Question

Figure 23-31 shows a closed Gaussian surface in the shape of a cube of edge length 2.5 m, with one corner at x1 = 5.0 m,y1 = 3.7 m. The cube lies in a region where the electric field vector is given by = -3.4 -4.4 y2 +2.8 N/C, with y in meters. What is the net charge (in Coulombs) contained by the cube?

Answer 1

length of the edge of the cube L = 2.5 m

distances x1 = 5 m and y1 = 3.7 m

electric field vector E = [-3.4 i^ - 4.4y² j^ + 2.8 k^] N/C

from Gauss's law ,

electric flux inside a closed surface is

          φ = ∫ E.dA   ........................ (1)

non zero component of electric field is

           E_{non zero} =  - 4.4y² j^ N/C

the face of a cube located at y1 = 3.7 m has an area is

           A = (2.5 m)(2.5 m)

               = 6.25 m²

this area directed along +j^ direction.

the flux through the face is

        φ =  E_{non zero} (A)

            = (- 4.4y² N/C)(6.25 m²)

            = (- 4.4(3.7 m)² N/C)(6.25 m²)

            = -376.475 Wb

..................................................................................

the face of cube located at y = 3.7 m - 2.5 m = 1.2 m

this area directed along -j^ direction.

the flux through the face is

        φ =  E_{non zero} (A)

            = (- 4.4y² N/C)(6.25 m²) cos180

            = (- 4.4(1.2 m)² N/C)(6.25 m²)(-1)

            = 39.6 Wb

hence ,

the net electric flux is

    φ = -376.475 Wb + 39.6 Wb

       = -336.875 Wb

using Gauss's law ,

      charge Q = ε₀φ

                     = (8.85*10^-12 C²/N.m²)(-336.875 Wb)

                     = -2.98*10^-9 C

                     = -2.98 nC