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While I was reading 'symmetry' from wikipedia, then I came to this statement:

...For example, an electric field due to a wire is said to exhibit cylindrical symmetry, because the electric field strength at a given distance r from the electrically charged wire of infinite length will have the same magnitude at each point on the surface of a cylinder (whose axis is the wire) with radius r. Rotating the wire about its own axis does not change its position or charge density, hence it will preserve the field. The field strength at a rotated position is the same. Suppose some configuration of charges (may be non-stationary) produce an electric field in some direction, then rotating the configuration of the charges (without disturbing the internal dynamics that produces the particular field) will lead to a net rotation of the direction of the electric field. These two properties are interconnected through the more general property that rotating any system of charges causes a corresponding rotation of the electric field.

Saw the statement? What is the proof afterall behind this statement? Is it applicable to translation of charges also??

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Rotation is a relative quantity. If you rotate by an angle $\theta$ relative to me that means I rotate by an angle $-\theta$ relative to you.

So rotating a system of charges by an angle $\theta$ is exactly the same as leaving the charges stationary and rotating yourself by an angle $-\theta$. If you leave the charges stationary then obviously the electric field also remains stationary, so if you rotate an angle $-\theta$ relative to the charges you must also rotate an angle $-\theta$ relative to the field they generate. Finally we note that if you rotate an angle $-\theta$ relative to the field then the field rotates an angle $\theta$ relative to you.

The conclusion is that if the charges are rotated by an angle $\theta$ then the field is also rotated by the same angle $\theta$, just as the Wikipedia article says.

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Look at the connection between the electric field and the charge distribution:

$$\nabla \vec{E}(\vec{r}) = 4\pi\rho(\vec{r}).$$

If you rotate the charge distribution with some $R\in SO(3)$, i.e. $\rho(\vec{r})\longrightarrow \rho(R^{-1} \vec{r})$ and the charge distribution stays the same, i.e. $\rho(R^{-1} \vec{r})=\rho(\vec{r})$, then for the electric field $\vec{E}{\,}^\prime$due to the rotated charge distribution, one finds

$$\nabla \vec{E}{\,}^\prime(\vec{r}) = 4\pi\rho(R^{-1} \vec{r})=4\pi\rho(\vec{r})=\nabla \vec{E}(\vec{r}),$$

so if the charge distribution is invariant under the operation, so is the electric field.

Now if you translate, i.e. $\rho(\vec{r})\longrightarrow \rho(\vec{r}-\vec{a})$, you get (with $\nabla\vec{E}(\vec{r})=4\pi\rho(\vec{r})$)

$$\nabla\vec{E}{\,}^\prime (\vec{r}) = 4\pi\rho(\vec{r}-\vec{a})$$ $$\Rightarrow \nabla_{\vec{r}+\vec{a}}\vec{E}{\,}^\prime (\vec{r}+\vec{a})=4\pi\rho(\vec{r})$$ $$\Rightarrow \vec{E}{\,}^\prime(\vec{r})=\vec{E}(\vec{r}-\vec{a}),$$

where the last step is to be consumed with caution. Still, you see that the field $\vec{E}{\,}^\prime$ generated by the translated charge density is in fact the translated field $\vec{E}$.

The same applies to rotations if the charge density is not necessarily invariant, it's a completely analogous computation.

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