Finding Electric Flux through a Cube? - electric flux cube
Can someone please help me with this?
http://img91.imageshack.us/img91/1149/physuo5.jpg
That's the problem ... I have to find the electric flux through the cube.
Wednesday, February 17, 2010
Electric Flux Cube Finding Electric Flux Through A Cube?
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You need the component (normal) on the floor on both sides of the cube to integrate. The normal component is the product of E-field vector and the normal area vector dA. You need to find the vector of the DA for each surface. For the surface in the xy-plane dA = dx * dy * k (k is the unit vector in z-direction). The scalar product is then E * DA [(a + b) + c * i * j] * dx * dy * k, which also must be zero, since the vector field is not k (Y-direction) has components. Only areas that the components in the directions X and Y, to help the movement in this area ..
Now look at the surface in the plane XZ. Here dA = d * d * D and E with the scalar [(a + b) + c * i * j] * dx * dy * j = c * dx * dy. Now take the integral over the surface ∫ dxdy = c * c * x * y evaluated at x 0 to L and Y from 0 to L, TThe answer is: c ^ 2 * L. It flows through this area. You will find all the others in a similar way (although the sign of matter to mind, but goes to the surface, and the output of the other). Note that the surface that is parallel, but at y = L, the flow will be the same as the field strength is not a function of y. However, the flux from the surface, thereby controlling the movement of surfaces in the XZ plane.
Finally, the surface in the YZ plane. The area element dA dy * dz * i and the scalar [(a + b) + c * i * j] * dy * dz * I = (a + b) * dy * dz at x = 0 the value of ∫ dy * d * is * L ^ 2 and x = L, the value of ∫ (a + bL) * dy * dz * is an L ^ 2 + b ^ 3 * L. The net flow is the difference, or b * L ^ 3
Once you have the whole river, are linked to costs, that the valueOnce e0 (permittivity).
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