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I have a point charge and a Gaussian sphere in a spherical coordinate system. The Gaussian sphere is centered at the origin. I need to verify the integral form of Gauss' law for a Gaussian sphere centered at the origin at the moment my point charge moving in the $\hat z$ direction with constant velocity $\vec v$ is also at the origin. I know that the electric field for a moving charge is different than a stationary one, however I don't see how this is relevant if I look at a snapshot of the point charge when it's at the origin along with the Gaussian sphere. The problem doesn't state it but I'm assuming that I'm not in the same reference frame as the charge.

Wouldn't the electric field at that specific moment in time be the same as it would be for a stationary charge? Or is the fact that it's moving before and after that moment change it's electric field no matter what time I look at it?

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Let me first answer this by making an analogy. Think of the sattelite images of hurricanes that you see on TV or on websites like hurricanes.gov. Here is an example. When a hurricane is moving, it tends to elongate a bit in the direction of motion (think of a comet tail), while when a hurricane stalls, it tends to be more circular.

Now imagine the weather person on TV pauses the image of a moving hurricane. Does the area effected by strong winds suddenly shrink to a circle? No! It stays whatever shape it was as it was moving. By taking a freeze frame, we do not get to change the physics in any way. (After all, what is a movie but a series of freeze frames!)

So, in this problem you need to treat the moving charge like it is moving charge at every instant in time. The charge has not stopped. You simply took a picture of a single instant in its motion.

HOWEVER, it is important to point out that you haven't mentioned whether the velocity in this case is relativistic. If not (i.e. $v \ll c$), then the electric field of the moving charge will look identical to that of a stationary charge because, e.g.: $$ E_y' = \frac{E_y}{\sqrt{1-(v/c)^2}}=E_y, $$ and $$ E_x' = \frac{E_x}{\sqrt{1-(v/c)^2}}=E_x, $$ since $(v/c)^2$ will be very close to $0$.

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