\$C=\frac{Q}{V}\$
where Q is the magnitude of charge on each plate (or conductor) and V is the potential difference between the conductors.
For an ideal parallel‑plate arrangement the capacitance is
\$C=\frac{\varepsilon A}{d}\$
If a dielectric of relative permittivity \(\varepsilon_{r}\) fills the space,
\$\$\varepsilon=\varepsilon{0}\varepsilon{r},\qquad
\varepsilon_{0}=8.85\times10^{-12}\;\text{F m}^{-1}\$\$
The capacitance is increased by the factor \(\varepsilon_{r}\).
The electric field between the plates is \(E=V/d\).
Surface charge density \(\sigma = Q/A\).
From Gauss’s law \(E=\sigma/\varepsilon\) ⇒ \(\sigma=\varepsilon V/d\).
Substituting \(\sigma=Q/A\) gives \(Q=(\varepsilon A/d)V\) ⇒ \(C=Q/V=\varepsilon A/d\).
An isolated conducting sphere of radius \(r\) behaves as a single‑plate capacitor whose other “plate’’ is at infinity. Its capacitance is
\$C{\text{sphere}}=4\pi\varepsilon{0}r\$
\$V{\text{tot}}=V{1}+V{2}+\dots+V{n}\$
\$\frac{1}{C{\text{eq}}}= \frac{1}{C{1}}+\frac{1}{C{2}}+\dots+\frac{1}{C{n}}\$
\$Q{\text{tot}}=Q{1}+Q{2}+\dots+Q{n}\$
\$C{\text{eq}}=C{1}+C{2}+\dots+C{n}\$
Solution
\$\frac{1}{C{s}}=\frac{1}{4.0\;\mu\text{F}}+\frac{1}{6.0\;\mu\text{F}}=\frac{5}{12\;\mu\text{F}}\;\Rightarrow\;C{s}=2.4\;\mu\text{F}\$
\$C{\text{eq}}=C{s}+C_{3}=2.4\;\mu\text{F}+3.0\;\mu\text{F}=5.4\;\mu\text{F}\$
\$Q=C_{\text{eq}}V=5.4\;\mu\text{F}\times12\;\text{V}=64.8\;\mu\text{C}\$
\$U=\tfrac12 C_{\text{eq}}V^{2}= \tfrac12(5.4\;\mu\text{F})(12\;\text{V})^{2}=3.89\times10^{-4}\;\text{J}=388.8\;\mu\text{J}\$
For two capacitors in series, the voltage across each is proportional to the *other* capacitance:
\$\$V{1}=V{\text{tot}}\frac{C{2}}{C{1}+C_{2}},\qquad
V{2}=V{\text{tot}}\frac{C{1}}{C{1}+C_{2}}\$\$
This is frequently used to obtain a required fraction of a supply voltage without resistors.
Starting from the definition \(U=\displaystyle\int_{0}^{Q}V\,\mathrm{d}Q\) and using \(V=Q/C\):
\$U=\frac12\frac{Q^{2}}{C}\$
Because \(Q=CV\), the same expression can be written in two equivalent forms:
\$U=\frac12CV^{2}= \frac12QV\$
When a capacitor of capacitance \(C\) is connected to a resistor \(R\), the circuit has a time constant
\$\tau = RC\$
After a time \(\tau\) the voltage (and charge) has fallen to \(\displaystyle\frac{1}{e}\approx36.8\%\) of its initial value.
Applying Kirchhoff’s loop rule to a discharging circuit (\(V{C}+V{R}=0\)) gives
\$\frac{Q}{C}+R\frac{\mathrm{d}Q}{\mathrm{d}t}=0\$
Integrating,
\$Q(t)=Q{0}\,e^{-t/RC},\qquad V(t)=V{0}\,e^{-t/RC}\$
where \(Q{0}=CV{0}\) is the initial charge.
A 10 µF capacitor is charged to 15 V and then discharged through a 200 kΩ resistor. Find the voltage after 0.5 s.
| Type | Dielectric Material | Typical Applications | Capacitance Range |
|---|---|---|---|
| Ceramic | Metal‑oxide ceramic | High‑frequency circuits, decoupling | pF – µF |
| Electrolytic | Aluminium oxide (wet) or tantalum | Power‑supply filtering, bulk storage | µF – mF |
| Film | Polypropylene, polyester | Audio, precision timing, low‑loss filters | nF – µF |
| Mica | Natural mica | RF circuits, temperature‑stable capacitance | pF – nF |
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