An electric field given by [ E = 6.8i - 6.3(y2 + 5.3)j ] pierces a Gaussian cube

Question

An electric field given by [ E = 6.8i - 6.3(y2 + 5.3)j ] pierces a Gaussian cube of edge length 5.2 m and positioned as shown in Fig. 23-56. (The magnitude E is in newtons per coulomb and the position x is in meters.) What net charge (in Coulombs) is enclosed by the Gaussian cube?

An electric field given by [ E = 6.8½i - 6.3(y2 + 5.3)½j ] pierces a Gaussian cube of edge length 5.2 m and positioned as shown in Fig. 23-56. (The magnitude E is in newtons per coulomb and the position x is in meters.) What net charge (in Coulombs) is enclosed by the Gaussian cube?

Explanation / Answer

a) Flux through the topmost surface=EA= [6.8·i - 6.3(5.2+ 5.3)·j ].5.22 k=183.87-1788.7=-1604.83

Flux through the bottom=E.A= 6.8·i *5.22-k=-183.87

The flux through other sides cancel each other.

Net flux out=-1788.7=q/0

q=0*Flux=1.584*10-8C

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