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Why is the voltage distribution a quarter wavelength out of phase with the current in a half-wavelength dipole antenna, as show in this image?

It makes sense that in a graph over time, the voltage at the ends of the antenna is a quarter wavelength behind the current measured at the feed. But voltage distribution is over x; isn't the voltage in phase with current over distance?

I was under the impression that at the feed point, current is in phase with voltage. What makes it go 90° out of phase in an antenna?

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OK. I think I've got the answer. For anybody who stumbles upon this question and has the same question: Current is proportional to direction and concentration of charge, whereas Voltage is only proportional to concentration of charge.

So when the electrons "bounce" off the end of the end of the antenna (the wave is reflected), the NEGATIVE of the wave is reflected for current (since the direction reverses!), while the POSITIVE of the wave is reflected for the voltage. For example: if the wave is cos(wt-kx), the reflected traveling wave is -cos(wt+kx) for current, while the reflected wave for voltage is +cos(wt+kx).) The resulting standing waves formed are one with peaks at the ends for voltage, while the current's resulting standing wave has its peak at the center and nodes at the ends.

The graph for current, cos(wt-kx) - cos(wt+kx) will be 2sin(wt)sin(t). This is a standing wave with nodes at the ends of the antenna. However, the graph for voltage, cos(wt-kx)+cos(wt+kx) will be 2cos(wt)cos(kx). This is a standing wave with the node at the center of the antenna and peaks at the ends (90 degrees shifted from the current standing wave).

Of course there are more complex solutions (such as solving maxwell's equations and whatnot) but this is the best answer I could come up with to describe the physical system itself.

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