From stored energy to voltage

Problem

Inverse: from stored energy to voltage. A capacitor with $C = 50\;\mu\text{F}$ holds energy $U = 1{,}25\;\text{J}$. Calculate: (a) the voltage across the plates; (b) the total charge; (c) the maximum sustainable voltage if the dielectric has a breakdown strength $E_\text{max} = 10\;\text{kV/mm}$ and thickness $d = 0{,}5\;\text{mm}$. Is the device being used to its full potential?

Solution

(a) Voltage. From $U = \tfrac{1}{2}CV^2$ we get $V$: $V = \sqrt{\frac{2U}{C}} = \sqrt{\frac{2\cdot 1{,}25}{50\cdot 10^{-6}}} = \sqrt{5{,}0\cdot 10^{4}} \approx \ev{V \approx 224\;\text{V}}.$

(b) Charge. $Q = CV = 50\cdot 10^{-6}\cdot 224 \approx \ev{Q \approx 11{,}2\;\text{mC}}.$

(c) Maximum voltage. The potential difference must never exceed the value that drives the field to the dielectric breakdown strength: $V_\text{max} = E_\text{max}\,d = 10\;\text{kV/mm}\cdot 0{,}5\;\text{mm} = \ev{V_\text{max} = 5000\;\text{V}}.$

Utilisation. The capacitor operates at just $224$ V against the $5000$ V it could withstand: it uses less than $5\%$ of the maximum voltage. Since energy grows as $V^2$, at the limiting voltage it would store $U_\text{max} = \tfrac{1}{2}C V_\text{max}^2 \approx 625$ J, i.e. about $500$ times the current energy. The device is therefore vastly under-used: there is a huge margin before the dielectric breaks down.

Links

Topics: Campo elettrico e potenziale Concepts: Capacità e condensatore · Densità di energia del campo Skills: Conservazione dell’energia Methods: Metodo del gradiente Objects: Condensatore piano