The electric field as a region in which a charged particle feels a force; field strength E = F/Q

Coulomb's law for the force between two point charges, and the inverse-square law

The radial field of a point charge, E = Q/4πε₀ r², and combining the fields of two charges

Electric potential, V = Q/4πε₀ r, and field strength as minus the potential gradient

Field lines and equipotentials for radial and uniform fields

The uniform field between parallel plates, E = V/d, and balancing a charged drop

Charged particles in a uniform field: energy QV and the parabolic path between deflecting plates

1. An electric field is a region where a charged particle experiences a force; its strength E = F/Q is the force per unit positive charge, in N C⁻¹ (= V m⁻¹), and it is a vector.

2. Coulomb's law: F = Q₁Q₂/4πε₀ r². Like charges repel, unlike attract; double the distance, quarter the force.

3. Point charge: E = Q/4πε₀ r² (radial, inverse square) and V = Q/4πε₀ r (a scalar, zero at infinity).

4. Field = −potential gradient. Field lines point from high to low potential and cross equipotentials at right angles; no work is done moving along an equipotential.

5. Parallel plates: uniform field E = V/d, with equally spaced parallel field lines and equally spaced equipotentials.

6. A charge Q moved through a p.d. V gains energy QV. Fired across a uniform field it follows a parabola, like a projectile.

A field replaces "action at a distance" with something you can map, measure and calculate. Once you can describe it by its strength (a force per unit charge) and its potential (an energy per unit charge), the same tools explain how a cathode-ray beam is steered, how a tiny oil drop can float and why nuclei are so hard to approach.

E is in N C⁻¹, F is the force on the charge (N) and Q the charge (C); the equation is in the formulae list. E is a vector, in the direction of the force on a positive charge. A negative charge, an electron for example, is pushed the opposite way to the field.

The test charge must be small, so that it does not disturb the charges that create the field.

Field lines show the field: the arrow gives the direction of the force on a positive charge, and lines closer together mean a stronger field.

Field lines start on positive charges and end on negative charges, and they never cross, because the field has only one direction at each point.

Units: 1 N C⁻¹ = 1 V m⁻¹ (shown below, in the uniform-field section).

WORKED EXAMPLE · QUESTION

A small sphere carrying a charge of +2.5 nC is placed at point P in an electric field. It experiences a force of 6.0 × 10⁻⁵ N towards the east. (a) Calculate the electric field strength at P and state its direction. (b) The sphere is replaced by an electron. Calculate the force on the electron and its acceleration.

SOLUTION

(a) E = F/Q = (6.0×10⁻⁵)/(2.5×10⁻⁹) = 2.4 ×10⁴ N C⁻¹ towards the east

(b) Same field, different charge: F = eE = 1.60×10⁻¹⁹ ×2.4×10⁴ = 3.84×10⁻¹⁵ N. 3.8 × 10⁻¹⁵ N towards the west

a = F/m = (3.84×10⁻¹⁵)/(9.11×10⁻³¹) = 4.2 ×10¹⁵ m s⁻²

The field at P does not change when the charge is swapped: E belongs to the point, while F depends on what you put there.

F is the force between the charges (N), Q₁ and Q₂ the two charges (C), r the distance between their centres (m) and ε₀ the permittivity of free space, 8.85 × 10⁻¹² F m⁻¹. Coulomb's law is in the formulae list. The constant 1/4πε₀ = 8.99×10⁹ N m² C⁻² is often written k. The law applies to point charges and to charged spheres, measured from their centres.

Like charges repel, unlike charges attract. The forces on the two charges are equal in size and opposite in direction (Newton's third law).

Inverse-square law: F ∝ 1/r², so doubling the separation divides the force by 4, and trebling it divides by 9.

If you substitute the charges with their signs, a negative answer means attraction and a positive one repulsion. Many people prefer to use sizes only and state the direction in words.

Electric forces are enormous compared with gravitational ones. For two protons, F_electric/F_gravity = ke²/Gmₚ² ≈1.2×10³⁶ at any separation, because both forces follow an inverse-square law.

WORKED EXAMPLE · QUESTION

Two small conducting spheres on insulating stands carry charges of +45 nC and −30 nC. Their centres are 12 cm apart. (a) Calculate the force between them and say whether it is attractive or repulsive. (b) Calculate the force when the separation is increased to 36 cm. (c) The spheres are touched together and then returned to 12 cm apart. Calculate the new force.

SOLUTION

(a) Sizes only, distance in metres: F = (8.99×10⁹ ×45×10⁻⁹ ×30×10⁻⁹)/0.12² = 8.43×10⁻⁴ N. 8.4 × 10⁻⁴ N, attractive

(b) Separation × 3, so force ÷ 3² = 9: F = 8.43 × 10⁻⁴ ÷ 9 = 9.4 × 10⁻⁵ N

(c) Charge is conserved and identical spheres share it equally: total = +45 − 30 = +15 nC, so each has +7.5 nC. F = (8.99×10⁹ ×(7.5×10⁻⁹)²)/0.12² = 3.5 × 10⁻⁵ N, now repulsive

Put a test charge q at distance r from a point charge Q. Coulomb's law gives the force, and dividing by q gives the field:

E is the field strength (N C⁻¹), Q the charge creating the field (C) and r the distance from its centre (m); the equation is in the formulae list. The field points radially outwards for a positive Q and inwards for a negative Q. Outside a charged sphere the field is the same as if all the charge were at its centre. It is radial and non-uniform: the lines spread out, and E ∝ 1/r².

With two or more charges, the field at a point is the vector sum of the fields of each charge. Along a straight line, give each field a sign for its direction and add. Between two like charges there is a neutral point where the two fields are equal and opposite, so E = 0. It is closer to the smaller charge.

WORKED EXAMPLE · QUESTION

A charge of +16 nC is fixed at point A and a charge of +4.0 nC at point B, 0.60 m from A. (a) Calculate the resultant field strength at the midpoint of AB and state its direction. (b) Determine the distance from A of the point on AB where the field strength is zero.

SOLUTION

(a) At 0.30 m: E_A = (8.99×10⁹ ×16×10⁻⁹)/0.30² = 1600 N C⁻¹ (towards B) and E_B = 400 N C⁻¹ (towards A).

Opposite directions, so subtract: E = 1600 − 400 = 1.2 × 10³ N C⁻¹ in the direction A → B

(b) At distance x from A the two fields have equal size: 16/x² = 4.0/((0.60 - x)²), so 4/x = 2/(0.60 - x), giving x = 0.40 m

Taking square roots of both sides keeps the algebra linear; the constant k and the 10⁻⁹ cancel.

V is in volts (J C⁻¹), Q is the charge creating the field (C) and r the distance from its centre (m); the equation is in the formulae list. Potential is zero at infinity. It is positive around a positive charge (work must be done to push a positive charge closer) and negative around a negative charge. Note 1/r, not 1/r².

The potential difference between two points is the work done per unit charge moving a charge between them, so moving a charge Q through a p.d. ΔV transfers energy W = QΔV.

The electric potential energy of a charge q at a point of potential V is qV.

For two point charges this is Qq/4πε₀ r.

An equipotential is a line (or surface) joining points at the same potential. No work is done moving a charge along it.

Field strength is the potential gradient

E is the field strength (V m⁻¹) and ΔV/Δr the potential gradient, the rate of change of potential with distance. This general form is not in the formulae list, so learn it; its uniform-field form, E = V/d, is. The minus sign says the field points the way the potential falls. On a graph of V against r, E = −(gradient of the tangent).

Because no work is done along an equipotential, the force, and so the field, has no component along it: field lines cross equipotentials at right angles.

Where equipotentials drawn at equal steps of V are close together, the potential gradient is steep and the field is strong.

Around a point charge, equipotentials are concentric spheres (circles on paper). For equal steps of V they get further apart with distance, because r = Q/4πε₀ V.

WORKED EXAMPLE · QUESTION

A small sphere carries a charge of +5.00 nC. Points X and Y are 0.500 m and 0.200 m from its centre. (a) Calculate the potential at X and at Y. (b) Calculate the work done to move a +3.00 nC charge from X to Y. (c) Calculate the average potential gradient between X and Y and compare it with the field strengths at X and Y.

SOLUTION

(a) V_X = (8.99×10⁹ ×5.00×10⁻⁹)/0.500 = 89.9 V; similarly V_Y = 225 V

(b) ΔV = 225 − 89.9 = 135 V, so W = QΔV = 3.00 × 10⁻⁹ × 135 = 4.05 × 10⁻⁷ J

(c) Over 0.300 m: ΔV/Δr ≈ 135 ÷ 0.300 = 450 V m⁻¹. The actual fields are E_X = 180 N C⁻¹ and E_Y = 1.12×10³ N C⁻¹.

The average lies between the two end values: in a radial field the gradient, and so E, changes along the way. Only in a uniform field does a single ΔV/Δr give the field everywhere.

WORKED EXAMPLE · QUESTION

A proton with kinetic energy 2.5 MeV is fired directly towards a tin nucleus (charge +50e), which can be treated as a fixed point charge. (a) Calculate the distance of closest approach. (b) Calculate the electric field strength at that distance and the force on the proton there.

SOLUTION

(a) At closest approach all the kinetic energy has become electric potential energy. Eₖ = 2.5×10⁶ ×1.60×10⁻¹⁹ = 4.0×10⁻¹³ J.

Eₖ = (50e)e/(4πε₀ r), so r = (8.99×10⁹ ×50 ×(1.60×10⁻¹⁹)²)/(4.0×10⁻¹³) = 2.9 ×10⁻¹⁴ m

(b) E = (8.99×10⁹ ×50 ×1.60×10⁻¹⁹)/((2.88×10⁻¹⁴)²) = 8.7 ×10¹⁹ N C⁻¹, and F = eE = 14 N

Quick check: the potential at closest approach is 2.5 MV (2.5 MeV per unit charge e), and for a point charge E = V/r = 2.5 × 10⁶ ÷ 2.88 × 10⁻¹⁴, the same answer. A force of about 14 N on a single proton shows how strong fields are close to a nucleus.

Two parallel metal plates connected to a supply produce a uniform field between them: the field strength has the same size and direction at every point (away from the edges).

Field lines are parallel, equally spaced and perpendicular to the plates, from the positive plate to the negative plate.

Equipotentials are planes parallel to the plates, equally spaced for equal steps of potential: the potential falls steadily from one plate to the other.

Near the edges the lines bulge outwards and the field is weaker (edge effects), so questions use the central region.

E is the field strength (V m⁻¹), V the p.d. between the plates (V) and d their separation (m); the equation is in the formulae list.

Why: moving a charge Q from one plate to the other, work = force × distance = QEd, and also = QV. So E = V/d, and 1 N C⁻¹ = 1 V m⁻¹.

Balancing a charged drop

A tiny charged drop between horizontal plates feels its weight mg downwards and an electric force QE = QV/d. Adjust V until the drop is stationary: QV/d = mg, so Q = mgd/V. A negative drop needs the upper plate positive; a positive drop needs it negative. When many drops are measured, every charge is a whole-number multiple of 1.60 × 10⁻¹⁹ C: charge comes in units of e (it is quantised). This is the idea behind Millikan's oil-drop experiment.

WORKED EXAMPLE · QUESTION

An oil drop of mass 3.2 × 10⁻¹⁵ kg is held stationary between two horizontal plates 5.0 mm apart when the p.d. between them is 245 V. The upper plate is positive. (a) Calculate the field strength between the plates. (b) Calculate the charge on the drop and deduce how many excess electrons it carries.

SOLUTION

(a) E = V/d = 245/(5.0×10⁻³) = 4.9 ×10⁴ V m⁻¹

(b) Balance, QE = mg: Q = (3.2×10⁻¹⁵ ×9.81)/(4.9×10⁴) = 6.4 ×10⁻¹⁹ C

n = 6.41 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 4.00, so 4 excess electrons (the drop is negative, pulled up towards the positive plate).

Accelerating a charge through a p.d.

A charge Q that moves through a p.d. V has work QV done on it. Starting from rest, ½ mv² = QV, so v = √(2QV/m). The energy gained depends only on Q and V, not on the plate separation. An electron accelerated through 1 V gains 1 eV = 1.60 × 10⁻¹⁹ J. An electron and a proton accelerated through the same p.d. gain the same energy, but the electron, being about 1800 times lighter, ends up much faster.

Firing a charge across a uniform field

A particle entering the field at right angles to the field lines feels a constant force QE parallel to the field lines and no force at right angles to them.

So, exactly like a projectile under gravity, it has constant velocity at right angles to the field and constant acceleration a = QE/m along the field: the path is a parabola.

Time between the plates: t = L/v. Sideways deflection: y = ½ at². After leaving the plates, the particle moves in a straight line.

Gravity on electrons and ions is negligible here: for the electron below, eE is about 10¹⁴ times its weight.

WORKED EXAMPLE · QUESTION

Electrons are accelerated from rest through a p.d. of 1.6 kV. They then enter the space midway between two horizontal parallel plates, travelling parallel to the plates. The plates are 6.0 cm long and 2.0 cm apart, with a p.d. of 150 V between them. (a) Calculate the speed of the electrons entering the plates. (b) Calculate the vertical deflection of an electron as it leaves the plates and show that it does not hit a plate. (c) Calculate the angle of the beam to its original direction as it leaves, and the kinetic energy gained between the plates.

SOLUTION

(a) ½ mv² = eV = 1.60×10⁻¹⁹ ×1600 = 2.56×10⁻¹⁶ J, so v = 2.37 ×10⁷ m s⁻¹

(b) E = 150 ÷ 0.020 = 7500 V m⁻¹; a = eE/m = 1.32×10¹⁵ m s⁻²; t = 0.060 ÷2.37×10⁷ = 2.53×10⁻⁹ s.

y = ½ at² = ½ ×1.32×10¹⁵ ×(2.53×10⁻⁹)² = 4.2 mm, which is less than 10 mm (half the gap), so the electron clears the plate.

(c) v_y = at = 3.33×10⁶ m s⁻¹; tan θ= (3.33×10⁶)/(2.37×10⁷), so θ = 8.0°

Energy gained = charge × p.d. moved through inside the plates: p.d. across 4.22 mm = 7500 × 0.00422 = 31.6 V, so ΔEₖ = e ×31.6 = 5.1 ×10⁻¹⁸ J (≈ 32 eV)

Combining the steps gives y = VₚₗₐₜₑₛL²/4dV_acc = (150 ×0.060²)/(4 ×0.020 ×1600) = 4.2 mm: the charge and mass cancel, so any particle accelerated from rest by the same p.d. follows the same path.

6 Current and p.d., 2.5 Photons, and 1.1 Projectiles

V = W/Q from electric circuits is the same idea as the potential difference between two points in a field, and the electronvolt is the energy gained crossing 1 V. A charge fired across a uniform field moves like a ball thrown horizontally: resolve, use constant velocity one way and suvat the other.

4 Capacitors, 4.5 Magnetic fields and 4.7 Particle accelerators

A capacitor is a pair of parallel plates storing charge in a uniform field. In a magnetic field a moving charge follows a circle, not a parabola: the basis of particle accelerators and detectors, along with the alpha-scattering evidence for the nucleus.

7 Gravitational fields

Newton's law of gravitation has the same inverse-square form as Coulomb's law, with field strength, potential and equipotentials defined the same way, but gravity only attracts.

1. Explain why two electric field lines can never cross.

2. Calculate the electric field strength 5.0 cm from a point charge of +2.0 nC.

3. The distance from a point charge is trebled. By what factors do the field strength and the potential change?

4. Two parallel plates 2.0 mm apart have a p.d. of 12 V between them. Calculate the field strength and the force on an electron between them.

5. An electron is accelerated from rest through a p.d. of 250 V. Calculate its kinetic energy in joules and its final speed.

1. The field has a single direction at each point, the direction of the force on a positive charge. Two lines crossing would give two directions at the same point.

2. E = (8.99×10⁹ ×2.0×10⁻⁹)/0.050² = 7.2×10³ N C⁻¹, directed away from the charge.

3. The field strength falls to 1/9 (inverse square); the potential falls to 1/3 (inverse proportion).

4. E = 12 ÷2.0×10⁻³ = 6.0×10³ V m⁻¹; F = eE = 9.6×10⁻¹⁶ N, towards the positive plate.

5. Eₖ = 250 ×1.60×10⁻¹⁹ = 4.0×10⁻¹⁷ J; v = √(2 ×4.0×10⁻¹⁷ ÷9.11×10⁻³¹) = 9.4×10⁶ m s⁻¹.