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?
Best Answer
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).