There is no "low resistance path from the battery directly to ground" in your circuit.
Current is always "looking to go to ground", but it cannot get there through the inside of the battery. Battery construction techniques are widely varied, but they all consist of some form of internal barrier that prevents the internal mobile charges of one polarity from being able to reach the internal net charges of the opposite polarity. If that were possible, the battery would not contain any potential energy (the ability to do useful work).
Because the positive and negative charges in the battery can't recombine internally, the only way they can get to each other is to travel through the external circuit (path) you provide. Thus the charges are "forced" to do useful work in your external pathway because they are "looking to go to ground".
Current (in EE convention) will flow from the positive terminal. So the only way down is through your resistor chain. Interestingly enough, a circuit attached to "ground" (Earth or otherwise) will not necessarily see no lower voltages.
Current does not "stop" at ground. This is Kirchhoff's Voltage Law. In your schematic it would make no difference if you turned your battery upside down. The voltages (relative to "ground") would just now be negative.
Ok, so lets say we flip the battery...
So lets say we flip the battery and current flows from the bottom of
the battery, approaching the first resistor. Wouldn't current choose
the ground path over the first resistor?
It might be interesting (if not amusing) to think of the electrons as little people. Each person is motivated by two things:
- Getting away from other similar people that it finds annoying (other electrons)
- Hanging out with (going towards) people that it finds attractive (positive charges).
That process is why currents circulate -- (must!) flow in loops. Forget "ground" for a minute. I think the circuit and concepts make sense to you.
You're getting stuck on the concept of ground. The reason is that electrons and "ground" are unintuitive because it's a mixed metaphor. "ground" is supposed to represent the "lowest potential" in the system. It doesn't always. Nothing in physics demands that -- it's just terminology.
What it really represents in practice in a simple circuit like this one (and why we teach the concept of "ground" or better "reference") is that it represents the place the mobile charges would most like to be. In this case it's the negative terminal of your battery.
In electrical engineering, we are stuck with the convention that current is defined in terms of the flow of positive charges. Therefore, the positive charges would most like to be at "ground" (the most negative -- lowest -- point in the system). Actual electron flows are reversed from this convention, which is, I suspect, the source of your confusion.
But even that isn't the whole story since "ground" is just a reference potential and nothing forces it to be the absolute lowest potential in the circuit. Really it is just an arbitrary point where we say the potential is 0 Volts. Voltage is a differential measurement and cannot be defined without a point of comparison.
KVL can be more generally stated as: the electric field is a conservative vector field. This can be expressed by the path integral:
\begin{align}
\int_{p1}^{p2} \vec{E} \cdot d\vec{l} = V_{p2} - V_{p1}
\end{align}
I.E. the integral is "path independent", and the result is the same no matter what path you take (so long as you start at p1 and end at p2). This can be 0, but it doesn't have to be. If \$p1=p2\$ (taking a closed path),
\begin{align}
\oint \vec{E} \cdot d\vec{l} = 0
\end{align}
This is why KVL even applies: KVL states that in any closed loop, the voltage across it is 0. It doesn't mean that the voltage must be 0 between any two points on that loop. It doesn't matter if there is any conductive path between two points. Having an ideal wire is just an extra statement that two distinct points are at the same voltage.
Best Answer
Two wires connected to a voltage source of 10V (the polarity is of no importance here) have the same voltage over the whole length under the following conditions.
Think also of this.
If there was a current you would need also the resistance of the wire expressed in Ohm/m and the distance of the 2 mm spot from the source.
Remember that the expression of fieldstrengh is V/m
The field strengh depends on the voltage and distance. Therefore in the question we have 10V/2mm = 5000 V/m.