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In the context of calculating electric flux, is there a difference between element of surface area versus vector element of surface area?

Thanks

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2 Answers 2

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The integral form of Gauss's law is $$\Phi_E=\int\mathbf E\cdot\text d\mathbf A$$ Where $\text d \mathbf A$ is a vector of magnitude $\text dA$ and direction $\hat n$ that is perpendicular to the surface of integration and points outward.

Therefore the difference between $\text dA$ and $\text d\mathbf A$ is that the latter has a direction associated with it. The vector is what you start out with in Gauss's law, but then you usually make some symmetry arguments in order to reduce it to the scalar form (usually by arguing that $\mathbf E$ and $\hat n$ are in either in the same direction or perpendicular at all points on the smartly chosen Gaussian surface so that $\mathbf E\cdot\text d\mathbf A=E\text dA$ or $\mathbf E\cdot\text d\mathbf A=0$ depending on the part of the surface you are looking at).

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The surface area element $dA$ represents a small area on the surface. $d\vec{A}$ represents a vector normal to $dA$ with magnitude $dA$.

Electric flux can be written two ways: $$\int \vec{E} \cdot d\vec{A} \ \ \ \ (1)$$ and $$\int \vec{E} \cdot \hat{n} dA \ \ \ \ (2) $$ These are different only in notation. In $(1)$, $d\vec{A}$ represents a vector of magnitude $dA$ pointing normally relative to the area element $dA$.

In $(2)$, $\hat{n}$ is a unit vector pointing normally relative to the the area element $dA$, but the dot product is scaled by the $dA$.

So you can think of it as $d\vec{A}=\hat{n}dA$.

When calculating the flux, the result is the same.

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