[[184_notes:eflux_curved|Return to Electric Flux through Curved Surfaces notes]]
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=====Example: Flux from a Dipole=====
=====Example: Flux from a Dipole=====
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Suppose you have a two charges, one with value 5μC, the other with value 5μC. There are at separate locations, a distance 1 m apart, and they can be modeled as a dipole. What is the flux through a cylinder with radius 4 m and length 16 m that encloses this dipole?
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Suppose you have a two charges, one with value 5μC, the other with value $-5 \mu\text{C}.Thereareatseparatelocations,adistance1 \text{ m}apart,andtheycanbemodeledasadipole.Whatisthefluxthroughacylinderwithradius4 \text{ m}andlength16 \text{ m}$ that encloses this dipole?
===Facts===
===Facts===
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===Representations===
===Representations===
* We represent the situation with the following diagram.
* We represent the situation with the following diagram.
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{{ 184_notes:5_dipole_cylinder.png?300 |Dipole and Gaussian cylinder}}
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[{{ 184_notes:5_dipole_cylinder.png?300 |Dipole and Gaussian cylinder}}]
====Solution====
====Solution====
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Before we dive into calculations, let's consider how we can simplify the problem. Think about the nature of the electric field due to a point charge, and of the d→Avector for a spherical shell. The magnitude of the electric field will be constant along the surface of a given sphere, since the surface is a constant distance away from the point charge. Further, →E will always be parallel to d→A on these spherical shells, since both are directed along the radial direction from the point charge. A more in-depth discussion of these symmetries can be found in the notes of [[184_notes:eflux_curved#Making_Use_of_Symmetry|using symmetry]] to simplify our flux calculation.
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First, notice that we probably do not want to do any calculations here, since the it will not be fun to take a dot-product of the dipole's electric field and the area-vector, and it will get very messy very quickly when we start integrating over the surface of the cylinder. Instead, we evaluate the situation more qualitatively. Consider the electric field vectors of the dipole near the surface of the cylinder:
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[{{ 184_notes:5_dipole_field_lines.png?300 |Dipole Electric Field Lines}}]
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Since $\vec{E}isconstantwithrespectto\text{d}\vec{A}(inthiscase,itissufficientthat\vec{E}isparallelto\text{d}\vec{A}$ and has constant magnitude), we can rewrite our flux representation:
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Φe=∫→E∙d→A=E∫dA
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We can rewrite E (scalar value representing →E as a magnitude, with a sign indicating direction along point charge's radial axis) since it is constant on the surface of a given spherical shell. We use the formula for the electric field from a point charge.
We can plug in values for q and r for each spherical shell, using what we listed in the facts. For the smaller shell, we have E=1.0⋅107 N/C. For the larger shell, we have E=2.5⋅106 N/C.
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To figure out the area integral, notice that the magnitude of the area-vector is just the area. This means that our integrand is the area occupied by dA. Since we are integrating this little piece over the entire shell, we end up with the area of the shell's surface:
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Notice that the vectors near the positive charge are leaving the cylinder, and the vectors near the negative charge are entering. Not only this, but they are mirror images of each other. Wherever an electric field vector points out of the cylinder on the right side, there is another electric field vector on the left that is pointing into the cylinder at the same angle. These mirror image vectors also have the same magnitude, though it is a little tougher to visualize.
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∫dA=A=4πr2
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The last expression, 4πr2, is just the surface area of a sphere. We can plug in values for r for each spherical shell, using what we listed in the facts. For the smaller shell, we have A=1.13⋅10−2 m2. For the larger shell, we have A=4.52⋅10−2 m2.
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Now, we bring it together to find electric flux, which after all our simplifications can be written as $\Phi_e=EA$. For our two shells:
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We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}−vectorsaremirrored,butopposite\left(\vec{E}_{left}=-\vec{E}_{right}\right)$. We tentatively write the equality,
We get the same answer for both shells! It turns out that the radius of the shell does not affect the electric flux for this example. You'll see later that electric flux through a closed surface depends //only// on the charge enclosed. The surface can be weirdly shaped, and we can always figure out the flux as long as we know the charge enclosed.
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Putting it together, we tentatively write:
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Φcylinder=Φleft+Φright=0
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We gain more confidence when we read the [[184_notes:gauss_ex|next section of notes]], where we define "Gauss' Law". This law states that the total flux through a close surface is the amount of charge enclosed divided by ϵ0, the [[https://en.wikipedia.org/wiki/Vacuum_permittivity|permittivity of free space]].
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Φtotal=∫→E⋅d→A=Qenclosedϵ0
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Since the total charge of the dipole is 0, then indeed the charge enclosed is 0, and we were correct with our reasoning about the electric field and flux above.