Electric fields govern everything from the structure of atoms to the operation of cathode ray tubes and inkjet printers. Understand how charges create fields and how fields act on charges — the foundation of all electromagnetism.
⚡Define electric field strength E = F/Q and apply it to point charges and uniform fields
📐Use E = kQ/r² (radial) and E = V/d (uniform) in calculations
🌀Draw and interpret electric field line diagrams for various charge configurations
💡Define electric potential V = kQ/r and explain its sign for positive and negative charges
⭕Describe equipotential surfaces and their relationship to field lines
↔️Compare gravitational and electric fields — similarities and differences
Electric Field Strength
An electric field is a region where a charge experiences an electric force. The electric field strength E at a point is defined as the force per unit positive charge at that point:
E = F / Q (N C⁻¹ or V m⁻¹)
E is a vector — it points in the direction of force on a positive test charge.
For a point charge Q: By Coulomb's law, the force on a test charge q at distance r is F = kQq/r². So:
E = kQ / r² where k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²
E due to a positive charge points radially outward; due to a negative charge, radially inward.
For a uniform field between parallel plates: A potential difference V across plates separated by distance d creates a uniform field:
E = V / d (V m⁻¹ ≡ N C⁻¹)
Permittivity of free space: ε₀ = 8.85 × 10⁻¹² F m⁻¹ (or C² N⁻¹ m⁻²). In a medium with relative permittivity ε_r: k → k/ε_r, or equivalently ε₀ → ε₀ε_r. This reduces the field strength in a dielectric medium.
Electric Field Lines
Field lines represent the direction and relative strength of an electric field:
Lines point from positive to negative charges (direction of force on positive charge)
Denser lines = stronger field
Lines never cross (unique field direction at every point)
Lines are perpendicular to conducting surfaces at equilibrium
Key field line patterns:
Configuration
Field line pattern
Field type
Isolated positive charge
Radial outward from charge
Non-uniform (radial)
Isolated negative charge
Radial inward to charge
Non-uniform (radial)
Two opposite charges (+/−)
Leave + , curve to −
Non-uniform (dipole)
Parallel plates (opposite charges)
Uniform, parallel, perpendicular to plates
Uniform (between plates)
Two like charges
Repel away from each other, neutral point between
Non-uniform
⚠️ A common misconception: electric field lines do NOT represent the path a charge would follow. A positive charge released from rest follows the field line only if E is in the direction of motion. If launched with velocity perpendicular to E, the charge follows a parabolic path (like projectile motion in a gravitational field).
Electric Potential
The electric potential V at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point:
V = W / Q (J C⁻¹ = V)
For a point charge Q at distance r:
V = kQ / r = Q / (4πε₀r)
Sign convention:
V = 0 at infinity (reference point)
Positive charge → positive V (work done against repulsion to bring +ve test charge in)
Negative charge → negative V (work done by field when bringing +ve test charge in — attractive)
Electric potential energy: E_p = QV = kQ₁Q₂/r. For a charge Q₁ in the field of Q₂, the potential energy is positive if charges have the same sign (they repel — energy stored), and negative if opposite signs (they attract — bound system).
Relationship between field and potential:
E = −dV/dr (or E = −ΔV/Δr for uniform fields: E = V/d)
This is analogous to the gravitational case: field points in the direction of steepest decrease in potential.
Equipotential Surfaces in Electric Fields
Equipotential surfaces (in 3D) or equipotential lines (in 2D cross-sections) are surfaces/lines of constant potential. Key properties (analogous to gravitational equipotentials):
No work is done moving a charge along an equipotential.
Electric field lines are always perpendicular to equipotentials.
For an isolated point charge: equipotentials are concentric spheres.
For uniform field (parallel plates): equipotentials are planes parallel to the plates.
Conducting surfaces are always equipotentials at electrostatic equilibrium.
The work done moving charge Q between two points at potentials V₁ and V₂:
W = Q(V₂ − V₁) = QΔV
If the charge moves to higher potential (for positive Q), work must be done on the charge. If it moves to lower potential, the field does positive work on the charge (KE increases).
The electron-volt (eV): 1 eV is the KE gained by an electron accelerated through a potential difference of 1 V. 1 eV = 1.6 × 10⁻¹⁹ J. Useful for atomic and nuclear energy scales: a 5 MeV alpha particle has KE = 5×10⁶ × 1.6×10⁻¹⁹ = 8×10⁻¹³ J.
Comparison: Gravitational and Electric Fields
Property
Gravitational
Electric
Force law
F = Gm₁m₂/r²
F = kQ₁Q₂/r²
Field strength
g = GM/r²
E = kQ/r²
Potential
V = −GM/r
V = kQ/r
Sign
Always attractive; V always negative
Can repel; V positive or negative
Constant
G = 6.67×10⁻¹¹ N m² kg⁻²
k = 8.99×10⁹ N m² C⁻²
Source property
Mass (always positive)
Charge (positive or negative)
Relative strength
Weakest force
~10³⁶ times stronger than gravity
The mathematical structures are identical — this is not a coincidence but reflects deep symmetry in the laws of nature. Many results derived for one field can be adapted for the other by substituting the appropriate constants and replacing mass with charge.
Two parallel plates are separated by 8.0 mm and have a potential difference of 400 V between them. Calculate (a) the electric field strength, (b) the force on a proton (e = 1.6×10⁻¹⁹ C) between the plates, (c) the acceleration of the proton (m_p = 1.67×10⁻²⁷ kg).
1E = V/d = 400/(8.0×10⁻³) = 50 000 V m⁻¹ = 5.0 × 10⁴ N C⁻¹
2Force on proton: F = EQ = 5.0×10⁴ × 1.6×10⁻¹⁹ = 8.0 × 10⁻¹⁵ N
3Acceleration: a = F/m = 8.0×10⁻¹⁵ / 1.67×10⁻²⁷ = 4.79 × 10¹² m s⁻² ≈ 4.8 × 10¹² m s⁻²
E = 5.0×10⁴ N C⁻¹; F = 8.0×10⁻¹⁵ N; a = 4.8×10¹² m s⁻² (enormous — atomic scales!)
Calculate the electric field strength and potential at a distance of 0.20 m from a point charge of +4.0 μC. (k = 8.99×10⁹ N m² C⁻²)
1Q = 4.0×10⁻⁶ C; r = 0.20 m
2E = kQ/r² = (8.99×10⁹ × 4.0×10⁻⁶)/(0.20)² = (3.596×10⁴)/0.04 = 8.99×10⁵ N C⁻¹ ≈ 9.0×10⁵ V m⁻¹
3V = kQ/r = (8.99×10⁹ × 4.0×10⁻⁶)/0.20 = 3.596×10⁴/0.20 = 1.798×10⁵ V ≈ 1.8×10⁵ V
E = 9.0×10⁵ V m⁻¹ (directed radially outward); V = +1.8×10⁵ V (positive for positive charge)
An electron is accelerated from rest through a potential difference of 5000 V. Find (a) its kinetic energy in joules and eV, (b) its final speed. (m_e = 9.11×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C)
1Work done on electron = QΔV = eV = 1.6×10⁻¹⁹ × 5000 = 8.0×10⁻¹⁶ J
2KE = 8.0×10⁻¹⁶ J = 8.0×10⁻¹⁶ / 1.6×10⁻¹⁹ eV = 5000 eV = 5 keV
3Speed: ½m_e v² = 8.0×10⁻¹⁶ → v² = 2×8.0×10⁻¹⁶/9.11×10⁻³¹ = 1.757×10¹⁵ → v = 4.19×10⁷ m s⁻¹
4v ≈ 4.2×10⁷ m s⁻¹ ≈ 0.14c — relativistic effects start to matter above ~10% of c.
KE = 8.0×10⁻¹⁶ J = 5000 eV = 5 keV; v = 4.2×10⁷ m s⁻¹ (14% speed of light)
Q1. Electric field lines indicate the direction of force on:
Q2. Parallel plates are 5.0 mm apart with 250 V across them. What is the electric field strength?
Q3. Which of the following is a key difference between electric and gravitational fields?
Q4. The electric potential due to a positive charge Q at distance r is V. At distance 2r, the potential is:
Q5. What is the work done moving a charge of +3.0 μC from a point at potential +200 V to a point at potential +800 V?
Challenge 1. An electron enters a uniform electric field horizontally with speed 3.0×10⁶ m s⁻¹. The field is 1.5×10⁴ V m⁻¹ directed vertically upward. The plates are 0.10 m long. (a) Find the vertical deflection as the electron passes through. (b) Find the angle at exit. (c) What happens to KE?
✓ Electron charge = −e, so force on electron is DOWNWARD (opposite to E): F = eE = 1.6×10⁻¹⁹×1.5×10⁴ = 2.4×10⁻¹⁵ N downward. Acceleration: a = F/m = 2.4×10⁻¹⁵/9.11×10⁻³¹ = 2.634×10¹⁵ m s⁻² downward. (a) Time in field: t = L/v_x = 0.10/(3.0×10⁶) = 3.33×10⁻⁸ s. Vertical deflection: y = ½at² = ½×2.634×10¹⁵×(3.33×10⁻⁸)² = ½×2.634×10¹⁵×1.11×10⁻¹⁵ = 1.46×10⁻³ m ≈ 1.5 mm (downward). (b) Vertical speed at exit: v_y = at = 2.634×10¹⁵×3.33×10⁻⁸ = 8.77×10⁷ m s⁻¹. angle = arctan(v_y/v_x) = arctan(8.77×10⁷/3.0×10⁶) = arctan(29.2) ≈ 88° — the electron is deflected almost vertically. (c) KE increases: work is done BY the electric force on the electron (force and displacement both downward). ΔKE = Fy = 2.4×10⁻¹⁵×1.46×10⁻³ = 3.50×10⁻¹⁸ J.
Challenge 2. Two point charges: Q₁ = +8.0 μC at the origin and Q₂ = −2.0 μC at x = 0.30 m. Find (a) the point on the x-axis where E = 0 (not between them), (b) the potential at the midpoint (x = 0.15 m).
✓ (a) The neutral point where E = 0 for unlike charges is NOT between them — it is beyond the smaller charge. Let the point be at x = 0.30 + d (beyond Q₂). At this point: E due to Q₁ = kQ₁/(0.30+d)² pointing right. E due to Q₂ = kQ₂/d² pointing right (toward −ve charge, i.e., left → toward Q₂ from the right means pointing left, i.e., −ve x direction). For E = 0: kQ₁/(0.30+d)² = k|Q₂|/d² → Q₁/Q₂ = (0.30+d)²/d² → 8/2 = (0.30+d)²/d² → 4 = ((0.30+d)/d)² → 2 = (0.30+d)/d → 2d = 0.30+d → d = 0.30 m. Point is at x = 0.30 + 0.30 = 0.60 m. (b) Potential at midpoint (x = 0.15 m): V = kQ₁/r₁ + kQ₂/r₂ = k(Q₁/0.15 + Q₂/0.15) = (k/0.15)(Q₁+Q₂) = (8.99×10⁹/0.15)(8.0−2.0)×10⁻⁶ = 5.993×10¹⁰ × 6.0×10⁻⁶ = 3.60×10⁵ V. (Note potentials add as scalars — no direction needed.)
Challenge 3 (Synoptic). Compare the ratio of electric to gravitational force between an electron and a proton in a hydrogen atom (r = 5.3×10⁻¹¹ m). Discuss why atomic structure is determined by electric forces rather than gravity.
✓ Electric force: F_E = ke²/r² = (8.99×10⁹ × (1.6×10⁻¹⁹)²)/(5.3×10⁻¹¹)² = (8.99×10⁹ × 2.56×10⁻³⁸)/2.809×10⁻²¹ = 2.302×10⁻²⁸/2.809×10⁻²¹ = 8.19×10⁻⁸ N. Gravitational force: F_G = Gm_e m_p/r² = (6.674×10⁻¹¹ × 9.11×10⁻³¹ × 1.67×10⁻²⁷)/(5.3×10⁻¹¹)² = (6.674×10⁻¹¹ × 1.521×10⁻⁵⁷)/2.809×10⁻²¹ = 1.015×10⁻⁶⁷/2.809×10⁻²¹ = 3.61×10⁻⁴⁷ N. Ratio F_E/F_G = 8.19×10⁻⁸/3.61×10⁻⁴⁷ = 2.3×10³⁹. Electric force is about 10³⁹ times stronger than gravity at atomic scale! Implication: gravity is completely negligible for atomic structure. The electron orbits the proton because of electric attraction, not gravity. This enormous ratio also explains why stars don't collapse under gravity more quickly — radiation pressure (from electromagnetic/nuclear forces) easily counteracts gravity at stellar scales. Gravity only dominates at large scales because it is always attractive and masses accumulate, while electric charges tend to cancel out (matter is overall neutral).