Supercapacitors Benefit from Large Effective Surface Areas and Small Electrode$100 \mathrm {~F}$

Question

Examlex · Accepted Answer

The Answer is The capacitance of a parallel plate capacitor is

\mathrm { C } = \varepsilon _ { \mathrm { r } } \varepsilon _ { 0 } A / d

Solving for the area gives

A = C d / \left( \varepsilon _ { \mathrm { r } } \varepsilon _ { 0 } ight)

So substituting values from the problem

A = \left( 0.01 imes 10 ^ { - 3 } \mathrm {~m} ight) imes ( 100 \mathrm {~F} ) / \left( 2 imes 8.854 imes 10 ^ { - 12 } \mathrm {~F} / \mathrm { m } ight) = 5.6 imes 10 ^ { 7 } \mathrm {~m} ^ { 2 }

or a square about

7.5 \mathrm {~km}

on a side.

Supercapacitors Benefit from Large Effective Surface Areas and Small Electrode $100 \mathrm {~F}$