📂Electrodynamics

Electric Field Curl

정리

Electric Field’s Curl is always $\mathbf{0}$.

$$\nabla \times \mathbf{E} = \mathbf{0}$$

Proof¹

We will derive a general result from the special case where a point charge is located at the origin. The electric field due to a point charge at a distance of $r$ from the origin is as follows.

$$\mathbf{E}=\dfrac{1}{4 \pi \epsilon_{0} } \dfrac{q}{r^2} \hat{\mathbf{r}}$$

If we perform a path integral of the electric field from point $\mathbf{a}$ to point $\mathbf{b}$ in a spherical coordinate system, we obtain the following.

$$\begin{align*} \int_\mathbf{a} ^\mathbf{b} \mathbf{E} \cdot d\mathbf{l} =&\ \int_\mathbf{a}^\mathbf{b} \left( \dfrac{1}{4 \pi \epsilon_{0}} \dfrac{q}{r^2} \hat{\mathbf{r}} \right) \cdot \left( dr \hat{\mathbf{r}} + rd\theta\hat{\boldsymbol{\theta}} + r\sin\theta d\phi \hat{\boldsymbol{\phi}} \right) \\ =&\ \int_\mathbf{a}^\mathbf{b} \dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}dr \\ =&\ \dfrac{q}{4 \pi \epsilon_{0}} \int_\mathbf{a}^\mathbf{b} \dfrac{1}{r^2} dr = \dfrac{q}{4 \pi \epsilon_{0}} \left[ -\dfrac{1}{r} \right]_{r_{a}}^{r_{b}} \\ =&\ \dfrac{q}{4 \pi \epsilon_{0}} \left( \dfrac{1}{r_{a}}-\dfrac{1}{r_{b}} \right) \end{align*}$$

Here, $r_{a}$ and $r_{b}$ are the distances from the origin to point $\mathbf{a}$ and point $\mathbf{b}$, respectively. As can be seen from the result of the above integration, the integral over a closed path is $0$.

$$\oint \mathbf{E} \cdot d \mathbf{l} = 0$$

Stokes’ Theorem

$$\int_{\mathcal{S}} \left( \nabla \times \mathbf{v} \right) \cdot d\mathbf{a} = \oint _{\mathcal{P} }\mathbf{v} \cdot d \mathbf{l}$$

Using Stokes’ Theorem

$$\int \left( \nabla \times \mathbf{E} \right) \cdot d\mathbf{a} =\oint \mathbf{E} \cdot d\mathbf{l}=0$$

therefore, it can be seen that it must be $\nabla \times \mathbf{E} = \mathbf{0}$. Since the integral over any arbitrary area must yield $\mathbf{0}$, it can only be $\nabla \times \mathbf{E} = \mathbf{0}$.

The electric field for several point charges is the same as adding up the electric fields for each point charge. For continuously distributed charges, only change $\sum$ to $\int$. Therefore, it is $\mathbf{E}=\mathbf{E}_{1} + \mathbf{E}_2+\mathbf{3}+\cdots$, and since the curl of each electric field is $\mathbf{0}$, their sum is naturally $\mathbf{0}$.

$$\begin{align*} \nabla \times \mathbf{E} =&\ \nabla \times (\mathbf{E}_{1} + \mathbf{E}_2+\mathbf{3}+\cdots ) \\ =&\ (\nabla \times \mathbf{E}_{1}) +(\nabla \times \mathbf{E}_2 )+(\nabla \times \mathbf{E}_{3})+\cdots \\ =&\ \mathbf{0} \end{align*}$$

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