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Suoppose I have a non uniform conductor which is kept in a uniform electric field maintaining a constant potential difference across its ends.Both electric field and current density are a properties of points in space.Since current density is defined as current per cross sectional area, inorder to maintain same current throught out the conductor the current density changes with cross sectional area.Now according to ohm's law J=sigma*E,as current density changes electric field should also changes with cross sectional area.How is this possible if electric field is a property of point in space and we maintain a uniform field across the conductor.How electric field changes with cross sectional area?

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Now according to Ohm's law $\mathbf J=\sigma\mathbf E$, as current density changes electric field should also change with cross sectional area. How is this possible if electric field is a property of point in space and we maintain a uniform field across the conductor.

We can't maintain uniform electric field across such a conductor, because its resistance is non-uniform.

How electric field changes with cross sectional area?

If you have cross section $A$ depending on distance $d$ from the origin as

$$A = A(d)$$

the resistance will depend on distance too:

$$\textrm dR(d) = \frac 1 \sigma \frac{\textrm d \ell}{A(d)}$$

where $\textrm d\ell$ is a linear element of the conductor and $\textrm dR(d)$ is its resistance.

Since

$$V = IR$$

on a linear element $\textrm d\ell$ there will be a drop of potential $\textrm d\varphi(d)$

$$\textrm d\varphi(d) = I \frac 1 \sigma \frac{\textrm d \ell}{A(d)}$$

Since

$$\int\limits_a^b E\,\textrm d\ell = V$$

we can write for the difference of potentials on an element $\textrm d\ell$

$$E\, \textrm d\ell = \textrm d\varphi(d)$$

Therefore

$$E = I \frac 1 {\sigma {A(d)}}$$

Hopefully this shows the intuition behind the electric field not being uniform.

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