# Electric field inside a wire

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. • $\Delta V$ is between the battery terminals rather than between two arbitrary points of the wire. Nov 25, 2021 at 9:27
• 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. Nov 25, 2021 at 9:59
• 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. Nov 25, 2021 at 10:16
• @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 ?? Nov 25, 2021 at 10:23

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 $$E=\frac{iR_l}{l}=\frac{i\rho}A=constant$$