1
$\begingroup$

Always when I study displacement Current it is zero outside the capacitor because the electric field is zero outside

For example this photo

enter image description here

Why this electric field on the surface s one is zero I wondering why is that .charges move in the circuit because of electric field

$\endgroup$
2
$\begingroup$

Always when I study displacement Current it is zero outside the capacitor because the electric field is zero outside

That is "mostly true". The field created by a charged capacitor is mostly contained between the plates of the capacitor. However there are "fringing" field lines, and a very small amount of field will go from the outside of one plate to the outside of the other.

Why this electric field on the surface s one is zero I wondering why is that .charges move in the circuit because of electric field

The electric field through the surface s is near 0, but not exactly. In particular, if there is current flowing through the wire, then there is an electric field corresponding to the microscopic version of Ohm's Law.

$$\vec{J} = \sigma\vec{E}$$

Where $\vec{J}$ is the current density, $\sigma$ is the conductivity of the wire material, and $\vec{E}$ is the electric field.

$\endgroup$
12
  • 1
    $\begingroup$ The OP asks a good question, and this answer is correct. It's not hard to show that the field in the wire is typically several orders of magnitude lower than the field in a capacitor. Why the textbooks gloss over this is a mystery. $\endgroup$ – garyp May 31 at 13:46
  • $\begingroup$ Also i forgot to write that the electric field is zero by assuming that the wire is a perfect conductor $\endgroup$ – homam hassn May 31 at 13:47
  • 1
    $\begingroup$ If the wire is a "perfect conductor" (but not a super-conductor) then the electric field internal to the wire will be 0 (again using microscopic Ohm's law). But there will still be a small field outside the wire due to "fringing". It will be small, but not exactly 0. $\endgroup$ – Math Keeps Me Busy May 31 at 13:58
  • $\begingroup$ Is there any electric field outside due to the wire $\endgroup$ – homam hassn May 31 at 14:42
  • 1
    $\begingroup$ @garyp. An ideal conductor is usually taken to mean a model of a resistive conductor as the resistivity goes to zero. (In this sense, a superconductor is not an ideal conductor.) For any uniform resistive conductor with no current, ohm's law gives a 0 electric field. Taking the limit of a zero electric field as resistivity goes to zero is still a zero electric field. Likewise, in a uniform resistive conductor, assuming ohmic linearity, for a given current density, E is proportional to resistivity. Taking the limit as resistivity approaches 0 gives E=0. You may have different idea of ideal wire $\endgroup$ – Math Keeps Me Busy May 31 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.