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.
Best Answer
Correct.
OK.
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.