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If the surface charges propagate an electric field to cause the electrons around the bulb in the wire flow(which is why it will light up even if the circuit is as long as 3*10^8 km, if the battery and bulb are close although at lower intensity), then if the bulb is closer to the battery, this field will be stronger and the bulb at the end should go glow brighter, while if it is far away, we can say that the poynting vectors that arise directly from the battery towards the bulb will be almost 0 and at some point the bulb will reach a minimum brightness, assuming the circuit has no resistance? Or will it maintain a constant brightness. Also doesn't this mean that a broken circuit will still light up an LED at it's end?

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For your hypothetical case of 0 resistance circuit, and lets assume also no inductive of capacitive effects, then in the steady state I think the bulb will be just as bright no matter how far it is from the battery. However this is not at all realistic or representative of the real world. In reality resistive losses would build up over the length of your circuit, though the effect may be minimal, it would result in a less bright bulb further from the battery I believe.

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