Electric Field Strength

Spec mapping: OCR H556 Module 6.2 — Electric fields (definition of electric field strength $E = F/Q$E=F/Q as force per unit positive charge, vector quantity; uniform field between parallel plates $E = V/d$E=V/d; radial field around a point charge $E = Q/(4\pi\varepsilon_0 r^2)$E=Q/(4πε0​r2); representation of electric fields by field lines; principle of superposition). Refer to the official OCR H556 specification document for exact wording.

We now leave capacitors for a while and spend four lessons on electric fields. A field is a region of space in which a charged object experiences a force. The central question of the topic is: "given a distribution of charges, what force would a small test charge experience at any point in space?" The answer is expressed as a vector field — the electric field strength — and just as with gravity, the field concept lets us separate the question of what a source does (produce a field) from the question of how matter responds (feel a force).

This lesson begins Module 6.2 — Electric fields, which parallels the Module 5.4 gravitational field content very closely. Many of the formulae look almost identical once you swap mass $m$m for charge $Q$Q and Newton's $G$G for the Coulomb constant $1/(4\pi\varepsilon_0)$1/(4πε0​). The deep structural analogy is one of the most beautiful results in undergraduate-level classical physics, and OCR examiners regularly probe it.

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Defining Electric Field Strength

Take a small positive test charge $+Q$+Q and place it at a point in the field. It experiences an electric force $F$F. The electric field strength $E$E at that point is defined as the force per unit charge:

$\boxed{\,E = \frac{F}{Q}\,}$E=QF​​

Symbol	Meaning	Unit
$E$E	Electric field strength	N C$^{-1}$−1 or V m$^{-1}$−1
$F$F	Force on test charge	newton (N)
$Q$Q	Magnitude of test charge	coulomb (C)

Because force is a vector, the electric field is also a vector. Its direction is the direction of the force on a positive charge. A negative charge placed in the same field experiences a force in the opposite direction, but the field itself is unchanged.

Key idea: The field is a property of the region of space; the force depends on both the field and the charge placed in it. $E = F/Q$E=F/Q is the definition of $E$E — it applies wherever there is a field, regardless of what produces it.

The two units N C$^{-1}$−1 and V m$^{-1}$−1 are equivalent. From the definition $E = F/Q$E=F/Q in N C$^{-1}$−1, and from the work-energy relation $W = qV$W=qV giving an alternative form $E = V/d$E=V/d in V m$^{-1}$−1:

$1\ \text{N C}^{-1} = 1\ \text{N J}^{-1}\,\text{J C}^{-1} = 1\ \text{V m}^{-1}.$1 N C−1=1 N J−1J C−1=1 V m−1.

Using V m$^{-1}$−1 is more common in practice and makes the uniform-field formula $E = V/d$E=V/d look natural.

Definition vs derived formula — the same pattern as $C = Q/V$C=Q/V

$E = F/Q$E=F/Q is the defining equation of electric field strength. It applies to any field, whether produced by parallel plates, a point charge, a complicated arrangement of charges, or even an external source of unspecified origin. The derived formulae $E = V/d$E=V/d (parallel plates) and $E = Q/(4\pi\varepsilon_0 r^2)$E=Q/(4πε0​r2) (point charge) are consequences of the definition applied to specific geometries.

This is the same conceptual pattern as $C = Q/V$C=Q/V in lesson 1: the definition is universal; the geometric formulae are special cases. OCR examiners regularly probe whether you understand the distinction.

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Uniform Fields — Parallel Plates

Between two parallel plates held at potential difference $V$V and separation $d$d, the electric field is uniform: the same magnitude and direction everywhere between the plates (ignoring edge effects). It is given by

$\boxed{\,E = \frac{V}{d}\,}$E=dV​​

The field points from the high-potential plate to the low-potential plate, and its magnitude in V m$^{-1}$−1 is the voltage drop per metre.

Worked Example 1 — Field Between Plates

Two parallel plates are $5.0$5.0 mm apart and have a potential difference of $200$200 V between them. Calculate the electric field strength and the force on a $2.0$2.0 nC charge placed between the plates.

$$\begin{aligned} E &= \frac{V}{d} = \frac{200}{0.005} = 4.0 \times 10^4\ \text{V m}^{-1}, \\ F &= EQ = (4.0 \times 10^4)(2.0 \times 10^{-9}) = 8.0 \times 10^{-5}\ \text{N}. \end{aligned}$$EF​=dV​=0.005200​=4.0×104 V m−1,=EQ=(4.0×104)(2.0×10−9)=8.0×10−5 N.​

The field is $40$40 kV m$^{-1}$−1 — substantial, but still well below the breakdown field of air (about $3$3 MV m$^{-1}$−1). Above that field strength, air becomes ionised and conducts in a violent flash, which is what makes a lightning bolt visible.

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Radial Fields — Point Charges

For a single point charge $Q$Q (or a spherical charge distribution, treated externally as if the charge were concentrated at the centre — exactly like the gravitational case), the field is radial: it points directly away from positive $Q$Q and directly towards negative $Q$Q. Its magnitude falls as $1/r^2$1/r2:

$\boxed{\,E = \frac{Q}{4\pi\varepsilon_0 r^2}\,}$E=4πε0​r2Q​​

This is the electric-field equivalent of Newton's gravitational $g = GM/r^2$g=GM/r2. The constant $1/(4\pi\varepsilon_0)$1/(4πε0​) plays the role of $G$G.

Constant	Value
$\varepsilon_0$ε0​	$8.85 \times 10^{-12}$8.85×10−12 F m$^{-1}$−1
$1/(4\pi\varepsilon_0)$1/(4πε0​)	$8.99 \times 10^9$8.99×109 N m$^2$2 C$^{-2}$−2

The inverse-square law $E \propto 1/r^2$E∝1/r2 is a direct consequence of the geometry of three-dimensional space: lines of force emanating from a point charge spread out over a spherical surface of area $4\pi r^2$4πr2, so their density (and hence the field strength) falls as $1/r^2$1/r2. The same geometric argument produces $1/r^2$1/r2 in gravity (lesson 7 of Module 5.4) and in the intensity of light from a point source.

Worked Example 2 — Radial Field

Calculate the electric field $10$10 cm from a point charge of $+5.0$+5.0 nC.

$$\begin{aligned} E &= \frac{Q}{4\pi\varepsilon_0 r^2} \\ &= \frac{5.0 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times (0.10)^2} \\ &= (5.0 \times 10^{-9})(8.99 \times 10^9)/0.01 \\ &\approx 4500\ \text{V m}^{-1}. \end{aligned}$$E​=4πε0​r2Q​=4π×8.85×10−12×(0.10)25.0×10−9​=(5.0×10−9)(8.99×109)/0.01≈4500 V m−1.​

Because the field is radial, this number is the same at every point on a sphere of radius $10$10 cm around the charge. Double the distance and the field becomes one quarter ($\sim 1125$∼1125 V m$^{-1}$−1 at $20$20 cm); halve the distance and it quadruples (to $\sim 18\,000$∼18000 V m$^{-1}$−1 at $5$5 cm). The $1/r^2$1/r2 scaling is sharp.

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Field Lines

Field lines are drawings that show the direction (by their arrowed sense) and the magnitude (by their spacing) of the electric field. Rules:

1. Start on positive charges, end on negative charges (or at infinity).
2. Never cross each other (because the field has a single direction at every point).
3. Are perpendicular to the surface of a conductor in electrostatic equilibrium (otherwise charges would flow along the surface until the parallel component of $E$E vanished).
4. Are closer together where the field is stronger; equally spaced for a uniform field.

A uniform field (parallel plates) has equally spaced, parallel lines pointing from $+$+ plate to $-$− plate. A radial field has lines fanning outward (or inward) from the point charge. Two opposite point charges give a "dipole" pattern with lines bowing from positive to negative. Two like charges give a saddle-shaped pattern with a region of zero field at the midpoint between them.

graph LR
    Pos((+)) -->|field lines| Neg((−))

The field-line picture is a qualitative tool — useful for sketching field directions and locating zero-field points by symmetry — but the quantitative field at any point is found by summing the inverse-square contributions of every source charge.

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Worked Example 3 — Field Due to Multiple Charges

Two point charges $+3$+3 nC and $-2$−2 nC are $20$20 cm apart. Find the electric field at a point midway between them (on the line joining them).

The midpoint is $10$10 cm from each charge.

Field from $+3$+3 nC: points away from $+3$+3 nC, towards the $-2$−2 nC charge.

$E_1 = \frac{(3 \times 10^{-9})(8.99 \times 10^9)}{(0.10)^2} \approx 2697\ \text{V m}^{-1}.$E1​=(0.10)2(3×10−9)(8.99×109)​≈2697 V m−1.

Field from $-2$−2 nC: points towards $-2$−2 nC, which is in the same direction as $E_1$E1​.

$E_2 = \frac{(2 \times 10^{-9})(8.99 \times 10^9)}{(0.10)^2} \approx 1798\ \text{V m}^{-1}.$E2​=(0.10)2(2×10−9)(8.99×109)​≈1798 V m−1.

Because both vectors point the same way, we simply add the magnitudes:

$E_\text{total} = 2697 + 1798 \approx 4500\ \text{V m}^{-1}.$Etotal​=2697+1798≈4500 V m−1.

Electric fields obey the principle of superposition — the total field at any point is the vector sum of the fields due to each charge present, treating each charge as if the others were absent. In this problem the two vectors happened to be parallel; in general they are not, and you must resolve them into components before summing.

Exam tip. For an off-axis point, resolve into perpendicular components, sum each component independently, and recombine using Pythagoras at the end. The vector arithmetic is identical to what you do for gravitational fields in Module 5.4 — only the constant in front of $1/r^2$1/r2 changes.

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Deflection of Charges in Fields

A charged particle in a uniform electric field experiences a constant force $F = EQ$F=EQ, hence a constant acceleration $a = EQ/m$a=EQ/m. This is exactly analogous to gravity on a projectile — a charged particle entering a horizontal field perpendicular to its initial velocity follows a parabolic path, exactly like a ball thrown horizontally.

Worked Example 4 — Cathode-Ray Deflection

An electron enters a uniform horizontal electric field of $5000$5000 V m$^{-1}$−1 at $2.0 \times 10^7$2.0×107 m s$^{-1}$−1 perpendicular to the field. The field region is $4.0$4.0 cm long. Find the vertical deflection as the electron exits the field.

Time in the field.

$t = \frac{L}{v} = \frac{0.040}{2.0 \times 10^7} = 2.0 \times 10^{-9}\ \text{s}.$t=vL​=2.0×1070.040​=2.0×10−9 s.

Acceleration.

$$\begin{aligned} F &= eE = (1.6 \times 10^{-19})(5000) = 8.0 \times 10^{-16}\ \text{N}, \\ a &= \frac{F}{m} = \frac{8.0 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 8.78 \times 10^{14}\ \text{m s}^{-2}. \end{aligned}$$Fa​=eE=(1.6×10−19)(5000)=8.0×10−16 N,=mF​=9.11×10−318.0×10−16​≈8.78×1014 m s−2.​

Deflection.

$y = \tfrac{1}{2} a t^2 = 0.5 \times 8.78 \times 10^{14} \times (2.0 \times 10^{-9})^2 \approx 1.76 \times 10^{-3}\ \text{m} \approx 1.8\ \text{mm}.$y=21​at2=0.5×8.78×1014×(2.0×10−9)2≈1.76×10−3 m≈1.8 mm.

This is how the old cathode-ray tube televisions built an image — deflecting electrons onto a phosphor-coated screen using a steerable parallel-plate field. Modern flat-panel displays use a different mechanism, but the CRT deflection equation is still the cleanest illustration of charged-particle motion in a uniform field.

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Synoptic Links

• Gravitational field strength (Module 5.4): The structural analogy is one of the most beautiful results in classical physics. Both fields obey an inverse-square law: $g = GM/r^2$g=GM/r2 and $E = Q/(4\pi\varepsilon_0 r^2)$E=Q/(4πε0​r2). Both define field strength as force-per-test-particle: $g = F/m$g=F/m and $E = F/Q$E=F/Q. Both obey superposition. The crucial differences are (i) gravity is always attractive while electric can be either, (ii) electric is vastly stronger per unit source, (iii) electric can be screened by conductors but gravity cannot, and (iv) only one sign of mass is known but both signs of charge exist.
• Parallel-plate capacitance (Module 6.1, lesson 1): The uniform field $E = V/d$E=V/d between capacitor plates is the same field that determines the capacitor's geometry: $C = \varepsilon_0 A/d$C=ε0​A/d follows from $E = V/d$E=V/d and $\sigma = \varepsilon_0 E$σ=ε0​E (the surface-charge density / field relation), giving $Q = \sigma A = \varepsilon_0 E A$Q=σA=ε0​EA and hence $C = Q/V = \varepsilon_0 A/d$C=Q/V=ε0​A/d. Capacitor and field topics interlock tightly.
• Charged-particle motion (Module 3.2): A charged particle in a uniform field undergoes constant-acceleration kinematics — the same equations you use for projectile motion, with $g$g replaced by $EQ/m$EQ/m. This is the basis of every electron-gun, mass-spectrometer source, and CRT deflection circuit.

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Specimen question modelled on the OCR H556 paper format

Question (10 marks): A horizontal parallel-plate capacitor has plates separated by $d = 8.0$d=8.0 mm and a potential difference of $V = 400$V=400 V applied across them, with the upper plate at the higher potential.

(a) (i) State the magnitude and direction of the electric field $E$E between the plates. (ii) Calculate the force on an electron placed midway between the plates. [3]

(b) The electron enters the field horizontally at $v = 1.5 \times 10^7$v=1.5×107 m s$^{-1}$−1, exactly mid-way between the plates. The plates are $L = 50$L=50 mm long in the direction of motion. (i) Calculate the time the electron spends in the field. (ii) Calculate the vertical component of velocity gained, and (iii) the vertical deflection at the exit. (iv) State whether the electron will hit the upper plate before exiting. Justify. [5]

(c) A second identical electron is fired into the field at the same position but at twice the speed. Without calculation, explain qualitatively how the vertical deflection at the exit compares with that of the first electron. [2]

AO breakdown