Electric Fields

Electric Fields

Info: Board Coverage AQA Paper 2 | Edexcel CP3 | OCR (A) Paper 2 | CIE P4

Explore the simulation above to develop intuition for this topic.

Intuition

Electric fields are how charges “feel” each other without touching: Imagine placing a positive charge in space — it creates an invisible field around it that tells other charges how to move. A positive test charge placed nearby would be pushed away; a negative test charge would be pulled in. The field is the intermediary that transmits the force across empty space.

Why it matters: Electric fields explain how capacitors store energy, how particle accelerators work, how lightning forms, and how your phone’s touchscreen detects your finger. Understanding fields is the foundation for all electronics and electromagnetism.

The key insight: Electric field lines always point from positive to negative charges. The density of lines indicates field strength — closely spaced lines mean a strong field. This visual representation makes it easy to see where forces will be large and where they’ll be small, without doing any calculations.

1. Coulomb’s Law

Coulomb’s Law. The electrostatic force between two point charges $q_1$ and $q_2$ separated by Distance $r$ in vacuum is:

$\boxed{F = \frac{q_1 q_2}{4\pi\varepsilon_0 r^2}}$

Where $\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$ is the permittivity of free space.

The force is attractive for opposite charges and repulsive for like charges. It acts along The line joining the charges.

Coulomb’s constant: $k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9$ N m$^2$ C$^{-2}$.

Comparison with Gravity

The electrostatic force is $\sim 10^{36}$ times stronger than gravity. We don’t notice this in Everyday life because most objects have nearly equal numbers of positive and negative charges.

Inverse Square Law Parallels

Both gravitational and electrostatic forces obey the inverse square law, $F \propto 1/r^2$. This Shared geometry arises because both fields propagate through three-dimensional space: the surface Area of a sphere grows as $4\pi r^2$So field lines spread out and their density (which determines Field strength) decreases as $1/r^2$.

The key difference is that gravitational potential is always negative (zero at infinity, becoming More negative as $r$ decreases), whereas electric potential can be positive or negative depending on The source charge. This means gravitational bound systems always have negative total energy, while Electrostatic bound systems can be either bound ($U \lt 0$) or unbound ($U \gt 0$).

Defining similarity. Both fields have a central $1/r$ potential, and both satisfy a Gauss’s-law-type conservation of flux. The field at radius $r$ from a point source depends only on The total enclosed “charge” (mass or electric charge), not on its distribution. This means a Spherically symmetric shell of charge produces the same external field as a point charge at its Centre — a result exploited in both gravitational and electrostatic problems.

Difference in screening. A significant practical difference is that electric fields can be Screened by conductors (Faraday cages), whereas gravitational fields cannot. This is because there Are two types of electric charge that can rearrange themselves to cancel external fields, but there Is only one type of mass.

2. Electric Field Strength

Definition. The electric field strength $\mathbf{E}$ at a point is the force per unit Positive charge placed at that point:

$\mathbf{E} = \frac{\mathbf{F}}{q}$

SI units: N C$^{-1}$ (equivalent to V m$^{-1}$).

Field of a Point Charge

$\boxed{E = \frac{Q}{4\pi\varepsilon_0 r^2}}$

Derivation. Place a test charge $q$ at distance $r$ from $Q$. By Coulomb’s law: $F = \frac{Qq}{4\pi\varepsilon_0 r^2}$. Therefore:

$E = \frac{F}{q} = \frac{Q}{4\pi\varepsilon_0 r^2}$

$\square$

The field points radially outward from a positive charge and radially inward towards a negative Charge.

Uniform Electric Field

Between two parallel plates with p.d. $V$ and separation $d$:

$\boxed{E = \frac{V}{d}}$

Derivation. A charge $q$ moving between the plates experiences force $F = qE$. The work done is $W = Fd = qEd$. But also $W = qV$. Therefore $qEd = qV$Giving $E = V/d$. $\square$

Field of an Infinite Plane of Charge

Consider an infinite plane with uniform surface charge density $\sigma$ (C m$^{-2}$). The field on Either side of the plane is:

$\boxed{E = \frac{\sigma}{2\varepsilon_0}}$

This is remarkable: the field is independent of distance from the plane. No matter how far away You move, the field strength does not decrease. This is because an infinite plane always subtends The same solid angle from any observation point.

Proof of the Infinite Plane Field

Use a cylindrical Gaussian surface of cross-sectional area $A$ piercing the plane, with equal-length Extensions on each side. By symmetry, $\mathbf{E}$ is perpendicular to the plane, has equal Magnitude on both sides, and has no component parallel to the plane.

Flux through both end caps: $\Phi_E = EA + EA = 2EA$. (No flux through the curved surface since $\mathbf{E}$ is parallel to it.)

Enclosed charge: $Q_{\mathrm{enc}} = \sigma A$.

By Gauss’s law, $\Phi_E = Q_{\mathrm{enc}}/\varepsilon_0$:

$2EA = \frac{\sigma A}{\varepsilon_0}$

$E = \frac{\sigma}{2\varepsilon_0}$

$\square$

Connection to parallel plates. This result explains why the field between two large parallel Plates is approximately uniform. Near the centre of the plates, each plate’s contribution resembles That of an infinite plane. Between the plates, the fields add: $E_{\mathrm{total}} = \sigma/\varepsilon_0$. Outside the plates, the fields cancel: $E_{\mathrm{total}} = 0$. This is the ideal parallel-plate capacitor field, and it justifies the Formula $E = V/d$ for the uniform field region.

3. Electric Potential

Definition. The electric potential $V$ at a point is the work done per unit positive charge To bring a small test charge from infinity to that point:

$\boxed{V = \frac{Q}{4\pi\varepsilon_0 r}}$

SI units: volts (V), where 1 V = 1 J C$^{-1}$.

Derivation. The work done to bring $q$ from $\infty$ to $r$ against the electric force:

$W = \int_{\infty}^{r} F\,dr' = \int_{\infty}^{r} \frac{Qq}{4\pi\varepsilon_0 r'^2}\,dr' = \frac{Qq}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r} = \frac{Qq}{4\pi\varepsilon_0 r}$

$V = \frac{W}{q} = \frac{Q}{4\pi\varepsilon_0 r}$

$\square$

Intuition. Potential is positive near a positive charge (work must be done against repulsion) And negative near a negative charge (work is done by the attraction). Potential decreases with Distance, approaching zero at infinity.

Electric Potential Energy

The potential energy of two charges $q_1$ and $q_2$ separated by $r$:

$U = \frac{q_1 q_2}{4\pi\varepsilon_0 r} = q_2 V_1$

Where $V_1 = q_1/(4\pi\varepsilon_0 r)$ is the potential due to $q_1$ at the location of $q_2$.

Energy Stored in a Charged Capacitor

When a capacitor of capacitance $C$ carries charge $Q$ at p.d. $V$ ($Q = CV$), the energy stored is:

$\boxed{U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}}$

Proof of Energy in a Capacitor

The energy stored equals the total work done in transferring charge onto the capacitor. At any Instant during charging, the p.d. Across the plates is $v = q/C$. To transfer a small increment of Charge $dq$ against this p.d., the work done is $dW = v\,dq = \frac{q}{C}\,dq$.

$W = \int_{0}^{Q} \frac{q}{C}\,dq = \frac{1}{C}\left[\frac{q^2}{2}\right]_{0}^{Q} = \frac{Q^2}{2C}$

Since $Q = CV$: $U = \frac{1}{2}CV^2$. And since $V = Q/C$: $U = \frac{1}{2}QV$.

$\square$

Intuition. The factor of $\frac{1}{2}$ arises because the average p.d. During charging is $V/2$ (the p.d. Grows linearly from 0 to $V$ as charge accumulates). The total work is therefore $W = Q \times V/2 = \frac{1}{2}QV$Which is the area of the triangle under the $V$—$q$ graph. This Graphical interpretation is frequently tested: sketch the $V$—$Q$ graph and find the area to Determine stored energy.

Where does the energy go? The energy is stored in the electric field between the plates. For a Parallel-plate capacitor of area $A$ and separation $d$The energy density (energy per unit volume) Is:

$u = \frac{U}{Ad} = \frac{\frac{1}{2}CV^2}{Ad} = \frac{\frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)\left(Ed\right)^2}{Ad} = \frac{1}{2}\varepsilon_0 E^2$

This result is general: any electric field of strength $E$ stores energy density $\frac{1}{2}\varepsilon_0 E^2$. Doubling the field strength quadruples the energy density.

Comparison with gravitational energy. Just as gravitational potential energy can be visualised As the area under the force—distance curve ($W = \int F\,dr$), the energy in a capacitor is the Area under the voltage—charge curve. Both represent stored field energy, and both can be recovered As useful work when the system returns to its equilibrium state.

4. Relationship Between $E$ and $V$: $E = -dV/dr$

The electric field is the negative gradient of the electric potential.

Proof

Consider a test charge $q$ moved by a small distance $dr$ in the direction of the field. The work Done by the field is:

$dW = F\,dr = qE\,dr$

This work comes from the change in potential energy: $dW = -q\,dV$ (negative because the field does Work, reducing potential energy).

$qE\,dr = -q\,dV$

$\boxed{E = -\frac{dV}{dr}}$

In 3D: $\mathbf{E} = -\nabla V$.

Verification for a point charge. $V = \frac{Q}{4\pi\varepsilon_0 r}$.

$-\frac{dV}{dr} = -\frac{Q}{4\pi\varepsilon_0}\left(-\frac{1}{r^2}\right) = \frac{Q}{4\pi\varepsilon_0 r^2} = E \quad \checkmark$

Intuition. The minus sign tells us that the electric field points in the direction of decreasing potential — a positive charge “rolls downhill” in the potential landscape.

5. Charged Particle Motion in a Uniform Electric Field

Consider a particle of charge $q$ and mass $m$ entering a uniform electric field $E$ between Parallel plates of length $L$With initial velocity $v$ perpendicular to the field.

Horizontal motion (perpendicular to field): uniform velocity.

$x = vt, \qquad t = \frac{L}{v}$

Vertical motion (parallel to field): uniform acceleration.

$F = qE, \qquad a = \frac{qE}{m}$

$y = \frac{1}{2}at^2 = \frac{qE}{2m} \cdot \frac{L^2}{v^2}$

Derivation of the parabolic trajectory. Eliminating $t$: $y = \frac{qE}{2mv^2}x^2$. This is a Parabola ($y \propto x^2$).

Vertical velocity at exit:

$v_y = at = \frac{qEL}{mv}$

Deflection angle:

$\tan\theta = \frac{v_y}{v} = \frac{qEL}{mv^2}$

6. Real-World Applications of Electric Fields

Van de Graaff Generator

A Van de Graaff generator uses a moving rubber or latex belt to transport charge to a hollow metal Dome. As charge accumulates on the outer surface of the dome, the electric potential rises (since $V = kQ/r$ and the dome has a large but finite radius). Potentials of several million volts can be Achieved in laboratory demonstrations.

Key physics:

• Charge resides entirely on the outer surface of the dome. A conductor in electrostatic equilibrium has zero electric field inside, so no charge is needed on the inner surface to shield the interior.
• The dome is approximately spherical, so the field just outside is $E = kQ/r^2$Directed radially outward for a positively charged dome.
• When the field at the surface exceeds the breakdown strength of air ($E \approx 3 \times 10^6$ V m$^{-1}$ at STP), spark discharge occurs to any nearby grounded object.
• The potential at breakdown is $V_{\mathrm{breakdown}} = E_{\mathrm{breakdown}} \times r$. Larger domes reach higher breakdown potentials because the same surface charge density produces a weaker field on a larger sphere ($E = \sigma/\varepsilon_0$ for a sphere).

Safety demonstration. A person touching the dome while standing on an insulating platform also Acquires a high potential. Their hair stands on end because each strand acquires the same sign of Charge, and strands repel each other via Coulomb’s law — a vivid demonstration of electrostatic Repulsion.

Inkjet Printers

Inkjet printers use electric fields to precisely control the trajectory of charged ink droplets. There are two main types:

1. Continuous inkjet (CIJ): A stream of ink is broken into uniform droplets by a vibrating piezoelectric crystal at a rate of roughly 100,000 droplets per second. Droplets pass through a charging electrode where each droplet is given a controlled charge. The charged droplets then pass between deflection plates creating a transverse electric field. The vertical deflection is $y = qEL^2/(2mv^2)$So different charges produce different deflections. Uncharged droplets are collected in a gutter and recycled; charged droplets strike the paper to form characters.

2. Drop-on-demand (DOD): Thermal or piezoelectric actuators fire individual droplets only where needed. Some designs still use electric fields for final trajectory correction.

Physics involved. The deflection of each droplet in the plates is exactly the parabolic Trajectory problem from Section 5. By varying the charge $q$ on each droplet (controlled by the Charging electrode voltage), the printer achieves different vertical positions on the paper. The Horizontal position is controlled by the timing of droplet ejection combined with paper motion.

Electrostatic Precipitator

Electrostatic precipitators remove fine particulate matter (dust, ash, smoke) from industrial Exhaust gases. They are widely used in coal-fired power stations, waste incinerators, and cement Plants, and can remove over 99% of particulate emissions.

How it works:

1. A thin high-voltage wire (at $\sim 50$ kV) runs along the centre of a duct or between large grounded collecting plates. The strong field near the wire ionises the surrounding air, creating a corona discharge.
2. The electric field (directed radially inward from the outer collecting plates towards the central wire, since the wire is negative) drives negative ions towards the collecting plates.
3. Particulate matter in the gas stream collides with these ions, acquires a negative charge, and is then attracted to the grounded collecting plates by the electric field.
4. The plates are periodically rapped (mechanically vibrated) to dislodge the collected dust into hoppers below for disposal or recycling.

Efficiency. Modern electrostatic precipitators can capture particles as small as $0.01\,\mu$M. The collection efficiency depends on the drift velocity of charged particles, which increases with Both the charge acquired by the particle and the electric field strength. The Deutsch—Anderson Equation relates collection efficiency $\eta$ to plate area, gas flow rate, and drift velocity.

Problem Set

Board-Specific Notes

AQA (7408)

• Electric fields are assessed in Paper 2 (Sections 6.1—6.2).
• Coulomb’s law and the inverse square law formula are provided on the data sheet.
• Required practical 10 investigates electric fields using conducting paper (mapping equipotentials and field lines).
• The comparison with gravitational fields is a common synoptic link, particularly in multi-mark questions.
• Graphical methods for determining energy stored (area under $V$—$Q$ curve) are frequently examined.

Edexcel (9PH0)

• Electric fields are covered in Core Practical 3 (CP3) and Topic 11.
• Edexcel places emphasis on practical applications and numerical problem-solving using the formula booklet.
• The energy stored in capacitors is covered in Topic 11 alongside charging and discharging through resistors.
• Deflection of charged particles (the “electron gun” and CRT context) is a recurring exam theme.
• Questions often require identifying the correct equation from the formula sheet for multi-step calculations.

OCR (A) (H556)

• Electric fields appear in Module 6 (Sections 6.1 and 6.3).
• OCR commonly tests full derivations from first principles (e.g., deriving $E = -dV/dr$ or the potential of a point charge from Coulomb’s law).
• The $V$—$Q$ graph area-under-curve argument for energy is a staple of OCR marking schemes.
• Gravitational—electrostatic comparison questions are a hallmark of OCR synoptic assessment.
• Multi-part questions often begin with a recall/derivation and build to an application or calculation.

CIE (9702)

• Electric fields are in Paper 4 (A2), Chapters 21—22 of the coursebook.
• CIE requires the most mathematical rigour: full derivations from Coulomb’s law to potential and field strength are expected.
• CIE commonly combines electric fields with mechanics (projectile motion of charged particles in fields).
• Permittivity calculations and relative permittivity ($\varepsilon_r$) of dielectric materials are explicitly examinable.
• Numerical answers are given to 2 or 3 significant figures, matching the precision of the data provided.

Common Pitfalls

1. Confusing EMF and potential difference. EMF is the total energy per unit charge supplied; PD is the energy per unit charge transferred to a component.

2. Forgetting that ammeters are connected in series and voltmeters in parallel.

3. Incorrectly applying Kirchhoff’s second law by missing components in a loop.

4. Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.

5. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

6. Incorrectly applying $\vec{F} = m\vec{a}$when forces are not collinear. Resolve into components first.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

• Magnetic Fields — Electric and magnetic fields are unified in electromagnetism; charged particles experience forces in both.
• Electromagnetism Unification — Maxwell’s equations unite electric and magnetic fields into a single coherent theory of electromagnetic waves.
• Current and Resistance — Electric fields drive current through conductors, connecting field theory to circuit behaviour.
• Gravitational Fields — Coulomb’s law parallels Newton’s law of gravitation, with both obeying the inverse square law.