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How does an electric field describe the force on a charge, and how do field strength and potential relate to one another?

Define electric field strength and potential, apply Coulomb's law and the field of a point charge, and analyse the uniform field between parallel plates

A focused answer to the H2 Physics learning outcome on electric fields. Coulomb's law, electric field strength, the field and potential of a point charge, and the uniform field between charged parallel plates.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to define electric field strength and electric potential, to apply Coulomb's law and the radial field of a point charge, and to analyse the uniform field between charged parallel plates. This mirrors the gravitational field thread, with the key difference that charges can be positive or negative, so forces can attract or repel.

The answer

Coulomb's law

Two point charges Q1Q_1 and Q2Q_2 a distance rr apart in a vacuum exert a force on each other:

F=14πε0Q1Q2r2F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}

where ε0=8.85×1012 F m1\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1} is the permittivity of free space and 14πε0=8.99×109 N m2C2\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9\ \text{N m}^2\text{C}^{-2}. Like charges repel; unlike charges attract. The force is inverse-square, like gravity.

Electric field strength

The electric field strength at a point is the force per unit positive charge placed there:

E=FqE = \frac{F}{q}

with units N C1\text{N C}^{-1} (equivalently V m1\text{V m}^{-1}). It is a vector, directed away from positive charge and toward negative charge. For a point charge QQ:

E=14πε0Qr2E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}

Electric potential

The electric potential at a point is the work done per unit positive charge to bring it from infinity to that point:

V=14πε0QrV = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

Potential is a scalar. It is positive near a positive charge and negative near a negative charge, with zero taken at infinity. The potential energy of a charge qq is Ep=qVE_p = qV.

The uniform field between parallel plates

Two parallel plates with a potential difference VV across a separation dd produce a uniform field between them:

E=VdE = \frac{V}{d}

directed from the positive to the negative plate. The field lines are parallel and evenly spaced, so a charged particle between the plates feels a constant force F=qEF = qE. This is the basis of cathode-ray deflection and many particle experiments.

Examples in context

Example 1. Deflecting an electron beam. In a cathode-ray tube, electrons pass between charged parallel plates where the uniform field E=V/dE = V/d exerts a constant force eEeE, deflecting them like a projectile under gravity. Changing the plate voltage steers the beam, the operating principle of older oscilloscopes and televisions.

Example 2. Millikan's oil-drop experiment. A charged oil drop is held stationary between parallel plates when the upward electric force qEqE balances its weight mgmg. Measuring the balancing voltage gives the charge qq, and Millikan found charge comes in multiples of the elementary charge ee, a landmark in physics.

Try this

Q1. Define electric potential at a point. [2 marks]

  • Cue. The work done per unit positive charge to bring a small charge from infinity to that point, a scalar with zero at infinity.

Q2. Find the field strength midway between two parallel plates 8.0 mm8.0\ \text{mm} apart with 120 V120\ \text{V} across them. [2 marks]

  • Cue. E=V/d=120/(8.0×103)=1.5×104 V m1E = V/d = 120/(8.0 \times 10^{-3}) = 1.5 \times 10^4\ \text{V m}^{-1} (uniform, so the same at every point between the plates).

Q3. State how the electric field strength and the electric potential of a point charge each depend on distance. [2 marks]

  • Cue. Field strength 1/r2\propto 1/r^2 (inverse-square); potential 1/r\propto 1/r (inverse).

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksTwo point charges of +3.0 nC+3.0\ \text{nC} and 5.0 nC-5.0\ \text{nC} are 0.20 m0.20\ \text{m} apart in a vacuum. Take 14πε0=8.99×109 N m2C2\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9\ \text{N m}^2\text{C}^{-2}. (a) Find the magnitude of the force between them. (b) State whether it is attractive or repulsive.
Show worked answer →

(a) Coulomb's law: F=14πε0Q1Q2r2F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_1 Q_2}{r^2}.

F=8.99×109×(3.0×109)(5.0×109)(0.20)2=8.99×109×1.5×10170.040F = 8.99 \times 10^9 \times \dfrac{(3.0 \times 10^{-9})(5.0 \times 10^{-9})}{(0.20)^2} = 8.99 \times 10^9 \times \dfrac{1.5 \times 10^{-17}}{0.040}.

F=8.99×109×3.75×1016=3.37×106 NF = 8.99 \times 10^9 \times 3.75 \times 10^{-16} = 3.37 \times 10^{-6}\ \text{N}.

(b) The charges have opposite signs, so the force is attractive.

Markers reward correct substitution into Coulomb's law using magnitudes, the answer in newtons, and the conclusion that opposite charges attract.

Original5 marksTwo parallel plates 5.0 mm5.0\ \text{mm} apart have a potential difference of 200 V200\ \text{V} across them. (a) Find the electric field strength between the plates. (b) Find the force on an electron between the plates (e=1.60×1019 Ce = 1.60 \times 10^{-19}\ \text{C}). (c) Find its acceleration (me=9.11×1031 kgm_e = 9.11 \times 10^{-31}\ \text{kg}).
Show worked answer →

(a) Uniform field: E=Vd=2005.0×103=4.0×104 V m1E = \dfrac{V}{d} = \dfrac{200}{5.0 \times 10^{-3}} = 4.0 \times 10^4\ \text{V m}^{-1}.

(b) Force: F=eE=1.60×1019×4.0×104=6.4×1015 NF = eE = 1.60 \times 10^{-19} \times 4.0 \times 10^4 = 6.4 \times 10^{-15}\ \text{N}.

(c) Acceleration: a=Fme=6.4×10159.11×1031=7.0×1015 m s2a = \dfrac{F}{m_e} = \dfrac{6.4 \times 10^{-15}}{9.11 \times 10^{-31}} = 7.0 \times 10^{15}\ \text{m s}^{-2}.

Markers reward the uniform-field relation E=V/dE = V/d, the force F=eEF = eE, and the acceleration from Newton's second law, each with units.

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