Imagining in space, there are two parallel conductors of different voltages that are say, 10 million km in length each, and are 'grounded' at one end, and connected in some other configuration you don't know about at the other. If you were to come to the middle of this setup (in your space suit that also happens to be very conductive), and place one hand on each, would current flow through you?

I'm having trouble figuring out when current does and does not flow. Would it take time for the circuit to 'realise' it could flow through you? (In a speed of information sense). If you had two parallel conductors that aren't connected at either end, but have a voltage difference between them, and you connect them, does it try to even out the charge and thus current flow through you? Like if you have two differently charged spheres that touch?

I have the same question about the parallel conductors except that they are 'infinitely' long, from your perspective, and that they may or may not be connected at both ends, one, or not at all.

Any help would be greatly appreciated.


1 Answer 1


It may hep if you change from long wires to large conductive sheets. That is, change to a parallel plate capacitor. Let us pick a capacitor where the plates are large, but not infinite. Parallel wires are also a capacitor, but paralellel plates may be a more familiar example.

When the two plates are at the same voltage, the capacitor is discharged. That is, there is no charge on either plate.

If you move some - charge from one plate to the other, one will have a negative charge and the other positive. If you touched both plates, charge would flow until the plates were both neutral.

It helps to understand what voltage is.

$$Volts = \frac{Joules}{Coulomb} = \frac{energy}{charge}$$

When you move - charge from one plate to the other, you leave behind + nuclei. Electrons and nuclei attract each other. It takes energy to separate them. You are also squeezing - charges together on the negative plate. It takes energy to force them together. It is like compressing springs.

Suppose you have a charged capacitor, with charge q and energy E.

$$V_1 = \frac{E_1}{q_1}$$

To double the charge, you have to supply 4 times the energy.

$$V_2 = \frac{E_2}{q_2} = \frac{4E_1}{2q_1} = 2V_1$$

So voltage goes up as you add charge.


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