5. Current density and surface integrals: Let J = -z²(î + ý + 2) A/m² denote the electrical

current density field - i.e., current flux per unit area in a region of space represented in Cartesian

coordinates. A current density of J = −2² (ê + ý + 2) A/m²2 implies the flow of electrical current in

the direction= 2 tỷ tệ with a magnitude of |J| = ²√3 amperes (A) per unit area.

a) Calculate the total current flux fs J. dS "out" of a closed surface S enclosing a cubic volume

V = 1m³ with vertices at (x, y, z) = (0,0,0) and (1,1,1) m.

Hint: Surface S of cube V consists of six surfaces of square shapes having equal areas S; = 1 m²,

i=1,2,..., 6. The flux fg J-dS is therefore the sum of six surface integrals fs J-dS taken over

surfaces S₁, where the infinitesimal area vectors dS are, in turn, ±ždxdy, ±âdydz, and ±ýdzd.x

- by convention dS; are taken as vectors pointing away from volume V (at each subsurface S;)

in flux calculations.

b) Given the result of part (a), do you expect the total amount of electrical charge Qy contained

in volume V to increase or decrease? In answering this question assume that electric charge

is conserved (as in real life!). Remember that electrical current represents electrical charges in

motion.

c) Given the result of part (b), do you expect the electric field flux fg E-dS out of the same closed

surface S to increase or decrease as a function of time? Explain in terms of Gauss' law.