Convert to a Differential equation
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Convert to a Differential equation
![**Begin with Gauss' law in integral form.**
\[
\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q}{\varepsilon_0}
\]
**Explanation:**
Gauss' Law in integral form relates the electric flux through a closed surface \( S \) to the charge \( Q \) enclosed by that surface. \( \mathbf{E} \) represents the electric field, \( d\mathbf{S} \) is a differential area on the closed surface \( S \), and \( \varepsilon_0 \) is the permittivity of free space. The left-hand side of the equation, \( \oint_S \mathbf{E} \cdot d\mathbf{S} \), symbolizes the integral of the electric field over the surface, indicating how much field flows through the surface.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F047a7e6a-f025-4b5b-ab83-4ffe14f69253%2F5131bd5d-25f2-44fc-9350-98a51adf7f0a%2Fthqloei_processed.png&w=3840&q=75)
Transcribed Image Text:**Begin with Gauss' law in integral form.**
\[
\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q}{\varepsilon_0}
\]
**Explanation:**
Gauss' Law in integral form relates the electric flux through a closed surface \( S \) to the charge \( Q \) enclosed by that surface. \( \mathbf{E} \) represents the electric field, \( d\mathbf{S} \) is a differential area on the closed surface \( S \), and \( \varepsilon_0 \) is the permittivity of free space. The left-hand side of the equation, \( \oint_S \mathbf{E} \cdot d\mathbf{S} \), symbolizes the integral of the electric field over the surface, indicating how much field flows through the surface.
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