For the situation in problem 23.2, find the flux (in Nm²/C) through the y = a (right) face of the cube of side a = 1.60 m using an electric field with components in N/C of Ex = 4.0, Ey = -2.0(x/a - 3.0), and E₂ = 0. (5 sig. figs.)

Physics for Scientists and Engineers: Foundations and Connections
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Chapter25: Gauss’s Law
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For the situation in problem 23.2, find the flux (in
Nm²/C) through the y = a (right) face of the cube
of side a = 1.60 m using an electric field with
components in N/C of Ex = 4.0, Ey = -2.0(x/a
- 3.0), and E₂ = 0. (5 sig. figs.)
Transcribed Image Text:For the situation in problem 23.2, find the flux (in Nm²/C) through the y = a (right) face of the cube of side a = 1.60 m using an electric field with components in N/C of Ex = 4.0, Ey = -2.0(x/a - 3.0), and E₂ = 0. (5 sig. figs.)
2 An electric field given by
Ē = 4.0î – 3.0(y² + 2.0)ĵ pierces a
Gaussian cube of edge length 2.0 m
and positioned as shown in Fig. 23-7.
(The magnitude E is in newtons per
coulomb and the position x is in me-
ters.) 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?
x
Transcribed Image Text:2 An electric field given by Ē = 4.0î – 3.0(y² + 2.0)ĵ pierces a Gaussian cube of edge length 2.0 m and positioned as shown in Fig. 23-7. (The magnitude E is in newtons per coulomb and the position x is in me- ters.) 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? x
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Side of cube=a=1.6m

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