Classical Electrodynamics |
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Page 86
... results can be obtained from this expansion. If we let x' → 0, only the m = 0 term
survives, and we obtain the integral representation: –––3 | cos kz Ko(kp) dk (
3.150) Vpo -- 22 to Jo If we replace p” in (3.150) by R* = p" + p” – 2pp'cos (q ...
... results can be obtained from this expansion. If we let x' → 0, only the m = 0 term
survives, and we obtain the integral representation: –––3 | cos kz Ko(kp) dk (
3.150) Vpo -- 22 to Jo If we replace p” in (3.150) by R* = p" + p” – 2pp'cos (q ...
Page 96
3.10 Solve for the potential in Problem 3.2, using the appropriate Green's function
obtained in the text, and verify that the answer obtained in this way agrees with
the direct solution from the differential equation. 3.11 A line charge of length 2d ...
3.10 Solve for the potential in Problem 3.2, using the appropriate Green's function
obtained in the text, and verify that the answer obtained in this way agrees with
the direct solution from the differential equation. 3.11 A line charge of length 2d ...
Page 402
To obtain E, we merely interchange ma and m, and change 6' into tr– 6' (cos 0' —
—cos 0'). The relation between angles 0' and 0s can be obtained from the
expression * - 5 – Ash a * = —to (12.49) Pan Yos(q'cos 0' + vow B3) Therefore we
...
To obtain E, we merely interchange ma and m, and change 6' into tr– 6' (cos 0' —
—cos 0'). The relation between angles 0' and 0s can be obtained from the
expression * - 5 – Ash a * = —to (12.49) Pan Yos(q'cos 0' + vow B3) Therefore we
...
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Contents
Introduction to Electrostatics | 1 |
BoundaryValue Problems in Electrostatics I | 26 |
BoundaryValue Problems in Electrostatics II | 54 |
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