitance, charge, voltage and field when it is disconnected without leakage and when a source maintains V₀. [4 marks]

1. Full insertion changes the capacitance to C′ = KC₀ in both cases. This depends on the dielectric and unchanged geometry.
2. When disconnected without leakage, the charge stays Q. Therefore V′ = Q/C′ = V₀/K.
3. For unchanged separation d, E′ = V′/d. The disconnected capacitor's field becomes E₀/K, where E₀ = V₀/d was the initial field.
4. When connected, V′ = V₀ and E′ = E₀. The source supplies charge until Q′ = C′V₀ = KQ.

Q7. Three initially uncharged 9 pF capacitors are connected in series across 120 V, with neutral isolated intermediate connections. Calculate equivalent capacitance, charge magnitude on each capacitor and voltage across each. Use 1 pF = 10⁻¹² F. [3 marks]

1. For series capacitors, reciprocals add: 1/Cₑ = 1/9 + 1/9 + 1/9 in inverse picofarads. Thus the equivalent capacitance is Cₑ = 3 pF.
2. All three capacitors acquire the same charge magnitude. It is Q = CₑV = (3 × 10⁻¹²)(120) = 3.6 × 10⁻¹⁰ C.
3. Each voltage is Q/C = (3.6 × 10⁻¹⁰)/(9 × 10⁻¹²) = 40 V. The three voltages add to the applied 120 V.

Q8. A 900 pF capacitor is charged to 100 V, disconnected from its battery, and connected in parallel to an identical uncharged capacitor. Assume no charge leakage. Calculate the initial energy, common final voltage, final total energy and energy transferred out of the electrostatic store. Use 1 pF = 10⁻¹² F. [5 marks]

1. The initial charge is Q = CV = (900 × 10⁻¹²)(100) = 9 × 10⁻⁸ C. This is conserved during sharing.
2. The initial stored energy is U = CV²/2 = (900 × 10⁻¹²)(100)²/2 = 4.5 × 10⁻⁶ J.
3. The final parallel capacitance is 1800 pF. The common voltage is Q/(2C) = (9 × 10⁻⁸)/(1800 × 10⁻¹²) = 50 V.
4. The final total energy is (2C)(50)²/2 = 2.25 × 10⁻⁶ J, counting the energy in both capacitors.
5. The energy transferred out is 4.5 × 10⁻⁶ − 2.25 × 10⁻⁶ = 2.25 × 10⁻⁶ J. During the transient current it becomes heat and electromagnetic radiation.

Key takeaways

• Potential difference measures external work per unit positive test charge, with motion slow enough to avoid changing kinetic energy.
• Point-charge potentials add algebraically, whereas electric fields require vector addition with the directions of their contributions included.
• A finite dipole has exactly zero equatorial potential, while its general short-dipole expression requires distances much larger than its separation.
• Equipotential surfaces are perpendicular to a nonzero electric field, which points towards the steepest decrease in potential.
• Charge-system energy counts each distinct pair once; dipole orientation energy is minimum when its moment points along a uniform field.
• Conductors redistribute free charges until their internal electrostatic field vanishes; dielectrics respond through molecular polarisation.
• Dielectric insertion increases capacitance, but charge and voltage changes depend on whether the capacitor remains connected to its source.
• Capacitor energy depends on charge, voltage and capacitance; charge sharing can conserve charge while reducing stored electrostatic energy.

Test yourself

Why is a zero reference needed before assigning a potential?

Potential is defined up to an additive constant. A chosen reference fixes its numerical value while leaving potential differences unchanged.

Why can a dipole have zero equatorial potential but a nonzero field there?

The scalar potentials of the equally distant opposite charges cancel. Their vector electric fields do not cancel on that plane.

What prevents a static conductor from maintaining a tangential surface field?

A tangential field would exert force on free surface charges and make them move, contradicting electrostatic equilibrium.

What distinction separates an induced dipole from a permanent molecular dipole?

An induced dipole develops through field-driven charge displacement. A permanent dipole exists within a polar molecule even without an external field.

Why must the connection to a battery be stated in a dielectric-insertion problem?

A disconnected capacitor retains charge without leakage, while a fixed-voltage source maintains potential difference and supplies additional charge.

What does the partial-slab formula predict when the slab thickness equals the full gap?

Setting t = d gives C′ = Kε₀A/d, the capacitance for a dielectric completely filling the space between the plates.

Why do equal series capacitors divide the applied voltage equally under the usual neutral-connection condition?

They carry the same charge magnitude and have equal capacitances. Their individual voltages Q/C are therefore equal.

Why does equal sharing with an identical uncharged capacitor reduce stored energy?

The total charge is unchanged but the equivalent capacitance doubles. Using Q²/(2C) for the combination therefore halves its electrostatic energy.