# Does the electric field strength along a simple circuit depend on the resistance of the conducting wire along its length?

A few hours ago I asked, if the electric field produced along and through a wire connected across the ends of a battery (a shorted out circuit) was uniform. The answer I was given was yes. But what if the wire doesn't have a uniform composition and therefore has different values of resistance along its length? Would the following calculations show anything?

$$V=IR → -\frac{dV}{dl}=E=-\frac{d}{dl} (IR)=-I\frac{dR}{dl}$$

Doesn't the above calculations imply that the electric field strength along a wire depends on the rate of change of resistance? If so, how is it possible that the battery produces different electric field strengths at different areas along the wire (on an atomic level)?

• A battery does not produce electric field; a battery produces an arbitrary amount of current between two fixed points (battery terminals) so that its rate of working (delivered electric power) is directly proportional to the produced current and the proportionality factor being nearly independent of that current. – hyportnex Apr 5 '18 at 18:01
• Your equations have a slight flaw. - d IR / dl is equal to - IdR/dl - RdI/dl. Both these terms add up to give zero (as IR or E is constant). That's because Electric field is determined by the potential difference. As long as that doesn't change, electric field remains constant. (sorry for not using latex, on mobile it is a slight pain) – Shreyansh Darshan Apr 5 '18 at 19:39

## 2 Answers

Rather than having a wire with variable resistance, simply imagine a circuit of several resistors connected in series. This is a simple model for such a variable resistance wire. Here it is clear that the voltage drop across the resistors will not be equal if the resistors do not all have equal resistance. The voltage drop across the battery, $\varepsilon$, is equal to the total voltage drop across all resistors: $\varepsilon = V_1+V_2+V_3...=IR_1+IR_2+IR_3+...$

Hopefully this concept should be familiar to you. The voltage drop across resistors of different resistance will be variable, and so the electric field across those resistors will also be variable. The battery only ensures that there is a constant voltage $\varepsilon$ across from its anode to its cathode.

how is it possible that the battery produces different electric field strengths at different areas along the wire (on an atomic level)?

Imagine first of all we have a simple wire with two regions: the first of relatively low resistance, and the second with relatively high resistance.

Now imagine that unlike what actually happens, all the charges are uniformly distributed, so the electric field is constant throughout the wire.

But since the resistance is not the same, the charges in the wire react differently to this (temporarily uniform) field. The charges in the low resistance section accelerate, and the charges in high resistance section decelerate. This causes charges to bunch up at the boundary. This bunching of charges changes the field strength there. In fact, the bunching continues until it creates a field that "pushes" the charge through both sections at the same speed. Once that's true, charges arrive and depart the "bunched" section with the same rate, and the charge distribution is stable.