4.3 Electric fields
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A2 content for Unit 4 (WPH14), specification points 92–99. Capacitors are on the next page (4.4); gravitational fields are in Unit 5.
What this page covers
7 points
- 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, \(E = Q/4\pi\varepsilon_0 r^2\), and combining the fields of two charges
- Electric potential, \(V = Q/4\pi\varepsilon_0 r\), 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
In one minute
6 points
- 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.
- Coulomb's law: \(F = Q_1Q_2/4\pi\varepsilon_0 r^2\). Like charges repel, unlike attract; double the distance, quarter the force.
- Point charge: \(E = Q/4\pi\varepsilon_0 r^2\) (radial, inverse square) and \(V = Q/4\pi\varepsilon_0 r\) (a scalar, zero at infinity).
- 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.
- Parallel plates: uniform field E = V/d, with equally spaced parallel field lines and equally spaced equipotentials.
- A charge Q moved through a p.d. V gains energy QV. Fired across a uniform field it follows a parabola, like a projectile.
Contents
12 sections
- What is an electric field, and how is its strength defined?
- How big is the force between two point charges?
- What is the field of a point charge, and how do two fields combine?
- What is electric potential, and how is it linked to field strength?
- What does a uniform field look like, and how strong is it?
- How does a charged particle move in a uniform field?
- Exam tips for 4.3
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is an electric field, and how is its strength defined?
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).
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.
(a) \(E = \dfrac{F}{Q} = \dfrac{6.0\times10^{-5}}{2.5\times10^{-9}} =\) \(2.4 \times 10^{4}\ \text{N C}^{-1}\) towards the east
(b) Same field, different charge: \(F = eE = 1.60\times10^{-19} \times 2.4\times10^{4} = 3.84\times10^{-15}\ \text{N}\). 3.8 × 10⁻¹⁵ N towards the west
\(a = \dfrac{F}{m} = \dfrac{3.84\times10^{-15}}{9.11\times10^{-31}} =\) \(4.2 \times 10^{15}\ \text{m s}^{-2}\)
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.
How big is the force between two point charges?
\[F = \frac{Q_1Q_2}{4\pi\varepsilon_0 r^2}\]F is the force between the charges (N), \(Q_1\) and \(Q_2\) the two charges (C), r the distance between their centres (m) and \(\varepsilon_0\) the permittivity of free space, 8.85 × 10⁻¹² F m⁻¹. Coulomb's law is in the formulae list. The constant \(1/4\pi\varepsilon_0 = 8.99\times10^{9}\ \text{N m}^2\ \text{C}^{-2}\) 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, \(F_{electric}/F_{gravity} = ke^2/Gm_p^2 \approx 1.2\times10^{36}\) at any separation, because both forces follow an inverse-square law.
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.
(a) Sizes only, distance in metres: \(F = \dfrac{8.99\times10^{9} \times 45\times10^{-9} \times 30\times10^{-9}}{0.12^2} = 8.43\times10^{-4}\ \text{N}\). 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. \(F = \dfrac{8.99\times10^{9} \times (7.5\times10^{-9})^2}{0.12^2} =\) 3.5 × 10⁻⁵ N, now repulsive
With r = 12 instead of 0.12 m, the answer in (a) would be 10⁴ times too small. Convert to metres, then square.
What is the field of a point charge, and how do two fields combine?
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:
\[E = \frac{Q}{4\pi\varepsilon_0 r^2}\]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.
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.
(a) At 0.30 m: \(E_A = \dfrac{8.99\times10^{9} \times 16\times10^{-9}}{0.30^2} = 1600\ \text{N C}^{-1}\) (towards B) and \(E_B = 400\ \text{N C}^{-1}\) (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: \(\dfrac{16}{x^2} = \dfrac{4.0}{(0.60 - x)^2}\), so \(\dfrac{4}{x} = \dfrac{2}{0.60 - x}\), giving x = 0.40 m
Taking square roots of both sides keeps the algebra linear; the constant k and the 10⁻⁹ cancel.
Field strength is a vector and potential is a scalar. Add fields with their directions (subtract when they oppose); add potentials as plain numbers, keeping the sign of each charge.
What is electric potential, and how is it linked to field strength?
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 \(Qq/4\pi\varepsilon_0 r\).
- 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
\[E = -\frac{\Delta V}{\Delta r}\]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 \(r = Q/4\pi\varepsilon_0 V\).
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.
(a) \(V_X = \dfrac{8.99\times10^{9} \times 5.00\times10^{-9}}{0.500} =\) 89.9 V; similarly \(V_Y = 225\ \text{V}\)
(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 \(E_X = 180\ \text{N C}^{-1}\) and \(E_Y = 1.12\times10^{3}\ \text{N C}^{-1}\).
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.
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.
(a) At closest approach all the kinetic energy has become electric potential energy. \(E_k = 2.5\times10^{6} \times 1.60\times10^{-19} = 4.0\times10^{-13}\ \text{J}\).
\(E_k = \dfrac{(50e)e}{4\pi\varepsilon_0 r}\), so \(r = \dfrac{8.99\times10^{9} \times 50 \times (1.60\times10^{-19})^2}{4.0\times10^{-13}} =\) \(2.9 \times 10^{-14}\ \text{m}\)
(b) \(E = \dfrac{8.99\times10^{9} \times 50 \times 1.60\times10^{-19}}{(2.88\times10^{-14})^2} =\) \(8.7 \times 10^{19}\ \text{N C}^{-1}\), 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.
Using \(Q/4\pi\varepsilon_0 r^2\) when the question is about energy or potential. Energy and potential go with 1/r; force and field strength go with 1/r².
What does a uniform field look like, and how strong is it?
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.
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.
(a) \(E = \dfrac{V}{d} = \dfrac{245}{5.0\times10^{-3}} =\) \(4.9 \times 10^{4}\ \text{V m}^{-1}\)
(b) Balance, QE = mg: \(Q = \dfrac{3.2\times10^{-15} \times 9.81}{4.9\times10^{4}} =\) \(6.4 \times 10^{-19}\ \text{C}\)
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).
Convert the plate separation to metres before using E = V/d: millimetres give an answer 1000 times too small. A sensible field between laboratory plates is typically 10³–10⁶ V m⁻¹.
How does a charged particle move in a uniform field?
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, \(\tfrac12 mv^2 = QV\), so \(v = \sqrt{2QV/m}\). 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: \(y = \tfrac12 at^2\). 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.
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.
(a) \(\tfrac12 mv^2 = eV = 1.60\times10^{-19} \times 1600 = 2.56\times10^{-16}\ \text{J}\), so \(v =\) \(2.37 \times 10^{7}\ \text{m s}^{-1}\)
(b) E = 150 ÷ 0.020 = 7500 V m⁻¹; \(a = eE/m = 1.32\times10^{15}\ \text{m s}^{-2}\); \(t = 0.060 \div 2.37\times10^{7} = 2.53\times10^{-9}\ \text{s}\).
\(y = \tfrac12 at^2 = \tfrac12 \times 1.32\times10^{15} \times (2.53\times10^{-9})^2 =\) 4.2 mm, which is less than 10 mm (half the gap), so the electron clears the plate.
(c) \(v_y = at = 3.33\times10^{6}\ \text{m s}^{-1}\); \(\tan\theta = \dfrac{3.33\times10^{6}}{2.37\times10^{7}}\), 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 \(\Delta E_k = e \times 31.6 =\) \(5.1 \times 10^{-18}\ \text{J}\) (≈ 32 eV)
Combining the steps gives \(y = \dfrac{V_{plates}L^2}{4dV_{acc}} = \dfrac{150 \times 0.060^2}{4 \times 0.020 \times 1600} = 4.2\ \text{mm}\): the charge and mass cancel, so any particle accelerated from rest by the same p.d. follows the same path.
Never mix the directions across and along the field lines in one suvat equation. Find the time from the direction with constant velocity, then use that time in the direction with constant acceleration.
Exam tips for 4.3
Field strength: force per unit positive charge on a small test charge. Potential: work done per unit charge bringing a small positive charge from infinity. "The force on a charge" or "the energy of a charge" is not a definition.
Before substituting, check which one the question asks for. Measure r from the centre of a charged sphere, in metres, and square it only for F and E.
Field lines run from + to −. A positive charge is pushed along the field, a negative charge (an electron) against it. For Coulomb forces, say "attractive" or "repulsive".
Radial field: straight lines, evenly spread, arrows outwards from + (inwards to −). Uniform field: parallel, equally spaced lines meeting the plates at 90°. Equipotentials always cross field lines at right angles.
If the question gives a p.d. and asks for a speed, use QV = ½mv². It works for any field shape, uniform or radial, and avoids finding the acceleration.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Field lines show the path a charge will follow. | They show the direction of the force. A charge that is already moving keeps its velocity across the lines. | The acceleration is along the field line; an electron fired across a uniform field follows a parabola. |
| Doubling the distance from a point charge halves the field. | The field follows an inverse-square law. | E falls to a quarter; it is the potential that halves (1/r). |
| Where the potential is zero, the field must be zero. | Field depends on how fast the potential changes, not on its value. | E = −ΔV/Δr. Midway between equal and opposite charges V = 0 but E is large. |
| The field between parallel plates is strongest close to the plates. | Field lines are equally spaced across the whole gap. | The field is uniform, E = V/d, everywhere away from the edges. |
| Equipotentials around a point charge are equally spaced for equal steps of potential. | V ∝ 1/r, so the same drop in V needs a longer distance further out. | They spread out with distance, showing the field getting weaker. |
| An electron and a proton accelerated through the same p.d. reach the same speed. | The energy gained is the same, eV, but their masses are very different. | Same kinetic energy; the lighter electron is about 43 times faster. |
How this topic connects
2.6 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.
4.4 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.
5.7 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.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| E = F/Q | definition of field strength; force on a charge in a field | Yes |
| \(F = Q_1Q_2/4\pi\varepsilon_0 r^2\) | Coulomb's law: force between two point charges | Yes |
| \(E = Q/4\pi\varepsilon_0 r^2\) | radial field of a point charge | Yes |
| \(V = Q/4\pi\varepsilon_0 r\) | potential in a radial field | Yes |
| E = V/d | uniform field between parallel plates | Yes |
| \(E = -\Delta V/\Delta r\) | field strength as minus the potential gradient | No — learn it |
| W = QΔV; ½mv² = QV | energy gained by a charge moving through a p.d. | No — derive it from V = W/Q and \(E_k = \tfrac12 mv^2\) |
| \(E_p = Qq/4\pi\varepsilon_0 r\) | electric potential energy of two point charges | No — derive it: \(E_p = qV\) |
| Q = mgd/V | charge on a drop balanced between horizontal plates | No — derive it: QV/d = mg |
Check your understanding
1. Explain why two electric field lines can never cross.
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. Calculate the electric field strength 5.0 cm from a point charge of +2.0 nC.
\(E = \dfrac{8.99\times10^{9} \times 2.0\times10^{-9}}{0.050^2} = 7.2\times10^{3}\ \text{N C}^{-1}\), directed away from the charge.
3. The distance from a point charge is trebled. By what factors do the field strength and the potential change?
The field strength falls to 1/9 (inverse square); the potential falls to 1/3 (inverse proportion).
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.
\(E = 12 \div 2.0\times10^{-3} = 6.0\times10^{3}\ \text{V m}^{-1}\); \(F = eE = 9.6\times10^{-16}\ \text{N}\), towards the positive plate.
5. An electron is accelerated from rest through a p.d. of 250 V. Calculate its kinetic energy in joules and its final speed.
\(E_k = 250 \times 1.60\times10^{-19} = 4.0\times10^{-17}\ \text{J}\); \(v = \sqrt{2 \times 4.0\times10^{-17} \div 9.11\times10^{-31}} = 9.4\times10^{6}\ \text{m s}^{-1}\).
Specification checklist
- (92) I can describe an electric field as a region in which a charged particle feels a force.
- (93) I can define electric field strength as force per unit positive charge and use E = F/Q, including the direction of the force on a negative charge.
- (94) I can use Coulomb's law, \(F = Q_1Q_2/4\pi\varepsilon_0 r^2\), and explain the inverse-square dependence on distance.
- (95) I can calculate the field strength of a point charge with \(E = Q/4\pi\varepsilon_0 r^2\) and combine the fields of two charges along a line.
- (96) I can explain that field strength is minus the potential gradient, find it from a graph of potential against distance, and explain why equipotentials are perpendicular to field lines.
- (97) I can use E = V/d for the uniform field between parallel plates, including balancing a charged drop.
- (98) I can define electric potential and use \(V = Q/4\pi\varepsilon_0 r\), including the work done moving a charge between two points.
- (99) I can draw and interpret field lines and equipotentials for radial and uniform fields.
- I can calculate the energy QV gained by a charge accelerated through a p.d., and analyse the parabolic path of a charged particle fired across a uniform field.