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Solved: The cube in Fig. E22.6 has sides of length L =
Chapter 22, Problem 6E(choose chapter or problem)
The cube in Fig. E22.6 has sides of length . The electric field is uniform, has magnitude \(E=400 \times 10^{3} \mathrm{~N} / \mathrm{C}\), and is parallel to the
-plane at an angle of \(53.1^{0}\) measured from the
-axis toward the
-axis. (a) What is the electric flux through each of the six cube faces \(S_{1}, S_{2}, S_{3}, S_{4}, S_{5}\) and \(S_{6}\)? What is the total electric flux through all faces of the cube?
Equation transcription:
Text transcription:
E=400 times 10^{3}{~N}{C}
53.1^{0}
S{1}, S{2}, S{3}, S{4}, S{5}
S{6}
Questions & Answers
QUESTION:
The cube in Fig. E22.6 has sides of length . The electric field is uniform, has magnitude \(E=400 \times 10^{3} \mathrm{~N} / \mathrm{C}\), and is parallel to the
-plane at an angle of \(53.1^{0}\) measured from the
-axis toward the
-axis. (a) What is the electric flux through each of the six cube faces \(S_{1}, S_{2}, S_{3}, S_{4}, S_{5}\) and \(S_{6}\)? What is the total electric flux through all faces of the cube?
Equation transcription:
Text transcription:
E=400 times 10^{3}{~N}{C}
53.1^{0}
S{1}, S{2}, S{3}, S{4}, S{5}
S{6}
ANSWER:Solution 6E
Step 1
(a)
The electric field in vector form can be written as,
…… (1)
Here a, b, c are components of in x, y, z direction respectively.
Now for given vector
Substitute value of a, b, c in (1)
Surface area of each side of the cube is,
Here L is length of side of cube.
Substitute for.
The normal vector of surface 1 of the cube is pointed along the negative y-axis. That is,
The electric flux through the negative y-axis (or through the surface of the cube) is,
