# Why the displacement current is zero outside the capacitor?

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

For example this photo

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

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.

• 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. – garyp May 31 at 13:46
• Also i forgot to write that the electric field is zero by assuming that the wire is a perfect conductor – homam hassn May 31 at 13:47
• 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. – Math Keeps Me Busy May 31 at 13:58
• Is there any electric field outside due to the wire – homam hassn May 31 at 14:42
• @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 – Math Keeps Me Busy May 31 at 22:24