Electric and Magnetic Fields

Physics· Unit 4 (WPH14) Spec 4.4, statements 92–110· 45 min read

1. Electric Fields (Radial and Uniform)★★☆☆☆⏱ 10 min

📘 Definition

Electric field strength

E

Force per unit positive test charge in a field, given by $E = F / Q$

Example:

A 2 nC charge experiences a force of 5×10⁻⁴ N in a field, so E = 2.5×10⁵ N C⁻¹

For point charge radial fields, use Coulomb's law to find force between two charges, and the derived radial field strength formula. For uniform parallel plate fields, field strength is constant and given by $E = V / d$ , where V is potential difference between plates and d is separation.

F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2} \quad \text{(Coulomb's law)}

E = \frac{Q}{4\pi\varepsilon_0 r^2} \quad \text{(Radial electric field strength)}

V = \frac{Q}{4\pi\varepsilon_0 r} \quad \text{(Radial electric potential)}

📐 Worked Example

Calculate the electric field strength 0.2 m from a point charge of 3 μC. Use $1/4 π ε_{0} = 8.99 \times 1 0^{9}$ N m² C⁻².

1
Substitute values into the radial field formula:
2
$E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{8.99 \times 10^9 \times 3 \times 10^{-6}}{(0.2)^2}$
3
Calculate the result: $E = 6.74 \times 1 0^{5}$ N C⁻¹, directed away from the positive point charge.

2. Capacitance and Energy Storage★★☆☆☆⏱ 8 min

📘 Definition

Capacitance

C

Charge stored per unit potential difference across a capacitor, $C = Q / V$ , measured in farads (F)

Example:

A capacitor storing 4 mC of charge at 12 V has capacitance 333 μF

Energy stored in a capacitor is equal to the area under the V-Q graph, leading to the base formula $W = \frac{1}{2} Q V$ . Two derived forms are given on the formula sheet for use when Q or V is unknown.

W = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

📐 Worked Example

A 220 μF capacitor is charged to a potential difference of 15 V. Calculate the energy stored in the capacitor.

1
Use the $W = \frac{1}{2} C V^{2}$ formula, as we know C and V:
2
$W = 0.5 \times 220 \times 10^{-6} \times (15)^2$
3
Calculate the result: $W = 0.0248$ J, or 24.8 mJ.

3. RC Circuit Charge and Discharge★★★☆☆⏱ 12 min

📘 Definition

Time constant

\tau

Time for charge/voltage/current in a discharging RC circuit to fall to 1/e (~37%) of its initial value, equal to $τ = R C$

For discharging circuits, charge, current and voltage follow a decaying exponential curve. For charging circuits, they follow a rising exponential curve. The log-linear form of the discharge equation can be plotted to get a straight line with gradient $- 1/ R C$ , used to find time constant experimentally (Core Practical 11).

Q = Q_0 e^{-t/RC} \quad \text{(Discharge charge equation)}

\ln Q = \ln Q_0 - \frac{t}{RC} \quad \text{(Log-linear discharge form)}

📐 Worked Example

A 470 μF capacitor discharges through a 10 kΩ resistor. Calculate the time taken for the charge to fall to 20% of its initial value.

1
First calculate the time constant: $τ = R C = 10 \times 1 0^{3} \times 470 \times 1 0^{- 6} = 4.7$ s
2
Rearrange the log equation to solve for t when $Q / Q_{0} = 0.2$ :
3
$t = RC \ln \left(\frac{Q_0}{Q}\right) = 4.7 \times \ln(5)$
4
Calculate the result: $t = 7.56$ s

4. Magnetic Fields and Forces★★★☆☆⏱ 8 min

📘 Definition

Magnetic flux density

B

Force per unit current per unit length of conductor perpendicular to the magnetic field, measured in tesla (T)

Example:

A 0.5 m wire carrying 2 A perpendicular to a 0.4 T field experiences force 0.4 N

Force on a current-carrying conductor is given by $F = B I l sin θ$ , where θ is the angle between the current and magnetic field. Force on a moving charged particle is $F = B q v sin θ$ , where θ is the angle between velocity and B field. Use Fleming's Left Hand Rule to find force direction: first finger = B field, second finger = conventional current/positive charge velocity, thumb = force direction.

📐 Worked Example

An electron travels at $2 \times 1 0^{6}$ m s⁻¹ perpendicular to a 0.3 T magnetic field. Calculate the magnetic force on the electron. Charge of electron = $1.6 \times 1 0^{- 19}$ C.

1
Since θ = 90°, $sin θ = 1$ . Substitute values into the force formula:
2
$F = Bqv = 0.3 \times 1.6 \times 10^{-19} \times 2 \times 10^6$
3
Calculate the result: $F = 9.6 \times 1 0^{- 14}$ N. Use Fleming's Left Hand Rule, reversing the current direction for the negative electron, to find force direction.

5. Electromagnetic Induction★★★★☆⏱ 7 min

Induced emf is generated when there is a change in flux linkage through a coil, either by relative motion between a magnet and coil, or change in current in a linked coil. Faraday's law gives the magnitude of induced emf, while the negative sign in the formula encodes Lenz's law, which states induced emf opposes the change that caused it.

📐 Worked Example

A 50-turn coil with area 0.02 m² is placed perpendicular to a 0.8 T magnetic field. The field is reduced to 0 T in 0.1 s. Calculate the average induced emf.

1
Calculate change in flux linkage: $Δ (N ϕ) = N \times Δ B \times A = 50 \times 0.8 \times 0.02 = 0.8$ Wb turns
2
Apply Faraday's law for average emf:
3
$|E| = \frac{\Delta(N\phi)}{\Delta t} = \frac{0.8}{0.1} = 8 \text{ V}$

6. Common Pitfalls

Wrong move:

Using $E = V / d$ for radial electric fields

Why:

$E = V / d$ only applies to uniform parallel plate fields; radial fields follow the inverse square law.

Correct move:

Use $E = Q / (4 π ε_{0} r^{2})$ for point charge radial fields, reserve $E = V / d$ exclusively for uniform fields.

Wrong move:

Forgetting the $sin θ$ term in magnetic force calculations

Why:

Force is zero when current/velocity is parallel to the B field, maximum only when perpendicular.

Correct move:

Always use θ = angle between current/velocity direction and magnetic field lines, even if θ = 90° (sinθ = 1) to avoid errors.

Wrong move:

Using the decaying exponential $Q = Q_{0} e^{- t / R C}$ for charging RC circuits

Why:

The decaying form applies only to discharge; charging follows a rising exponential $Q = Q_{0} (1 - e^{- t / R C})$ .

Correct move:

Match the exponential form to the process: falling curves = discharge, rising curves = charging.

Wrong move:

Calculating capacitor energy as $W = Q V$ instead of $W = \frac{1}{2} Q V$

Why:

Energy stored is the area under the V-Q graph, not the product of final charge and voltage values.

Correct move:

Use only the three given energy formulae: $\frac{1}{2} Q V$ , $\frac{1}{2} C V^{2}$ , or $\frac{Q ^{2}}{2 C}$ .

Wrong move:

Ignoring the negative sign in Faraday's law as irrelevant

Why:

The negative sign encodes Lenz's law, which requires you to state that induced emf opposes the change in flux linkage.

Correct move:

Use the magnitude for calculation, but explicitly reference Lenz's law to explain emf direction when asked.

7. Quick Reference Cheatsheet

Formula	Context	Units	Key Note
$E = F / Q$	All electric fields	N C⁻¹ / V m⁻¹	Definition of electric field strength
$F = Q_{1} Q_{2} /4 π ε_{0} r^{2}$	Force between two point charges	N	Repulsive for like charges, attractive for opposite
$E = V / d$	Uniform parallel plate fields	N C⁻¹	Only valid for constant field between plates
$C = Q / V$	Capacitance definition	F	1 F = 1 C V⁻¹
$W = \frac{1}{2} Q V$	Capacitor stored energy	J	Equal to area under V-Q graph
$τ = R C$	RC circuit time constant	s	Time for discharge to 37% of initial value
$ln Q = ln Q_{0} - t / R C$	Log-linear RC discharge	Unitless	Gradient of plot = $- 1/ R C$
$F = B I l sin θ$	Force on current-carrying conductor	N	θ = angle between I and B
$F = B q v sin θ$	Force on moving charged particle	N	Reverse direction for negative charges
$∣ E ∣ = N Δ ϕ /Δ t$	Induced emf (Faraday's law)	V	Average emf over time interval Δt

8. Frequently Asked

Do I need to derive the RC exponential discharge equation?

No, the exponential and log forms of the RC discharge equation are given on the formula sheet. You only need to apply them, and interpret log-linear plots to find the time constant.

Can I use Fleming's Left Hand Rule for both moving charges and current-carrying conductors?

Yes. For positive charges, use conventional current direction; for negative charges like electrons, reverse the current direction before applying the rule.

Is gravitational field content included in this topic?

No, gravitational fields are exclusively part of Unit 5 content for Edexcel IAL Physics, so no comparison between electric and gravitational fields is required here.

Going deeper

official_specEdexcel IAL Physics 2018 SpecificationRefer to Unit 4 section 4.4 for full topic requirements
practical_guideCore Practical 11: Capacitor Charge/DischargeIncludes methodology, variables, and error analysis guidance

What's Next

Once you have mastered electric and magnetic fields, you will be ready to tackle the final Unit 4 topic: Nuclear and Particle Physics, where you will apply field concepts to particle acceleration, deflection, and detection. This topic also forms a critical foundation for Unit 5 content including gravitational fields, astrophysics, and nuclear decay, so solidifying your understanding of field line, potential, and energy rules now will reduce your revision load later. Prioritize practicing past paper questions focused on RC log-linear plots and electromagnetic induction direction questions, as these are high-frequency exam questions that often trip up students. You should also review Core Practical 11 methodology to prepare for practical exam questions on capacitor time constant measurement.