By J. E. Parton, S. J. T. Owen, M. S. Raven (auth.)

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**Example text**

Dl round the two closed paths (1, 1, 0)(-1, 1, 0)(-I, -I, O)(I, -I, O)(I, I, 0) and (0, 0, O)(I, 0, 0) (I, I, 0)(0, I, 0)(0, 0, 0). Is Fa conservative field? Check by curl F. 6 Find the electric field strength E and volume charge density Pv at the point (a, a, a) for each of the fields in which scalar Vis given as follows. 7 Find the energy stored in the region r >a for a point charge Qat the origin. 8 Given an electric field strength£= (p/4ne 0 ilX2 cos Oa, +sin Oa6 ) V/m for a dipole p in the z direction at the origin, find the energy stored in the region r >a.

For the rectangular column of charge shown, the normal electric field strength En= Ex at a symmetrical point P can be found by considering the elementary wide plate, dx thick, having 'surface' charge density Pv dx. From the previous case we have Pv dx dEx = - - (rt- 2'Y)Ox :bteo and using x =c - b (tan 'Y) the integral can be evaluated between the limits 'YF and 'YB· The result is 'Yn)] ax Ex =Pv - [ a('YF + 'Ya) + c('Ya - 'YF) +bIn (cos -1t€o COS'YF Of course in all these cases PL, p 8 , Pv have been taken as constants; they could all be functions of position and the integrals would be correspondingly more difficult.

I1 S = '\72S = 0 a2 s a2 s a2 s ax 2 ay 2 az 2 -+-+-=0 1. 2 laplace equation div Fi= 0, curl F = 0 The field is not solenoidal but it is irrotational or lamellar and "i/2 S a2 s a2 s a2 s ax 2 ay 2 az 2 -+-+-i=-0 *0 Poisson equation 19 VECTOR ANALYSIS 1. 3 div F= 0, curl F=F 0 This is solenoidal but it is also rotational and cannot have been derived from a scalar potentialS. But suppose it has been obtained from a vector potential A so chosen that F = curl A = V' x A then we have curl F =curl (curl A)= grad divA - V' 2A We can also stipulate that the vector potential A is solenoidal so that divA = 0 giving finally grad divA = 0 and curl F=- V' 2A =I= 0 This is Poisson's equation using vector potential A instead of scalarS.