Capacitor

A capacitor consists of two conducting surfaces separated by an insulator. It stores energy by building up charge.

Good refresher video by Engineering Mindset: https://www.youtube.com/watch?v=X4EUwTwZ110

Capacitance is given in Farads, where $1F=\frac{1C}{1V}$

Interpretation of capacitance: The electric charge that must be added to the conducting body to increase its electric potential by $1V$.

Energy Stored in a Capacitor

$U_C=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}=\frac{QV}{2}$

The usefulness of a capacitor stems from the fact that it can be charged slowly but then can release energy very quickly. A defibrillator is an example of an application where capacitors are used to discharge current through a patient’s chest.

ECE140

An idea linear capacitor has the following linear relation for charge: $q(t)=Cv(t)$

Therefore, taking the derivative on both sides, we have $i(t)=C\frac{dv(t)}{dt}$

Some Remarks

1. Notice that current $i$ is 0 if the voltage is constant over time
2. A capacitor voltage must be a continuous function of time, else current will suddenly be infinity which is impossible, i.e. $v(t)$ cannot jump.

Finding $v(t)$ from $i(t)$ $v(t)=\frac{1}{C}{\int }_{t_0}^{t}i(\tau )d\tau +v(t_0)$

Power $p(t)=v(t)i(t)=Cv(t)\frac{dv(t)}{dt}$ Energy (derived from power, since $p(t)=\frac{dw(t)}{dt}$) $w(t)=\frac{1}{2}C{v}^2(t)=\frac{q^2(t)}{2C}$

Capacitor and DC Voltage (Open Circuit)

If the voltage $v(t)$ is constant (DC voltage), then the capacitor acts like an open circuit.

Combinations of Capacitors

Capacitors can be combined to form parallel capacitors and series capacitors.

For parallel capacitors, the formula is $C_{eq}=C_1+C_2+C_3+\dots$

For series capacitors, the formula is $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\dots$

ECE106 Stuff

Breaks things down at an even lower level.

The basic formula is $C=\frac{Q}{\Delta V_{2\to 1}}$ Steps for Calculating Capacitance:

1. pick coordinate system
2. Assume $+Q$/$-Q$ charge on each conductor
3. Find $\vec{E}$ with Gauss Law or other methods
4. Find $\Delta V$ by going from $-Q$ to $+Q$
5. Find $C$ $C=\frac{Q}{\Delta V_{2\to 1}}$

$Q$ always cancel out, the only parameters that remain are

• Geometric properties
• Permittivity
• Separation Distance

Capacitance

Make sure to understand WHY (ex: why $D=\rho$ ) $E=\frac{D}{ϵ}=\frac{Q}{Aϵ}$ $\Delta V_{2\to 1}=Ed=\frac{Qd}{Aϵ}$ so capacitance is $C=\frac{{ϵ}_0{ϵ}_{r}A}{d}$

Other formulas, you see them in ECE140 but this is cleaner:

Energy $U=\frac{1}{2}C{V}^2=\frac{QV}{2}=\frac{Q}{2C}$

Related

• Dielectrics
• Dielectric Breakdown