By Jack Vanderlinde

ISBN-10: 1402026994

ISBN-13: 9781402026997

ISBN-10: 1402027001

ISBN-13: 9781402027000

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**Additional resources for Classical electromagnetic theory**

**Sample text**

2–11) i=1 j=1 leading to V (r ) = 1 4πε0 ρ(r ) 3 d r |r − r | (2–12) 36 Classical Electromagnetic Theory = 1 4πε0 r· ρ(r )d3 r + r xi xj + = 1 4 πε0 q r·p + 3 + r r ρ(r )r d3 r r3 ρ(r ) 3xi xj − δij r 2r5 2 d3 r + ... (2–13) xi xj Qij + ... 2r5 (2–14) Here q is the total charge of the source, p= ρ(r )r d3 r (2–15) is the dipole moment of the distribution and Qij ≡ ρ(r )(3xi xj − δij r 2 )d3 r (2–16) is the ij component of the Cartesian (electric) quadrupole moment tensor. In principle this expansion could be continued to higher order, but it is rarely fruitful to do so.

1-10 Find the electric potential for the 1-6 Find the electric ﬁeld produced by charge distribution of problem 1-8. a spherically symmetric charge distribu- 1-11 Find the electric potential for the tion with charge density charge distribution of problem 1-2. r 2 ρ0 1 − for r ≤ a and 1-12 Find the electric potential for the ρ= a charge distribution of problem 1-3. 0 for r ≥ a Chapter 1 - Static Electric and Magnetic Fields in Vacuum 31 1-13 A ﬁne needle emits electrons isotropically at a steady rate.

We will show the ﬁrst of the integrals in (1–45) to vanish whereas the second gives, using (26), − µ0 4π J(r )∇2 1 r−r J(r )δ(r − r )d3 r = µ0 J(r ) d3 r = µ0 (1–46) We return now to the ﬁrst integral in (1–45). We observe that ∇f (r − r ) = −∇ f (r − r ) and invoke identity (7) to transform the integral to one part which can be integrated with the divergence theorem and a second that has the divergence acting on J(r ) : µ0 4π J(r ) · ∇ 1 r−r =− d3 r = − µ0 4π ∇ · µ0 4π 1 r−r J(r ) · ∇ J(r ) r−r d3 r − d3 r ∇ · J(r ) 3 d r r−r (1–47) In the case of static charge densities, as we are discussing here, the continuity equation (1–24) states that ∇ · J(r ) = 0.

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