# Exactly how does an oscillating electric field produce an oscillating magnetic field?

Let's say we have a capacitor which is connected to a sinusoidal voltage source, that means that the electric field within the capacitor is a sinusoidal function(assuming that the capacitor is a parallel plate capacitor ), but how does this model generate any magnetic field? If it does, then doesn't that mean that light is generated within the capacitor?

• en.wikipedia.org/wiki/Displacement_current – Count Iblis Oct 17 '16 at 20:53
• @CountIblis So, doesn't that totally mean that light is produced inside a capacitor? – Shahe Ansar Oct 17 '16 at 20:55
• Yes, but that doesn't depend on that capacitor per se, all you need is the oscillating current in the wire. – Count Iblis Oct 17 '16 at 20:59
• @CountIblis But that's a different thing, since the light is produced from an intentional magnetic field – Shahe Ansar Oct 17 '16 at 21:00

## 2 Answers

There is a correction to the Ampère Law that says that a variation of the flux of the electric field in a surface produce a circulation of a magnetic field around the surface. This is translated to: $$\oint_{\partial A}B\,d\ell= \varepsilon_0\mu_0 \,\frac{d}{dt}\int\int_A \hat{n}_A \cdot \vec {E} \,dA$$

As the electric field oscilates, if you look at a surface parallel to the plates you would see a circulation of the magnetic field. This circulation oscilates in the same frequency. And there would be production of light, dissipating the energy delivered into the capacitor. To make this system stationary you need to provide this energy all the time. Is impossible for this light being a visble light though. The capacitor need to be very thin and oscilating very fast. To produce visible light you always new to oscilate small things, like excitations of electron-hole pair in semiconductors or vibrations of electrons in atoms and molecules.

According to the Maxwell equation $$curl{\vec H} = \vec j + \epsilon \frac{∂\vec E}{∂t}$$ An oscillating electric $\vec E∝ exp{(i\omega t)}$ field creates a displacement current density $j_D=\epsilon \frac{∂\vec E}{∂t}=i\omega\vec E$ that is equivalent to a current in creating a magnetic field that encircles the sum of convection and displacement current. Therefore, in a capacitor with an oscillating electric field producing a displacement current $j_D$, a magnetic field encircles the displacement current in the capacitor just like the corresponding conduction current in the wire leads to the capacitors. In a normal capacitor it is practically impossible to have field oscillations in the light frequency range. Therefore you will have no light generation. But you could have radio frequency wave generation. This is, e.g., the case in an oscillating dipole antenna which is similar to an open capacitance-inductance electric oscillator.

• Radio waves are still light – Shahe Ansar Oct 18 '16 at 18:41