Capacitor electron flow

Well I was just wondering, a capacitor is made of 2 metallic plates separated by insulating material. During the charging of the capacitor electrons flow towards the opposite direction the battery's electric field.

The electrons flow through the insulator at a very very slow speed causing some of the charge, which was supposed to be stored, to be lost?

More specifically the electrons from the metallic plate which accepts the electrons start to flow ( more slowly) and it shouldn't be negatively charged.

A capacitor is made of two metallic plates separated by insulating material.

Correct.

During the charging of the capacitor electrons flow towards the opposite direction the battery's electric field.

OK.

Shouldn't the electrons flow through the insulator at a very very slow speed so that means some of the charge, which was supposed to be stored, be lost?

In a perfect insulator there will be no charge flow (current) under static voltage conditions. For a less than perfect insulator there will be a leakage current and the capacitor will lose its charge.

Figure 1. Extract from a random electrolytic capacitor series datasheet.

The datasheet shows that for these large Vishay electrolytics that the leakage current, ILS, is measured after the rated voltage, UR, has been applied for 5 minutes. (We can take it from this that there is something that will change a little with "soakage" time.)

If we look at the first entry, a 10,000 μF, 16 V model the IL is listed as 1.2 mA. The charge on the capacitor is given by $$\ Q = C V = 0.01 \ \text F \times 16 \ \text V = 0.16 \ \text C\$$.

$$\ 1.2 \ \text {mA} = 0.0012\ \text {C/s} \$$ so the capacitor is leaking at a rate of $$\ \frac {0.0012}{0.16} = 0.0075 \ \text {/s} = 0.75\%\text{/s} \$$ while fully charged.

Just by using the UR and IL figures we can calculate the equivalent leakage resistance as $$\ R_L = \frac {U_R}{I_L} = \frac {16}{1.2m} = 13.3\ \text k\Omega \$$.

simulate this circuit – Schematic created using CircuitLab

Figure 2. Equivalent circuit at 16 V.

Correct. A perfect insulator would allow no electrons to pass through it, and charge would accumulate at the electrodes - an increase of electrons at the negative side. This produces an electric field across the insulator, which causes charged particles (mostly electrons, but possibly ions of the material too) to experience a force, and some do break free and allow a small leakage current to flow, discharging the capacitor over time.

Yes a capacitor is made of two conductive plates separated by a dielectric (or insulator). When you apply a DC voltage (or charge) the capacitor, the plate connected to the negative terminal of the battery builds up a negative charge by the accumulation of electrons while the plate attached to the positive terminal of the battery builds up a positive charge by losing electrons. The buildup of positive and negative charge on each plate creates an electric field in the dielectric which is how energy is stored. Yes a tiny amount of electrons flow through the insulator and this is called the leakage current. Any good capacitor datasheet should tell you how much to expect.

Maybe more modern concepts would help. The energy transfer from one plate of a capacitor to the other is caused by photon flow (not electron flow) through the capacitor dielectric, known as displacement current. Photons move at the speed of light in the medium but electron drift velocity is very slow. All speed of light electromagnetic activity involves photons.

• Deeper physics is good to know, but can sometimes get in the way -- the question seemed to be about leakage current, and that does involve electrons moving through the dielectric. – amI Apr 15 '19 at 2:57
• @aml If the question was about leakage current, I completely missed it. – Cecil - W5DXP Apr 15 '19 at 4:25