An electric field given by right arrow E = 1.6 cap I ? 9.1(y^2 + 8.8) cap j pier

Question

An electric field given by right arrow E = 1.6 cap I ? 9.1(y^2 + 8.8) cap j pierces the Gaussian cube of edge length 0.120 m and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) what is the net electric flux through the cube? (a) Number -1.14584 Units N.m cap 2/C (b) Number 1.15315 Units N.m cap 2/C (c) Number -1.7472 Units N.m cap 2/C (d) Number 1.7472 Units N.m cap 2/C (e) Number -0.00731 Units N.m cap 2/C

Explanation / Answer

area = 0.12 *0.12 = 0.0144 m^2

a) fluxa = integration E . A

   = integration1.6 i - 9.1(y^2+8.8)j . 0.0144j

   = -9.1*0.0144*(y^3/3+8.8*y)

   = -0.1384 Nm^2/C

b) flux b = integration E . A

   = integration1.6 i - 9.1(y^2+8.8)j . 0.0144-j

   = 9.1*0.0144*(y^3/3+8.8*y)

   = +0.1384 Nm^2/C

c fluxc = integration E . A

   = integration1.6 i - 9.1(y^2+8.8)j . 0.0144-i

   = -1.6*0.0144*y
   = 0.00276 Nm^2/C

d) flux d= integration E . A

   = integration1.6 i - 9.1(y^2+8.8)j . 0.0144-k

   = 0 Nm^2/C

e) flux = 0

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