For a uniform (constant) electric field, we have the relation $E = - \Delta V/\Delta r$. Now, if the electric field provided by a battery is constant over a constant potential difference and if we calculate the field between two points on a wire taking the same value of $\Delta V$ (as of battery), the electric field will increase as we reduce the distance between the points on the wire, which contradicts the field being constant throughout the wire? Please explain.


  • $\begingroup$ $\Delta V$ is between the battery terminals rather than between two arbitrary points of the wire. $\endgroup$ Nov 25, 2021 at 9:27
  • $\begingroup$ @RogerVadim Check the image please. $\endgroup$
    – Spencer
    Nov 25, 2021 at 9:47
  • $\begingroup$ As Roger said, if your battery outputs 5V for example, that doesn't mean that between any two points on a wire $\Delta V = 5V$, but it's actually $\Delta V = 5V * l/L$ where l is the length of the segment between two selected points, and L is the total length of the wire (assuming uniform resistivity of the wire). EDIT : Hence it follows that your electric field is 5V/L, i.e. constant throughout the wire. $\endgroup$ Nov 25, 2021 at 9:59
  • $\begingroup$ The confusion is that you use the symbol V to mean the battery voltage at the same time as the voltage drop over any length of wire or element of the circuit. Also when you say 'wire' you really mean resistor. $E=\sigma J$ so unless you change the current or the conductivity it remains constant, independent of the length considered. $\endgroup$
    – my2cts
    Nov 25, 2021 at 10:16
  • $\begingroup$ @my2cts Means (potential drop across any resistor) divided by (length of that resistor) is always constant and is equal to the original electric field produced by the voltage source ?? $\endgroup$
    – Spencer
    Nov 25, 2021 at 10:23

1 Answer 1


if we calculate the field between two point on a wire taking the same value of ΔV (as of battery)

You cannot choose to take the potential between two points of a wire. It can be however be calculated if one knows the resistance and the current flowing through the two points. So if a current $i$ passes through the wire and the two points under consideration have distance $l$ with resistance between them as $R_l$ then the potential difference between the points is $iR_l$. If $\rho$ is the resistivity and $A$ is the cross-sectional area then $$R_l=\frac{\rho l}A$$ and consequently the electric field between the points is $$E=\frac{iR_l}{l}=\frac{i\rho}A=constant$$

Edit: As mentioned by @jensen paull resistance does not determine potential difference. I wrongly stated that it did and I fixed it in my edit.

  • $\begingroup$ Thank you so much! Now I completely get it $\endgroup$
    – Spencer
    Nov 25, 2021 at 10:53
  • $\begingroup$ Potential difference between a wire is not determined via resistance. You CAN take the potential difference between 2 points in the wire using ANY PATH. this is due to then fact that E is CONSERVATIVE and therefore PATH INDEPENDANT obviously finding E with this inside the wire is no good, if the path I chose isn't actually in the wire. This answer using Ohms law isn't correct per say it is complete circular reasoning. the microscopic ohms law says that current density is proportional to the electric field. to get to the form V = IR you have to assume that E is constant along the wire. $\endgroup$ Nov 25, 2021 at 11:07
  • $\begingroup$ So I'd untick this answer. The REAL answer is due to surface chargew being induced when there's an electric field inside wire , these induced surface charges then move to make the field equal. hard to explain in the comments so search it up . obviously in the presence of no surface charges then E field is OBVIOUSLY a function of distance. $\endgroup$ Nov 25, 2021 at 11:09
  • $\begingroup$ and Resistance doesnt inherently determine potential difference, Resistance along with current does, as this equation states the potential difference needed to Maintain a current under a Resistance. $\endgroup$ Nov 25, 2021 at 11:10
  • $\begingroup$ electronics.stackexchange.com/questions/532541/… $\endgroup$ Nov 25, 2021 at 11:13

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