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#1 We were given the following information involving Gauss' law in a math assignment, just as an application of the surface integral of a vector field:
[tex] Q = \epsilon_{0} \iint_{S} \vec{E} \cdot d \vec{S} [/tex]
[tex] \vec{E}(x,y,z) = x \hat{i} + y \hat{j} + z \hat{k} [/tex]
[itex] S [/itex] is the cube with vertices
[itex](\pm 1, \pm 1, \pm 1)[/itex]
I approached this problem by finding an expression for each of the faces (e.g. z = 1, within the appropriate bounds), calculating the surface integral over each one, and adding them together. I have two problems with this:
1. The textbook says that the integral given represents the charge [itex] Q [/itex] enclosed by a closed surface. So what does it mean to calculate this integral for a surface that isn't closed, like the face of a cube?
2. I got an answer of [itex] Q = 0 [/itex]. Is this correct?
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[tex] Q = \epsilon_{0} \iint_{S} \vec{E} \cdot d \vec{S} [/tex]
[tex] \vec{E}(x,y,z) = x \hat{i} + y \hat{j} + z \hat{k} [/tex]
[itex] S [/itex] is the cube with vertices
[itex](\pm 1, \pm 1, \pm 1)[/itex]
I approached this problem by finding an expression for each of the faces (e.g. z = 1, within the appropriate bounds), calculating the surface integral over each one, and adding them together. I have two problems with this:
1. The textbook says that the integral given represents the charge [itex] Q [/itex] enclosed by a closed surface. So what does it mean to calculate this integral for a surface that isn't closed, like the face of a cube?
2. I got an answer of [itex] Q = 0 [/itex]. Is this correct?
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