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The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Electric Charge and Electric Fields – Lecture 2 (University Physics II)

Overview

This lecture builds the foundation of electrostatics: how charged objects exert forces on one another and how those forces can be repackaged as an electric field that fills space. It opens with Coulomb's law — the inverse-square force between two point charges — and contrasts it with Newton's law of gravity. It then introduces the electric field as force-per-unit-charge, along with the field-line picture used to visualize it. The second half applies calculus to find the field produced by continuous charge distributions: a charged ring, a finite rod, a charged disc, and finally a pair of parallel plates. A recurring theme is checking each result against limiting cases (very near, very far) to confirm it behaves sensibly.

Key Concepts

Charge basics (recap). All charge comes in integer multiples of the elementary charge e=1.602×10−19 Ce = 1.602 \times 10^{-19}\ \text{C}; protons carry +e+e, electrons −e-e. Charge is conserved — never created or destroyed, only separated and recombined by moving electrons. Conductors let electrons flow freely; insulators do not. One coulomb is enormous: about 6×10186 \times 10^{18} electron charges. Real static charges are typically microcoulombs (μC\mu\text{C}) or nanocoulombs (nC\text{nC}).

Coulomb's law. The electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them: F=k q1q2r2,k=8.988×109 N⋅m2/C2.F = k\,\frac{q_1 q_2}{r^2}, \qquad k = 8.988 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2.

Electric field. A charge creates a field in the space around it; any other charge placed there feels a force. The field is the force per unit charge measured with a small test charge q0q_0: E⃗=F⃗q0.\vec{E} = \frac{\vec{F}}{q_0}.

Field lines. A way to visualize E⃗\vec{E}: the field is tangent to the line at each point, pointing the way a positive charge would move. Lines emanate from positive charges and terminate on negative ones. Where lines crowd together the field is strong; where they spread out it is weak. The lines show direction and strength, not the trajectory a particle would follow.

Continuous charge distributions. When charge is spread over a ring, rod, surface, or volume, slice it into infinitesimal point-like pieces dqdq, find each piece's field, and integrate. Symmetry usually cancels all components except one (e.g. the axial xx-component), so include a cos⁡α\cos\alpha factor and integrate that component alone. Charge density bridges dqdq to geometry:

Key distribution results (on the axis, distance xx):

Worked Example

Electron accelerated between charged plates. Two large plates produce a uniform field E=1×104 N/CE = 1 \times 10^{4}\ \text{N/C} and sit d=1 cm=0.01 md = 1\ \text{cm} = 0.01\ \text{m} apart. An electron is released from rest at the negative plate. Find its acceleration, speed at the positive plate, and crossing time.

Start from F⃗=qE⃗\vec F = q\vec E and Newton's second law, then the work–energy theorem W=qEd=12mv2W = qEd = \tfrac{1}{2}mv^2:

F=eE=(1.602×10−19)(1×104)=1.6×10−15 Na=Fme=1.6×10−159.11×10−31=1.76×1015 m/s2v=2Fdme=2(1.6×10−15)(0.01)9.11×10−31=5.94×106 m/st=va=5.94×1061.76×1015≈3.4×10−9 s\begin{aligned} F &= eE = (1.602\times10^{-19})(1\times10^{4}) = 1.6\times10^{-15}\ \text{N} \\ a &= \frac{F}{m_e} = \frac{1.6\times10^{-15}}{9.11\times10^{-31}} = 1.76\times10^{15}\ \text{m/s}^2 \\ v &= \sqrt{\frac{2Fd}{m_e}} = \sqrt{\frac{2(1.6\times10^{-15})(0.01)}{9.11\times10^{-31}}} = 5.94\times10^{6}\ \text{m/s} \\ t &= \frac{v}{a} = \frac{5.94\times10^{6}}{1.76\times10^{15}} \approx 3.4\times10^{-9}\ \text{s} \end{aligned}

The electron reaches about 2% of the speed of light in roughly 3.4 nanoseconds.

Summary