IT(b)-T(a).1 VT (2x4y)x(4x2Z3)y(6yz2)z;VT.
4.7: replace the (defective) solution with the following: If the potential is zero at innity, the energy of a point charge Q is (Eq.If the magnetic eld does not go to zero at innity, one must stipulate that the dipole starts out oriented perpendicular to the eld.Conclusion:Ax(BxC) (AxB)xC :eitherA isparalleltoC, orB isperpendiculartoA andC.Therefore /2 cos /2, total: Page 25, Prob.
Iaiib CI sin03n iaiibi sin01n iaiici sinO2n.
In line 1 change a r R t o s R ; in the same line change dr to ds; in the next line change dr to ds (twice and change r to s; in the last line change r to s, dr.
8.5(c there should be a minus sign in front of 2 in the box.7.32(c last line: in the nal two equations, insert an I immediately after.6.3; to move the dipole in from innity we must exert an opposite force, so the work done croc 2 pc iso is (I us state code lookup used the gradient theorem,.10.14: in the rst line, change (9.98) to (10.42).Problem.20 J (xyz)g(UxUyZz)J(Vg)g(VI).7.5, penultimate displayed equation: tp should.In that case W QV (r) V (r0 and r should be.Qed Problem.21 ( )z.So:Ax(BxC) - (AxB)xC -Bx(CxA) A(B.C) - C(A.B).