Mathematical Physics Vol 1

Chapter 4. Field theory

190

where A =( x 2 + y − 4 ) i + 3 xy j +( 2 xz + z 2 ) k , and the surface S is: a) the half of the sphere x 2 + y 2 + z 2 = 16 above the xy plane, b) the paraboloid z = 4 − ( x 2 + y 2 ) above the xy plane.

Solution a) − 16 π , b) − 4 π .

Exercise 140 The potential φ ( P ) , at point P ( x , y , z ) , in a system of charge particles q 1 , q 2 ,..., q n with position vectors r 1 , r 2 ,..., r n is given by the formula

n ∑ m = 1

q m r m

φ =

.

Prove Gauss’s law

x S

E · d S = 4 π Q , where E = − ∇ φ is the strength of the electric field, S the surface encompassing all particles, and Q = ∑ n m = 1 q m the total charge covered by S .

Exercise 141 Let V be a region bounded by S , ρ the fluid density, and φ ( P ) the potential at point P defined by φ = y V ρ d V r . Prove that a) x S E · d S = 4 π y V ρ d V , where E = − ∇ φ b) ∇ 2 φ = − 4 πρ (Poisson’s equation), at point P where there is fluid, and ∇ 2 φ = 0 (Laplace’s equation),

where there is no fluid.