A-Level — physicsFoundational3 min read

Current, Charge, and Potential Difference

I = ΔQ/Δt; V = W/Q; drift velocity; mean drift speed equation

Tutor AI Team
Tutor AI Team

# Current, Charge, and Potential Difference — A-Level Physics

At A-Level, we deepen our understanding of electrical quantities, introducing the drift velocity model and exploring the microscopic basis of current flow.

1. Electric Current

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}

Current is the rate of flow of charge. 1 ampere = 1 coulomb per second.

Charge Carriers

  • Metals: free (delocalised) electrons
  • Electrolytes: positive and negative ions
  • Semiconductors: electrons and holes

Elementary Charge

The charge on one electron: e=1.6×10−19e = 1.6 \times 10^{-19} C

For NN electrons: Q=NeQ = Ne

2. Drift Velocity

Electrons don't zoom through wires — they drift slowly while bouncing off ions.

I=nAve\boxed{I = nAve}

Where:

  • II = current (A)
  • nn = number density of charge carriers (m⁻³)
  • AA = cross-sectional area (m²)
  • vv = mean drift velocity (m/s)
  • ee = charge per carrier (C)

Key Insights

  • Typical drift velocity in copper: ~0.1 mm/s (very slow!)
  • The electric field propagates at near light speed → the circuit responds almost instantly
  • Higher nn → lower vv for same current (metals have high nn, so low vv)
  • Semiconductors have low nn, so drift velocity is higher for same current

Effect of Cross-Sectional Area

If a wire narrows: AA decreases → vv must increase (for same II and nn). This is like water speeding up in a narrow pipe.

3. Potential Difference

V=WQV = \frac{W}{Q}

1 volt = 1 joule per coulomb.

Potential difference across a component = energy transferred per unit charge as charge flows through it.

Electromotive force (EMF) of a source = energy transferred per unit charge by the source (chemical → electrical).

Worked Example: Charge and Current

Problem

A current of 3 A flows for 5 minutes. Calculate the charge and number of electrons.

Q=It=3×300=900Q = It = 3 \times 300 = 900 C N=Q/e=900/1.6×10−19=5.625×1021N = Q/e = 900/1.6 \times 10^{-19} = 5.625 \times 10^{21} electrons

Solution

Worked Example: Drift Velocity

Problem

A copper wire (n=8.5×1028n = 8.5 \times 10^{28} m⁻³) has diameter 1 mm and carries 2 A. Find vv.

A=π(0.5×10−3)2=7.85×10−7A = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} m² v=I/(nAe)=2/(8.5×1028×7.85×10−7×1.6×10−19)v = I/(nAe) = 2/(8.5 \times 10^{28} \times 7.85 \times 10^{-7} \times 1.6 \times 10^{-19}) =2/(1.067×104)=1.87×10−4= 2/(1.067 \times 10^4) = 1.87 \times 10^{-4} m/s = 0.19 mm/s

Solution

5. Practice Questions

    1. Calculate the number of electrons passing a point when 0.5 A flows for 2 minutes. (2 marks)
    1. A wire has n=5×1028n = 5 \times 10^{28} m⁻³, diameter 0.5 mm, and carries 1.5 A. Calculate drift velocity. (3 marks)
    1. Explain why drift velocity increases when a wire becomes narrower. (2 marks)

    Answers

    1. Q=0.5×120=60Q = 0.5 \times 120 = 60 C. N=60/1.6×10−19=3.75×1020N = 60/1.6 \times 10^{-19} = 3.75 \times 10^{20}.
    1. A=π(2.5×10−4)2=1.96×10−7A = \pi(2.5 \times 10^{-4})^2 = 1.96 \times 10^{-7} m². v=1.5/(5×1028×1.96×10−7×1.6×10−19)=1.5/1568=9.57×10−4v = 1.5/(5 \times 10^{28} \times 1.96 \times 10^{-7} \times 1.6 \times 10^{-19}) = 1.5/1568 = 9.57 \times 10^{-4} m/s.

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Summary

  • I=ΔQ/ΔtI = \Delta Q/\Delta t; Q=NeQ = Ne; e=1.6×10−19e = 1.6 \times 10^{-19} C
  • Drift velocity: I=nAveI = nAve — electrons move slowly (~mm/s)
  • V=W/QV = W/Q; EMF = energy per charge from source
  • High nn (metals) → low drift velocity
Tags:
#a-level#physics#electricity#current#charge#potential-difference#drift-velocity

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