## Conceptual doubts about capacitors and dipole antennas

I have some conceptual questions regarding capacitors and dipole antennas.

Consider a parallel plate capacitor made with two conducting plates, initially uncharged. If I connect each plate to a different terminal of the battery, charge will start to flow from one plate to another.

Consider also a dipole antenna that consists of two conducting rods connected to both terminals of an AC (alternating current) supply. In this case, current will also flow from one rod to the other one through the battery, and both rods will become charged with charges of opposite sign, but, since the power supply supplies AC, the signs of each rod will alternate in time.

My questions are, basically:

1) Is there a connection between the charging up of a capacitor and the dipole antenna? It seems so, because both consist of a power supply that, when connected two conducting objects, charges up both conductors with equal charges of opposite sign.
2) Why exactly does charge flow from a plate (in the case of the capacitor) or rod (in the case of the dipole antenna) to the other one? For the case of capacitors, the book that I'm reading only mentions that it is because "the plates are equipotential", but I'm not sure why this would force charges to flow from one plate to the other. For the case of antennas, the book doesn't give any further clarification.
3) What is special about a capacitor that makes it different from an open circuit? Suppose I connect two copper wires to the terminals of a battery, but I don't close the circuit. Will both wires get equal charges of opposite polarities, like a capacitor?

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Mentor
A dipole antenna is like a capacitor with a very small capacitance.

 2) Why exactly does charge flow from a plate (in the case of the capacitor) or rod (in the case of the dipole antenna) to the other one?
You change the potential on those plates/antennas. You create a potential difference, which attracts charges.

 3) What is special about a capacitor that makes it different from an open circuit?
The larger capacitance (can be several orders of magnitude).
 To further discuss the connections between the two: Also note that the simple dipole antenna (approx. $L < \frac{\lambda}{10}$ where L is the length of the antenna) constructed with an open-circuited wire with battery at the center actually produces a triangular current distribution. If the dipole axis is in the 'z' direction, the formula for the current would be: $$I(z) = I_{o} (1 - \frac{2 |z|}{L})$$ To get the constant current often used in short dipole calculations in many texts what you would actually want to do would be to construct an antenna with these capacitor end caps in order to produce that constant current configuration. This slight distinction of antenna is covered in entry level E.E. books, but I don't recall it in Griffith's Electrodynamics (the more or less comparable physics text on the subject.) I think the resultant fields of the open-circuited versus capacitor nuance is only different by like a factor or two or something (if I remember correctly), but I didn't check.