Section 3.3 · Topic 8 of 8

3.3.8 Problems and solutions

Each solution below starts with a short strategy, works through labelled steps, boxes the result and ends with a check. Throughout this subsection the potential and the field are related by

V (P) = \frac{1}{4 π ε _{0}} \int \frac{d q ^{'}}{R}, E = - \nabla V, V_{A} - V_{B} = \int_{A}^{B} E \cdot d l, Δ V = - \frac{ρ _{v}}{ε _{0}},

with $V (\infty) = 0$ unless another reference is stated. At the surface of a conductor the normal component of $D = ε_{0} E$ equals the surface charge density, $D_{n} = ρ_{S}$ , the normal pointing out of the conductor into the surrounding space (Section 3.2).

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Start with Problem 3.3.1 →

3.3.8 Problems and solutions

In this part

Problems · 3

Home exercises · 9

Further exercises · 10

Worked examples · 2