===== Defining capacitance =====

Consider two parallel conducting plates of area $A$ separated by a distance $d$. Such a device is called a //capacitor//. If we connect each plate to a voltage source like a battery or power supply, a charge will appear on each plate, as shown in Fig. 3.

<div figure wide>
| {{phylabs:lab_courses:phys-120_130-wiki-home:winter-experiments:capacitance:parallel_plate_capacitor_4x.png|}} |
| **Figure 3:** Parallel plate capacitor connected to a voltage source |
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The amount of charge $q$ held in a capacitor depends on the capacitor’s //capacitance// $C$ and on the voltage $V$ applied as

|  $ q = CV$  |  $(1)$  |

It can be shown that the capacitance is given by

|  $C = \dfrac{\epsilon_0 A}{d}$, |  $(2)$  |

where $\epsilon_0 = 8.85418782\times 10^{-12}\,\,\dfrac{\mathrm{s}^4\mathrm{A}^2}{\mathrm{m}^3 \,\mathrm{kg}}$ is the //permittivity of free space//.

The SI unit of capacitance is the //farad//. A capacitor is said to have a capacitance of one farad when one coulomb of charge produces a potential difference of one volt. The farad is an //enormous// unit, however. It is more typical to find capacitors on the order of pico-, nano-, or microfarads.

===== Discharging a capacitor =====

If a resistor is connected across a charged capacitor by closing a switch as shown in Fig. 4, the excess electrons on the negative plate will move through the resistor to the positively charged (electron-deficient) plate. In other words, the capacitor will discharge.

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{{ phylabs:lab_courses:phys-120_130-wiki-home:winter-experiments:capacitance:capacitor_discharge_circuit.png?direct&300 |}}
**Figure 4:** Capacitor discharging circuit
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Using Kirchhoff's rules, the voltage across the capacitor – which by $(1)$ is $V = q/C$ – must be equal to the voltage across the resistor – which is $V=IR$. Therefore,

|  $\dfrac{q}{C} = IR$.  |  $(3)$  |

Since $q$ and $I$ are related by the equation

|  $I = -\dfrac{dq}{dt}$,  |  $(4)$  |

we can substitute $(4)$ into $(3)$ and rearrange to find

|  $\dfrac{1}{q}dq = -\dfrac{1}{RC}dt$.  |  $(5)$  |

Integrating this and putting in the initial condition $q(0) = q_0$, we find

|  $\displaystyle q(t) = q_0 e^{-t/RC}$,  |  $(6)$  |

or, since by $(1)$, $q_0 = C V_0$ and $q=CV$,

|  $\displaystyle V_C(t) = V_0 e^{-t/RC}$.  |  $(7)$  |

Thus, the voltage across the capacitor (or resistor) must decay exponentially. The time for the voltage to drop to $1/e \approx 37\%$ of its initial value will be when $t/RC = 1$, i.e., when $t=RC$. We define this time to be the //time constant// of a resistance-capacitance or "$RC$" circuit: $\tau = RC$

A graph showing the prediction of $(7)$ for a discharging capacitor is shown in Fig. 5.

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{{ phylabs:lab_courses:phys-120_130-wiki-home:winter-experiments:capacitance:cap_discharge.png?direct&400 |}}
**Figure 5:** Discharge of a capacitor
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===== Charging a capacitor =====

Suppose instead of starting with the capacitor charged, we start with it discharged and proceed to charge the capacitor through a resistor using a circuit such as that shown in Fig. 6.

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{{ phylabs:lab_courses:phys-120_130-wiki-home:winter-experiments:capacitance:capacitor_charge_circuit.png?direct&400 |}}
**Figure 6:** Capacitor charging circuit
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Using Kirchhoff's rules, $(1)$, and Ohm's law, the equation of the circuit is

|  $V_0 = IR + \dfrac{q}{C}$.  |  $(8)$  |

Substituting $(4)$ for $I$, integrating, and putting in the boundary conditions, we find

|  $q(t) = CV_0 \left(1-e^{-t/RC}\right)$,  |  $(9)$  |

where $V_0$ is the voltage of the battery. In terms of the voltage across the capacitor,

|  $V_C(t) = V_0\left(1-e^{-t/RC}\right)$.  |  $(10)$  |

A graph showing the charging of a capacitor according to the predictions of $(10)$ is shown in Fig. 7. Again, the time constant $\tau= RC$ has meaning. This time, it is the time to reach $1-e^{-1} \approx 63\%$ of its final charge.

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{{ phylabs:lab_courses:phys-120_130-wiki-home:winter-experiments:capacitance:cap_charge.png?direct&500 |}}
**Figure 7:** Voltage vs. time for a charging capacitor
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