PX157 - B10 - capacitors

capacitance

C = \frac{Q}{V}

units: $C V^{- 1}$ or $f a r a d (F)$

parallel plate capacitor

electric field:

\vec{E} = {\vec{E}}_{t o p} + {\vec{E}}_{b o t t o m} = - \frac{σ}{ϵ_{0}} \hat{z}

where, $σ = \frac{Q}{A}$ : surface charge density

potential difference between the plates:

V = E d = \frac{σ}{ϵ_{0}} d = \frac{Q d}{ϵ_{0} A}

for parallel plates:

C = \frac{Q}{V} = \frac{ϵ_{0} A}{d}

cylindrical capacitor

\begin{aligned} V_{a b} & = \frac{λ}{2 π ϵ_{0}} \ln (\frac{R_{b}}{R_{a}}) \\ = \frac{Q}{2 π ϵ_{0} L} \ln (\frac{R_{b}}{R_{a}}) \end{aligned}

∴ C = \frac{2 π ϵ_{0} L}{\ln (\frac{R_{b}}{R_{a}})}

spherical capacitor

q = σ_{a} 4 π r_{a}^{2} = - σ_{b} 4 π r_{b}^{2}

V_{a b} = \frac{q}{4 π ϵ_{0}} (\frac{1}{r_{a}} - \frac{1}{r_{b}})

C = \frac{4 π ϵ_{0}}{(\frac{1}{r_{a}} - \frac{1}{r_{b}})}

single-plate spherical capacitor: van der graaf generator

capacitors in series

conservation of charge: both capacitors have the same charge, $Q :$

V_{1} = \frac{Q}{C_{1}}, V_{2} = \frac{Q}{C_{2}}

⟹ V = V_{1} + V_{2} = Q (\frac{1}{C_{1}} + \frac{1}{C_{2}})

this is equivalent to a circuit with one capacitor, $C_{e q}$ :

\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

assume both the capacitors have the same $A$ and $ϵ_{0} :$

\frac{1}{C_{e q}} = \frac{d_{1}}{ϵ_{0} A} + \frac{d_{2}}{ϵ_{0} A} = \frac{d_{1} + d_{2}}{ϵ_{0} A}

capacitors in parallel

both capacitors have the same potential drop

Q_{1} = C_{1} V, Q_{2} = C_{2} V

total charge: $Q = Q_{1} + Q_{2} = (C_{1} + C_{2}) V$
equivalent capacitor, $C_{e q} :$

C_{e q} = C_{1} + C_{2}

\1\n\2\n