# Is the strength of an electric field the same throughout a conductor connected to a battery?

Is an electric field the same all over a conductor?

As electrons move further in a conductor they start coming from high potential to low when moved from high potential end of conductor(cathode) to low potential end of conductor(anode,) so the potential difference between points keeps on decreasing thus causing a decrease in strength of electric field.

If this is the case then the current should be different through out a conductor (different strength of electric field between different points in conductor thus different accelaration.)

Where am I wrong?

The red dot is an electron. The conductor is connected to a10V battery. Section A is the midpoint of the conductor, so the voltage between section A and C should be 5V while between B and C is 10V so according to it the electric field between A and C should be weaker than the field between B and C so accearation should be lower between A and C than B and C, so doesn't it change the current at different points in the conductor as electrons at point B experience a higher accelaration thus more current than electrons at A?

• Electric field tries very hard to be zero everywhere in the interior of any good conductor. If there's no other element in your circuit capable of dropping the voltage, then you are worried about a non-ideal conductor and you will need details on the specific ways your conductor isn't ideal in order to do any analysis. Commented Oct 21, 2021 at 16:34
• A realistic real-world possibility is that most of the resistance (and therefore the voltage drop and electric field strength) occurs in the connection between the battery and the conductor. But certainly not the only case you could study. Commented Oct 21, 2021 at 16:37
• I think you Don't get what I am actually asking, here conductor is connected to battery of 10 V , please read my question carefully sir Commented Oct 21, 2021 at 16:43
• Electric field is Voltage difference divided by the distance between the two point (assuming the medium is homogeneous). There is twice the voltage for twice the distance and so the same electrical field. Commented Oct 21, 2021 at 17:20
• The electric field or e&m wave actually "travels" mostly through the insulator or free space around the conductor, not the conductor itself besides a very shallow depth. Commented Oct 21, 2021 at 18:23

For a conservative electric field, $$\E\$$ is equal to the (negative of the) gradient of the potential. That is:

$$E = -\nabla V$$

If the electric field is also uniform, then in the direction of the electric field, $$\E\$$ is equal to the change in potential per unit change in distance.

$$E = -\frac{\Delta V}{\Delta x}$$

If a wire has constant width and uniform resistivity, then the above equation will apply.

So, although the potential will drop along the length of the wire, the electric field will be uniform.

Note that the potential drop along a length of wire is given by

$$V = IR$$

The resistance of normal conductors is "low" and is often treated as negligible. When the resistance is treated as negligible, the voltage drop across the length of wire is negligible as well, hence the oft repeated mantra that the electric field inside a wire is 0. (If the current is non-zero, and the resistance is non-zero, then the electric field will be non-zero. It is just that for many purposes, we can ignore it.)

• Does it same means as @G.Bergeron ? So electric field at any point in conductor = - V / D (v is potential difference between two points and D is distance between 2 point) ??? Right? Therefore uniform in a conductor ? And therefore as length of conductor increase current become less as -V/D , in that at uniform V , if D increase cause weak electric field thus low current? Am I right sir? Here in your answer "x" is distance between 2 points?? Commented Oct 22, 2021 at 9:21
• "x" in my answer is the distance between two points. It is the same as "D" in your comment. As the length of a wire increases, its total resistance will increase, and if the battery voltage is constant, the total current will decrease. However, within a given wire, the current is the same at different points on the wire. G. Bergeron's comment is correct. Commented Oct 22, 2021 at 10:56
• OK sir I got it just one more normal doubt! I recently thinking about ur answer ...I got stuck at one point help Me out , a resistor say Length or distance between its end is 'X' , and I make a cross section just its middle distance say 'X/2' . Now I apply some calculation but I got stuck. Let say that resistor is of 10 ohm and battery of 10 V (current=1A from ohms law) now I calculated voltage drop BETWEEN 'X/2' and resistor end connected to anode. So obviously the voltage drop between that 2 points is 5V (half of voltage drop across 'X' that is 10V ) further more... Commented Oct 22, 2021 at 12:06
• Remember!! battery is still connected to resistor 'X' length I am just theoretically trying to calculate the voltage drop between half of resistors length . Moving on, I apply ohms law (I think it is where i am wrong) V=RI, I took Voltage drop =5V across those point (half of 10V across X ) and current is same through out conductor as you know so I = 1A but just calculated and I got R=5 ohm (how it is possible that resistor is of 10 Ohm, i haven't change its length nothing) Where I am wrong?? I think ohms law is applied only to know voltage drop 'ACROSS' a resistor not between in it??!!! Yes? Commented Oct 22, 2021 at 12:11
• If the resistor is 10 ohms, then the resistance of half the resistor is 5 ohms. Commented Oct 22, 2021 at 12:13

Current is by definition the flow of electrons. If you had different currents in different parts of a circuit then you would quickly get an excess or a deficit of electrons in one part, and this would change the voltages around the circuit, so the currents would quickly equalise again. In reality this does happen but the voltage differences are usually tiny.