Electrostatic potential and potential of a point charge
Water flows from a higher tank to a lower one; in the same way positive charge moves from a higher electric potential to a lower one. Link to earlier classes : work and potential energy from Class 11.
The electrostatic potential V at a point is the work done per unit positive charge in bringing it from infinity to that point without acceleration. The potential difference between two points is the work per unit charge needed to move a charge between them. The unit is the volt (1 V = 1 J C⁻¹).
Rule 1: Potential of a point charge — V = k Q ÷ r (Q = source charge with its sign, r = distance). The work done in moving a charge q between two points is W = q (V₂ − V₁).
- Easy : Q = 4 μC, r = 0.60 m ⇒ V = 9 × 10⁹ × 4 × 10⁻⁶ ÷ 0.60 = 6 × 10⁴ V.
- Medium : Moving 2 μC from a point at 2 × 10⁴ V to a point at 6 × 10⁴ V needs W = 2 × 10⁻⁶ × 4 × 10⁴ = 0.08 J.
- Exam : For +3 μC at 0.30 m and −3 μC at 0.60 m from the point, V = 9 × 10⁹ × 3 × 10⁻⁶ × (1 ÷ 0.30 − 1 ÷ 0.60) = 4.5 × 10⁴ V. Potential is a scalar, so the values are simply added with their signs.
Rule 2: The electron-volt — 1 eV = 1.6 × 10⁻¹⁹ J, the energy gained by an electron moving through 1 V.
- Easy : An electron moved through 100 V gains 100 eV = 1.6 × 10⁻¹⁷ J.
- Medium : Starting from rest, it gets the speed v = √(2 × 1.6 × 10⁻¹⁷ ÷ 9.1 × 10⁻³¹) = 5.9 × 10⁶ m s⁻¹.
- Exam : 1 eV is a unit of energy, not of potential.
V = kQ ÷ r (scalar, sign of Q matters); W = qΔV; 1 eV = 1.6 × 10⁻¹⁹ J.
Adding potentials as vectors, or forgetting the sign of the charge.
Memory clue : "Potential is a scalar: add the numbers, keep the signs."
Spot the difference
| Quantity | Formula | Falls with distance as |
|---|---|---|
| Field of a point charge | kQ ÷ r² | 1 ÷ r² |
| Potential of a point charge | kQ ÷ r | 1 ÷ r |
Practice questions
- 1Write the SI unit of potential in terms of J and C.
- 2Is the potential near a positive charge positive or negative?
- 3How much work is done in moving a charge along an equipotential surface?
- 4Find the potential at 0.10 m from a 5 nC point charge.
- 5Find the work needed to move 2 μC from a point at 0 V to a point at 150 V.
- 6Find the speed gained by an electron accelerated from rest through 100 V.
- 7Four charges of +2 μC sit at the corners of a square whose diagonal is 0.20 m. Find the potential at the centre.
- 8At the midpoint between equal and opposite charges, what are V and E?
Show answers (try them yourself first)
- 11 V = 1 J C⁻¹.
- 2Positive.
- 3Zero.
- 4V = 9 × 10⁹ × 5 × 10⁻⁹ ÷ 0.10 = 450 V.
- 5W = 2 × 10⁻⁶ × 150 = 3 × 10⁻⁴ J.
- 6½mv² = 100 eV = 1.6 × 10⁻¹⁷ J ⇒ v ≈ 5.9 × 10⁶ m s⁻¹.
- 7Each corner is 0.10 m from the centre: V = 4 × 9 × 10⁹ × 2 × 10⁻⁶ ÷ 0.10 = 7.2 × 10⁵ V.
- 8V = 0 (the two potentials cancel) but E ≠ 0 (the two fields point the same way).
Potential of a dipole, equipotential surfaces and the field–potential link
Contour lines on a hill map join points of equal height; equipotential surfaces join points of equal potential. Link to earlier classes : contour lines in geography and the dipole of Chapter 1.
An equipotential surface is a surface on which the potential is the same everywhere. No work is done in moving a charge along it, so the electric field is always perpendicular to it and points from higher to lower potential.
Rule 1: Potential of a dipole and of many charges — On the axis of a dipole (far away), V = k p ÷ r²; on its equatorial line V = 0. For a system of charges, V = k q₁ ÷ r₁ + k q₂ ÷ r₂ + …
- Easy : For p = 2 × 10⁻⁸ C m at r = 0.50 m on the axis, V = 9 × 10⁹ × 2 × 10⁻⁸ ÷ 0.25 = 720 V.
- Medium : On the equatorial line of the same dipole V = 0, even though the field there is not zero.
- Exam : A dipole potential falls as 1 ÷ r², faster than the 1 ÷ r of a single charge.
Rule 2: Relation between E and V — E = − dV ÷ dr. For a uniform field, E = ΔV ÷ d (d = separation along the field). The field points in the direction in which the potential falls fastest.
- Easy : 200 V across plates 4 cm apart ⇒ E = 200 ÷ 0.04 = 5000 V m⁻¹.
- Medium : If V falls by 30 V over 6 cm along the field, E = 500 V m⁻¹.
- Exam : If V = 5x² (volts, x in metres) then E = −10x, so at x = 2 m the magnitude is 20 V m⁻¹ directed towards −x.
In the figure : Equipotential surfaces of a point charge are concentric spheres; the field lines cut them at right angles.
E ⟂ equipotential; E = −dV/dr; V = 0 on the equator of a dipole.
Thinking that E = 0 wherever V = 0.
Memory clue : "Field lines cross equipotentials at 90°, downhill."
Spot the difference
| Source | Equipotential surfaces |
|---|---|
| Point charge | concentric spheres |
| Uniform field | parallel planes |
| Dipole | curved surfaces; the equatorial plane is at 0 V |
Practice questions
- 1What is the potential on the equatorial line of a dipole?
- 2What is the shape of the equipotential surfaces of a point charge?
- 3What is the angle between E and an equipotential surface?
- 4The potential difference across two plates 4 cm apart is 200 V. Find the field between them.
- 5Find the potential on the axis of a dipole p = 2 × 10⁻⁸ C m at 0.50 m.
- 6V falls by 30 V over 6 cm along the field direction. Find E.
- 7In a region V = 5x² volts (x in m). Find E at x = 2 m.
- 8Why can two equipotential surfaces never cross?
Show answers (try them yourself first)
- 1Zero.
- 2Concentric spheres.
- 390°.
- 4E = 200 ÷ 0.04 = 5000 V m⁻¹.
- 5V = kp ÷ r² = 720 V.
- 6E = 30 ÷ 0.06 = 500 V m⁻¹.
- 7E = −dV/dx = −10x = −20 V m⁻¹, so magnitude 20 V m⁻¹ towards −x.
- 8At the crossing the potential would have two values at the same point, which is impossible.
Potential energy of charges and of a dipole
A stretched spring stores energy; so does a pair of like charges pushed close. Link to earlier classes : potential energy of Class 11.
The electric potential energy of a system of charges is the work done in assembling the charges from infinity to their positions. It is positive for like charges and negative for unlike charges.
Rule 1: Potential energy of charges — For two charges, U = k q₁ q₂ ÷ r. For a charge in an external potential, U = q V. For three or more charges, add the energy of every pair.
- Easy : q₁ = 3 μC, q₂ = −2 μC, r = 0.20 m ⇒ U = 9 × 10⁹ × 3 × 10⁻⁶ × (−2 × 10⁻⁶) ÷ 0.20 = −0.27 J.
- Medium : Three charges of 1 μC at the corners of a triangle of side 0.30 m have U = 3 × 9 × 10⁹ × 10⁻¹² ÷ 0.30 = 0.09 J.
- Exam : Bringing 2 μC from infinity to a point where V = 3 × 10⁴ V needs W = qV = 0.06 J.
Rule 2: Energy of a dipole in a uniform field — U = − p E cos θ. The work done in turning the dipole from θ₁ to θ₂ is W = p E (cos θ₁ − cos θ₂).
- Easy : p = 2 × 10⁻⁸ C m, E = 4 × 10⁴ N C⁻¹, θ = 60° ⇒ U = −2 × 10⁻⁸ × 4 × 10⁴ × 0.5 = −4 × 10⁻⁴ J.
- Medium : Turning it from 0° to 90° needs W = pE = 8 × 10⁻⁴ J.
- Exam : Turning it from 0° (stable) to 180° (unstable) needs 2pE = 1.6 × 10⁻³ J.
U = kq₁q₂ ÷ r; U = qV; dipole U = −pE cos θ, least at θ = 0°.
Dropping the sign of the charges, which changes the sign of U.
Memory clue : "Like charges: positive energy. Unlike: negative. Dipole likes θ = 0."
Spot the difference
| Case | Sign of U |
|---|---|
| Two like charges | positive |
| Two unlike charges | negative |
| Dipole along the field (θ = 0°) | −pE, minimum |
| Dipole against the field (θ = 180°) | +pE, maximum |
Practice questions
- 1Is the potential energy of two like charges positive or negative?
- 2Which orientation of a dipole in a field is stable?
- 3Write the formula for the energy of a charge q in a potential V.
- 4Find U for charges 3 μC and −2 μC placed 0.20 m apart.
- 5Three charges of 1 μC sit at the corners of an equilateral triangle of side 0.30 m. Find the total potential energy.
- 6Find U of a dipole p = 2 × 10⁻⁸ C m at 60° to a field of 4 × 10⁴ N C⁻¹.
- 7Find the work needed to rotate the dipole of question 6 from its stable position to 180°.
- 8Find the work in bringing a 2 μC charge from infinity to a point at 3 × 10⁴ V.
Show answers (try them yourself first)
- 1Positive.
- 2θ = 0° (along the field).
- 3U = qV.
- 4U = −0.27 J.
- 5U = 3kq² ÷ a = 0.09 J.
- 6U = −pE cos 60° = −4 × 10⁻⁴ J.
- 7W = 2pE = 2 × 2 × 10⁻⁸ × 4 × 10⁴ = 1.6 × 10⁻³ J.
- 8W = qV = 2 × 10⁻⁶ × 3 × 10⁴ = 0.06 J.
Conductors, dielectrics and polarisation
A car with a metal body protects the people inside from lightning. Link to earlier classes : conductors and insulators from Chapter 1 of this book.
In electrostatics a conductor has free charges, which move until the field inside is zero. A dielectric is an insulator whose atoms or molecules get polarised in a field: it develops opposite charges on its two faces, and these weaken the field inside it.
Rule 1: Conductor in electrostatics — Inside, E = 0; the potential is the same throughout; the charge sits only on the outer surface; just outside the surface the field is perpendicular to it with E = σ ÷ ε₀.
- Easy : A conducting surface with σ = 2.655 × 10⁻⁶ C m⁻² has E = 2.655 × 10⁻⁶ ÷ 8.85 × 10⁻¹² = 3 × 10⁵ N C⁻¹ just outside.
- Medium : For a conducting sphere of radius R at potential V the surface field is E = V ÷ R. For R = 0.10 m and V = 900 V, E = 9 × 10³ N C⁻¹ and σ = ε₀E = 8.0 × 10⁻⁸ C m⁻².
- Exam : A hollow conductor shields its inside from outside fields (electrostatic shielding), which is why a closed metal car is safe in lightning.
Rule 2: Dielectric in a field — The induced charges reduce the field to E = E₀ ÷ K (K = dielectric constant, E₀ = field without the dielectric). A conductor can be viewed as having K = ∞.
- Easy : E₀ = 3000 N C⁻¹ and K = 5 ⇒ E = 600 N C⁻¹ inside the dielectric.
- Medium : A polar molecule (water, HCl) has a permanent dipole moment; a non-polar molecule (O₂, CO₂) becomes a dipole only in a field.
- Exam : Polarisation makes the dielectric slab behave like a pair of oppositely charged faces whose field opposes the applied field.
E inside a conductor = 0; E just outside = σ/ε₀; dielectric: E = E₀/K.
Saying that a conductor has a zero field outside as well. The field is zero only inside.
Memory clue : "Conductor: zero inside, charge outside. Dielectric: field shrinks by K."
Spot the difference
| Material | Reaction to an electric field |
|---|---|
| Conductor | free charges move; E inside = 0 |
| Dielectric | atoms polarise; E inside reduced by K |
Practice questions
- 1What is the electric field inside a charged conductor in electrostatics?
- 2What is the direction of the field at the surface of a conductor?
- 3What is the dielectric constant of a conductor?
- 4A conductor has surface density σ = 2.655 × 10⁻⁶ C m⁻². Find the field just outside it.
- 5A field of 3000 N C⁻¹ in vacuum is reduced inside a dielectric with K = 5. Find the new field.
- 6Why is a person inside a metal car safe from lightning?
- 7Name one polar and one non-polar molecule.
- 8A conducting sphere of radius 0.10 m is at 900 V. Find σ and the field at its surface.
Show answers (try them yourself first)
- 1Zero.
- 2Perpendicular to the surface.
- 3Infinite.
- 4E = σ ÷ ε₀ = 3 × 10⁵ N C⁻¹.
- 5E = 3000 ÷ 5 = 600 N C⁻¹.
- 6The metal body acts as a shield; the field inside is zero and the charge flows over the outer surface.
- 7Polar : H₂O (or HCl); non-polar : O₂ (or CO₂).
- 8E = V ÷ R = 900 ÷ 0.10 = 9 × 10³ N C⁻¹; σ = ε₀E = 8.85 × 10⁻¹² × 9 × 10³ ≈ 8.0 × 10⁻⁸ C m⁻².
Capacitors and the parallel plate capacitor
The flash of a camera is powered by a capacitor that stores charge and releases it all at once. Link to earlier classes : charge and potential difference from the earlier sections.
A capacitor is two conductors separated by an insulator, used to store charge and energy. Its capacitance is the charge stored per unit potential difference. The unit is the farad (1 F = 1 C V⁻¹).
Rule 1: Capacitance — C = Q ÷ V. For a parallel plate capacitor, C = ε₀ A ÷ d (A = area of each plate, d = plate separation). With a dielectric that fills the space, C = K ε₀ A ÷ d.
- Easy : A = 0.020 m², d = 1.0 mm ⇒ C = 8.85 × 10⁻¹² × 0.020 ÷ 0.0010 = 1.77 × 10⁻¹⁰ F = 177 pF.
- Medium : Filling it with a dielectric of K = 6 gives C = 6 × 177 = 1062 pF ≈ 1.06 nF.
- Exam : Connected to a 12 V battery, the air capacitor above holds Q = CV = 1.77 × 10⁻¹⁰ × 12 = 2.1 × 10⁻⁹ C.
In the figure : A parallel plate capacitor connected to a battery: charge +Q and −Q on the plates and a uniform field between them.
C = Q/V depends on size, shape and the dielectric, not on Q or V; C = Kε₀A/d.
Writing that C increases when d increases. C is inversely proportional to d.
Memory clue : "Big area, small gap, good dielectric: big capacitance."
Spot the difference
| Change | Effect on C |
|---|---|
| Plate area doubled | doubles |
| Separation doubled | halves |
| Dielectric K inserted | multiplied by K |
Practice questions
- 1Write the SI unit of capacitance in terms of C and V.
- 2Does C of a capacitor depend on the charge given to it?
- 3What happens to C when a dielectric slab fills the gap?
- 4A capacitor stores 60 μC at 12 V. Find C.
- 5Find C for plates of 0.020 m² area 1.0 mm apart in air.
- 6Find C when the gap of question 5 is filled with K = 6.
- 7How much charge does the air capacitor of question 5 hold at 12 V?
- 8A battery stays connected while the plate separation is doubled. What happens to C and Q?
Show answers (try them yourself first)
- 11 F = 1 C V⁻¹.
- 2No.
- 3It becomes K times larger.
- 4C = 60 μC ÷ 12 V = 5 μF.
- 5C = ε₀A ÷ d = 1.77 × 10⁻¹⁰ F (177 pF).
- 6C = 6 × 177 pF ≈ 1.06 nF.
- 7Q = CV ≈ 2.1 × 10⁻⁹ C.
- 8C becomes half; V is fixed by the battery, so Q = CV also becomes half.
Combination of capacitors
Electronic circuits join small capacitors in groups to get the capacitance they need. Link to earlier classes : series and parallel resistors from Class 10.
Capacitors can be joined in series (one after another, same charge on each) or in parallel (across the same two points, same potential difference on each).
Rule 1: Series — 1 ÷ C = 1 ÷ C₁ + 1 ÷ C₂ + … The equivalent capacitance is smaller than the smallest one, and the potential difference divides in the inverse ratio of the capacitances.
- Easy : 3 μF and 6 μF in series : C = 3 × 6 ÷ (3 + 6) = 2 μF.
- Medium : Across 12 V, Q = 2 μF × 12 V = 24 μC on each; V₁ = 24 ÷ 3 = 8 V and V₂ = 24 ÷ 6 = 4 V.
- Exam : 2 μF and 4 μF in series across 90 V : C = 4/3 μF, Q = 120 μC, so the voltages are 60 V and 30 V.
Rule 2: Parallel — C = C₁ + C₂ + … The equivalent capacitance is larger than the largest one, and the charge divides in the ratio of the capacitances.
- Easy : 3 μF and 6 μF in parallel : C = 9 μF.
- Medium : Three 6 μF capacitors give 2 μF in series and 18 μF in parallel.
- Exam : 2 μF and 4 μF in parallel, joined in series with 3 μF : (6 × 3) ÷ (6 + 3) = 2 μF.
Series : same Q, 1/C adds. Parallel : same V, C adds.
Using the resistor rules the wrong way round. For capacitors it is the parallel case that adds directly.
Memory clue : "Capacitors are opposite to resistors: parallel adds."
Spot the difference
| Quantity | Series | Parallel |
|---|---|---|
| Equivalent C | 1/C = Σ(1/Cᵢ) | C = ΣCᵢ |
| Charge | same | divides |
| Voltage | divides | same |
Practice questions
- 1Which quantity is the same on every capacitor in series?
- 2Which quantity is the same on every capacitor in parallel?
- 3Find the equivalent of two 4 μF capacitors in parallel.
- 4Find the equivalent of 3 μF and 6 μF in series.
- 5The pair of question 4 is connected to 12 V. Find the charge and the voltage across each.
- 6Find the equivalent of three 6 μF capacitors in series and in parallel.
- 7Capacitors of 2 μF and 4 μF are in series across 90 V. Find the voltage across each.
- 8A 2 μF and a 4 μF in parallel are joined in series with a 3 μF. Find the equivalent.
Show answers (try them yourself first)
- 1The charge.
- 2The potential difference.
- 38 μF.
- 42 μF.
- 5Q = 24 μC on each; 8 V across the 3 μF and 4 V across the 6 μF.
- 62 μF in series, 18 μF in parallel.
- 7C = 4/3 μF; Q = (4/3) × 90 = 120 μC; V = 60 V (2 μF) and 30 V (4 μF).
- 8(2 + 4) = 6 μF; then 6 × 3 ÷ 9 = 2 μF.
Energy stored in a capacitor
The flash bulb of a camera lights up for a moment because the capacitor gives out its stored energy at once. Link to earlier classes : energy and work from Class 11.
Work is done in charging a capacitor, and it is stored as energy in the electric field between the plates.
Rule 1: Stored energy — U = ½ C V² = Q² ÷ 2C = ½ Q V (C = capacitance, V = potential difference, Q = charge).
- Easy : C = 2 μF and V = 100 V ⇒ U = ½ × 2 × 10⁻⁶ × 10⁴ = 0.010 J.
- Medium : Q = 50 μC and C = 5 μF ⇒ U = (50 × 10⁻⁶)² ÷ (2 × 5 × 10⁻⁶) = 2.5 × 10⁻⁴ J.
- Exam : If V is doubled at fixed C, the energy becomes four times as large.
Rule 2: Energy density — u = ½ ε₀ E² (energy per unit volume of the field).
- Easy : E = 3 × 10⁵ N C⁻¹ ⇒ u = ½ × 8.85 × 10⁻¹² × 9 × 10¹⁰ = 0.40 J m⁻³.
- Medium : Energy density depends only on the field, so the same formula works for any field region.
- Exam : A 2 μF capacitor charged to 100 V is joined to an uncharged 3 μF capacitor. The common voltage is 200 μC ÷ 5 μF = 40 V. The energy falls from 0.010 J to ½ × 5 × 10⁻⁶ × 1600 = 4.0 × 10⁻³ J. The loss of 6.0 × 10⁻³ J appears as heat and radiation in the connecting wires.
U = ½CV²; u = ½ε₀E²; energy is lost when capacitors share charge.
Forgetting the ½ in U = ½CV².
Memory clue : "Half C V squared: half, because charging gets harder as it fills."
Spot the difference
| Form | Use when |
|---|---|
| ½CV² | C and V are known |
| Q² ÷ 2C | Q and C are known |
| ½QV | Q and V are known |
Practice questions
- 1Where is the energy of a charged capacitor stored?
- 2Write the three forms of the energy formula.
- 3What is the unit of energy density?
- 4Find the energy stored in a 2 μF capacitor at 100 V.
- 5A 5 μF capacitor carries 50 μC. Find its energy.
- 6Find the energy density in a field of 3 × 10⁵ N C⁻¹.
- 7A 2 μF capacitor charged to 100 V is connected to an uncharged 3 μF capacitor. Find the common voltage and the energy lost.
- 8By what factor does the stored energy change if V is doubled at constant C?
Show answers (try them yourself first)
- 1In the electric field between the plates.
- 2½CV², Q² ÷ 2C and ½QV.
- 3J m⁻³.
- 4U = 0.010 J.
- 5U = Q² ÷ 2C = 2.5 × 10⁻⁴ J.
- 6u = ½ε₀E² ≈ 0.40 J m⁻³.
- 7V = 200 ÷ 5 = 40 V; energy lost = 0.010 − 0.004 = 6.0 × 10⁻³ J.
- 8It becomes four times as large.