I have seen the equation $V = \frac {V_0}{K}$, but also the equation $U=\frac{1}{2}CV^2$. The values of C and V increase in the same linear ratio with $K$ (because $C=KC_0$). However, as the energy is proportional to $C$ and $V^2$, the energy stored by the capacitor actually DECREASES with the employment of a dielectric.

Am I correct in this interpretation? Do I take it that merely knowing the capacitance is NOT enough to compute the energy stored - I must also know this about it's construction?

(I think this may explain the problem with one of my electronics projects in the past. I see nothing to prevent two of the same capacitors from having a different energy store!)

Is it appropriate to summarize anything else one should be worried about when substituting capacitors?

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    $\begingroup$ Where have you seen this equation $V = \frac{1}{2}CV^2$? $\endgroup$ Mar 29 '14 at 5:14
  • $\begingroup$ Equation changed to $U=\frac12 CV^2$. As Pallavi Roy points out, the 1st $V$ must be electrostatic energy. $\endgroup$ Aug 2 '18 at 17:48

First of all that's $U=\frac{1}{2}CV^2$. But there is also another relation : $U=\frac{1}{2C}Q^2$. And I think your confusion will be gone if you realize when to use which relation. If you have a capacitor with a constant voltage source connected across it, then you use the first relation. But when a capacitor is already charged and is not part of a circuit, then the total charge $Q$ is constant in this case. So you use the second relation there. Accordingly, the relation between energy and capacitance will be found(since you already know how the dielectric affects the voltage and capacitance).


The capacitance value of parallel-plate capacitor can be obtained by the formula $$ C = {x\epsilon[(N-1)A]\over d}$$ where $A$ is the area of the plates, $d$ is the distance between them, $\epsilon$ is the dielectric constant of permittivity, $x = 0.0885$ when $A$ and $d$ are in centimeters, and $N$ is the number of plates. The capacitance is of course dependent on the dielectric as the energy stored is dependent on how much the electric field is allowed to disperse through the material and affect the parallel plate. Here are some values of relative permittivity for different materials

Air or vacuum 1.0
Paper 2.0–6.0
Plastic 2.1–6.0
Mineral oil 2.2–2.3
Silicone oil 2.7–2.8
Quartz 3.8–4.4
Glass 4.8–8.0
Porcelain 5.1–5.9
Mica 5.4–8.7
Aluminum oxide 8.4
Tantalum pentoxide 26
Ceramic 12–400,000
  • $\begingroup$ Your values are off by a $4*\pi*10^{-7}$ factor. $\endgroup$
    – TZDZ
    Dec 22 '14 at 7:36
  • $\begingroup$ How did you come to that conclusion? $\endgroup$
    – AlanZ2223
    Dec 22 '14 at 13:46
  • 1
    $\begingroup$ Well it's just that the values you gave are the relative permittivities (classical mistake when calculating capacitances). $\endgroup$
    – TZDZ
    Dec 22 '14 at 14:06

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