We know that the capacitance of a capacitor depends on the dielectric material in between the plates as $$C=\dfrac{K\epsilon_0 A}{d}\,.$$

But what if the electrode material is changed? For example suppose that the electrodes were built of Iron (Fe) and had a capacitance of $C$. Now I change it and make the electrodes with another material, say Copper (Cu). How will its capacitance change? What will be its new value?

Please do not assume that the capacitor is ideal. I am asking for a Real capacitor.

  • 1
    $\begingroup$ Vacuum is a dielectric. The material of the plates does not matter (as long as they conduct) as far as the dc capacitance goes. Only the dielectric. Now for ac capacitance it might make an impact... $\endgroup$
    – Jon Custer
    Nov 21, 2014 at 19:16
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    $\begingroup$ At DC the electrode material won't matter at all. Can you give more context? What frequency range and voltages do you care about? $\endgroup$
    – DanielSank
    Nov 21, 2014 at 19:27

1 Answer 1


Since many metals form a thin oxide layer of relatively high resistivity, they can be used to affect the dielectric properties of a capacitor made from that material.

For example, aluminium, tantalum and even copper all form oxide layers with high resistivity (compared to the respective metals) and high relative permittivity (dielectric constant). Since each of these oxides has a different dielectric constant and dielectric strength (withstand voltage), a capacitor made from one material would have different characteristics to a capacitor of similar geometry made from another material.

This effect is already used in practice in the fabrication of electrolyte capacitors, tantalum capacitors, where one or both of the plates is made from aluminium or tantalum in order to exploit the very high dielectric constants of the respective oxide.

aluminium electrolytic capacitor

It is also the basis of forming capacitors in semiconductor microchips as well as metal oxide semiconductor field effect transistors (MOSFETs).

More recent advances in fabrication of multilayer nanosheets has allowed the development of supercapacitors, made from very thin layers of copper oxide formed from heating the copper metal in a controlled atmosphere, resulting in extremely high dielectric constants and high devices with capacitance.

An ideal capacitor is a device which can store electrical charge indefinitely. The relationship between voltage $v(t)$ and current $i(t) through a capacitor is given by:


enter image description here

As you have already stated, the capacitance ($C$ in Farads, F) of a parallel plate capacitor is given by:

$$C=K\epsilon_0 \frac{A}{d}$$

where $K$ is the of the dielectric constant of the material (also called the relative permittivity), $\epsilon_0$ is the permittivity of a vacuum ($\epsilon_0 \approx 8.854 \times 10^{-12}F/m$ in free space), A is the area of overlap of the two plates [in $m^2$] and $d$ is the distance between the plates [in $m$].

So by utilising the fact that a very thin (small $d$) oxide layer with high dielectric constant (large $K$) can be formed on a metal plate used for a capacitor, capacitors with very large capacitances can be formed.

In practice, a 'real' capacitor a finite withstand voltage (dielectric strength, $D$), as well as some (non-zero) resistance due to connections, leads and plate material (total $R$), as well as resistance of the anodic oxide film ($r$) and inductance of the plate foils ($L$). A 'real' capacitor can therefore typically be represented as:

enter image description here

Practical values for L are typically very small at low frequencies (50Hz-1kHz) so inductance can sometimes be ignored, although not so at higher frequencies such as RF and microwaves. The dielectric strength (or withstand voltage) is represented by an ideal zenner diode with breakdown voltage $D$. That is, if the applied voltage exceeds $D$, the dielectric will 'break down' (conduct).

Some practical values for dielectrics are given below:

enter image description here

Since most metals are good electrical conductors, they have very low electrical resistivity), so the additional resistance offered by one metal over another when selecting the plate material of a capacitor usually has negligible affect on the capacity of the capacitor.

The reason for this is that the total resistance of a capacitor is the sum of the electrical resistance of the connections, leads and plates of the capacitor. The capacitance of the capacitor however is a function of the properties of the dielectric between the plates of the capacitor.

Tantalum and magnesium oxide capacitors are useful for their high capacity and small size, however, since they are typically fabricated from sintered metal powders, the effective series resistance of these capacitors can be significant.

  • $\begingroup$ I think you need two opposing Zener diodes back to back to represent the dielectric strength in your diagram. $\endgroup$ Dec 1, 2014 at 13:02
  • $\begingroup$ That's probably true, in general, but I've only shown one here to highlight the fact that metal oxide electrolytic capacitors are typically polar (require a DC bias) and will usually damage when reverse biased. $\endgroup$
    – theo
    Dec 1, 2014 at 16:36

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