= Dielectric capacitor response
{title2=$C=\epsilon_r C_0$}
Compare two initially identical <parallel-plate capacitors> with plate <area> $A$, fixed gap $d$, <vacuum> <capacitance> $C_0=\epsilon_0A/d$, and initial <electric potential difference> $V_0$. Completely fill both gaps with the same uniform linear <dielectric> of static <relative permittivity> $\epsilon_r>1$, neglecting the <fringing field of a capacitor>. Keep one connected to an ideal voltage source and isolate the other. The geometric <capacitance> becomes $C=\epsilon_r C_0$ in both cases; the electrical constraint determines what else changes.<University Physics Volume 2 by Samuel J. Ling and others/8.4 Capacitor with a Dielectric>[§8.4, Figure 8.17 and equation (8.11): isolated capacitor, unchanged free plate charge and increased capacitance]{ref}
Write $Q_0=C_0V_0$, $E_0=V_0/d$ and $U_0=C_0V_0^2/2$. Applying $Q=CV$, $E=V/d$ and the <capacitor energy> formula gives the complete comparison:
$$
\begin{array}{c|cc}
&\text{fixed voltage}&\text{isolated, fixed charge}\\ \hline
C/C_0&\epsilon_r&\epsilon_r\\
Q/Q_0&\epsilon_r&1\\
V/V_0&1&1/\epsilon_r\\
E/E_0&1&1/\epsilon_r\\
U/U_0&\epsilon_r&1/\epsilon_r
\end{array}
$$
Thus the statement that a <dielectric> reduces the <electric field> needs the fixed-free-charge condition. With fixed <electric potential difference>, the voltage source supplies additional free plate <electric charge>, maintaining $E=V_0/d$ despite the new bound <electric charges>.
\Image[/dielectric-capacitor-voltage-and-charge-constraints.png]
{title=<Dielectric capacitor response> at fixed voltage and fixed charge}
{description=Original normalized comparison for the stipulated model $\epsilon_r=3$: every ratio is relative to the same initially charged <vacuum> <capacitor>. The dashed line marks the initial value. The bars show final values, not a time history or measured data. Plate geometry is unchanged and the <dielectric> completely fills the gap.}
For the stipulated $\epsilon_r=3$ example, the voltage source supplies $\Delta Q=2Q_0$. At constant $V_0$, its delivered electrical <work> is $V_0\Delta Q=4U_0$, whereas the increase in <capacitor energy> is $2U_0$. In an ideal reversible quasistatic insertion, the remaining $2U_0$ is mechanical <work> delivered by the assembly, as accounted for in <fixed-voltage capacitor energy balance>. For the isolated assembly, <capacitor energy> falls from $U_0$ to $U_0/3$, allowing $2U_0/3$ of mechanical <work> in the same ideal limit. Actual losses reduce recoverable <work>; energy increases in the connected case do not arise from the <dielectric> supplying free <energy>.
Back to article page
