Thursday, February 18, 2010

Cube Field 3 What Is The Correct Gauss's Law Integral For A Point Charge At The Origin And 1 Face Of A Cube?

What is the correct Gauss's law integral for a point charge at the origin and 1 face of a cube? - cube field 3

There is 1 coulomb plus point at the origin [0,0,0 submitted]. A unit cube is in the first octant whose sides are parallel to XY, XZ, and YZ planes, with its center in [3 / 2, 3 / 2, 3 / 2] and the angle is closest to the origin [1 , 1.1 found]. The normal to this side of the cube is -1 in the z direction, I would really determine the cubic Gauss integral unit face parallel to the xy plane with z = 1 The electric field at a point where [x, y, z] kn [xi, yi, ZK] / (x ^ 2 + y ^ 2 + z ^ 2) ^ (3 / 2), where Q is the charge and i , j, k-vectors in the x, y, z axis directions. (Select i, j, k, z to avoid confusion with the use of x, y, because both scalar and vector) gives me the double integral of -1 / (x ^ 2 + y ^ 2 +1) ^ (3 / 2) dy dz, u0026lt with a limit of 1 \\ \\ \\ \\ \\ \\ \\ \\, = x, y u0026lt; = 2, but my value is not resolved.

Can someone show how to use the work on the surface of fixed point E ds = n where n-1K and ds = dx dy and how to successfully create the double integral?

Mathematica Integrator is arctan (x / (x ^ 2 + y ^ 2 +1)) for the result, but do not know how to find the right results are limits to both X and Y. run

Thank you!

1 comment:

Tabula Rasa said...

Hello,

No need to show how to traverse the surface of the double integral, since you did great!

[1,2] ∫ [1,2 ∫ -1 / (x ^ 2 + y ^ 2 +1) ^ (3 / 2) dx dy

It's just that the integration of the complicated double integral.
It can be integrated, but really complicated.

If you can use Mathematica integrator of solutions for this problem use, to individual and overall limits for themselves. It solves the double integral.

You must be achieved through the integration and WRT

[1,2] ∫ -1 / (x ² + y ² +1) ^ (3 / 2) dy = - y / [(x ² + 1) √ (x ² + y ² + 1)] 1-2 =

= -2 / [(X ² + 1) √ (x ² + 5)] + 1 / [(x ² + 1) √ (x ² + 2)]

Then, integrating wrt x

[1,2] ∫ (-2 / [(x ² + 1) √ (x ² + 5)] + 1 [(x ² + 1) √ (x ² + 2)]) dx =

= - Atan (2x / √ (x ² + 5)) | [1, 2] + atan (x / √ (x ² + 2)) | [1, 2] =

= - Atan (4 / 3) + atan (2 / √ 6) + atan (2 / √ 6) - atan (1 / √ 3) =

= - Atan (4 / 3) + 2 atan (√ 6 / 3) - atan (√ 3 / 3) =

= - Atan (4 / 3) + 2 atan (√ 6 / 3) - pi / 6 = - 0081

If you use your hand, you can follow the steps on how to get the result of substitution, please refer. Give me a message.

Hope this helps:) Good luck!

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