OpenStax University Physics/E&M/Capacitance

${\mathcal {E}}\&{\mathcal {M}}$ (Vol. 2):    5:Electric Charges and Fields    6:Gauss's Law    7:Electric Potential    8:Capacitance    9:Current and Resistance    10:Direct-Current Circuits    11:Magnetic Forces and Fields    12:Sources of Magnetic Fields    13:Electromagnetic Induction    14:Inductance    15:Alternating-Current Circuits    16:Electromagnetic Waves

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Chapter 8

Capacitance

$Q=CV$ defines capacitance. For a parallel plate capacitor, $C=\varepsilon _{0}{\tfrac {A}{d}}$ where A is area and d is gap length.

▭ $4\pi \varepsilon _{0}{\tfrac {R_{1}R_{2}}{R_{2}-R_{1}}}$ and ${\tfrac {2\pi \varepsilon _{0}\ell }{\ln(R_{2}/R_{1})}}$ for a spherical and cylindrical capacitor, respectively
▭ For capacitors in series (parallel) ${\tfrac {1}{C_{S}}}=\sum {\tfrac {1}{C_{i}}}\left(C_{P}=\sum C_{i}\right)$
▭  $u={\tfrac {1}{2}}QV={\tfrac {1}{2}}CV^{2}={\tfrac {1}{2C}}Q^{2}$ ▭ Stored energy density $u_{E}={\tfrac {1}{2}}\varepsilon _{0}E^{2}$
▭ A dielectric with $\kappa >1$ will decrease the capacitor's electric field $E={\tfrac {1}{\kappa }}E_{0}$ and stored energy $U={\tfrac {1}{\kappa }}U_{0}$, but increase the capacitance $C=\kappa C_{0}$ due to the induced electric field ${\vec {E}}_{i}=\left({\tfrac {1}{\kappa }}-1\right){\vec {E}}_{0}$

For quiz at QB/d_cp2.8

$Q=CV$ defines capacitance.

$C=\varepsilon _{0}{\tfrac {A}{d}}$ where A is area and d<<A^1/2 is gap length of parallel plate capacitor

${\text{Series}}:\;{\tfrac {1}{C_{S}}}=\sum {\tfrac {1}{C_{i}}}.$   ${\text{ Parallel:}}\;C_{P}=\sum C_{i}.$

$u={\tfrac {1}{2}}QV={\tfrac {1}{2}}CV^{2}={\tfrac {1}{2C}}Q^{2}$ = stored energy

$u_{E}={\tfrac {1}{2}}\varepsilon _{0}E^{2}$ = energy density