15.1: Electric Circuits:Capacitor
When current flows through wires and resistors in a circuit as a result of a difference in electric potential, charge does not build up significantly anywhere on its path. Capacitors are devices placed in electric circuits where charge can build up. The amount of charge a capacitor can store before it “fills up” depends on its geometry and how much electric potential is applied. The ratio of charge stored in a capacitor to the voltage applied is called its capacitance, measured in Farads \begin{align*}(F)\end{align*}
Key Equations and Definitions
\begin{align*}\begin{cases} \epsilon_0 = 8.85 \times 10^{12} \text{F/m} & \text{Constant called the permittivity of free space}\\ Q = CV & \text{Charge stored in a capacitor}\\ U = \frac{1}{2}CV^2 & \text{Potential energy stored in a capacitor}\\ C = \frac{\kappa \epsilon_0 A}{d} & \text{For two parallel metal plates, area A} \text{,separation} d, \text{dielectric} \kappa\\ Q(t) = Q_0 e^{\frac{t}{\tau}} & \text{ Discharge rate of a capacitor, where } \tau= RC \\ Q(t) = Q_0 (1e^{\frac{t}{\tau}}) & \text{ Charge rate of a capacitor, where } \tau= RC \; \text{and} \; Q_0 = V \dot C \\ I(t) = I_0 e^{\frac{t}{\tau}} & \text{ Discharge and Charge rate for current, where } \tau= RC \; \text{and} \; I_0 = \frac{V}{R} \\ C_{\text{pNet}} = \sum_{i} C_{\text{p}i} & \text{Capacitors in parallel add like resistors in series}\\ \frac{1}{C_{\text{sNet}}} = \sum_{i} \frac{1}{C_{\text{s}i}} & \text{And vice versa} \end{cases}\end{align*}
Electric Circuit Symbol
The symbol for a capacitor is two flat plates, mimicking the geometry of a capacitor, which typically consists of two flat plates separated by a small distance. The plates are normally wrapped around several times to form a cylindrical shape.
Key Concepts
 Current can flow into a capacitor from either side, but current doesn’t flow across the capacitor from one plate to another. The plates do not touch, and the substance in between is insulating, not conducting.
 One side of the capacitor fills up with negative charge, and the other fills up with positive charge. The reason for the thin, close plates is so that you can use the negative charge on one plate to attract and hold the positive charge on the other plate. The reason for the plates with large areas is so that you can spread out the charge on one plate so that its selfrepulsion doesn’t stop you from filling it with more charge.
 Capacitors wired in series all contain the same amount of charge (they have to because the opposite plates of the different capacitors are directly connected. However, they have different voltages.
 Capacitors wired in parallel all have the same voltage (they have to because all the positive plates are directly connected together and all the negative plates are directly connected together. However, they do have different charge.
 Typical dielectric constants κ are roughly 5.6 for glass and 80 for water. What these “dielectric” substances do is align their electric polarity with the electric field in a capacitor (much like atoms in a magnetic material) and, in doing so, reduces the electric field for a given amount of charge. Thereby allowing for more charge to be stored for a given Voltage.
 When a capacitor is initially uncharged, it is very easy to stuff charge in. As charge builds, it repels new charge with more and more force. Due to this effect, the charging of a capacitor follows a logarithmic curve. When you pass current through a resistor into a capacitor, the capacitor eventually “fills up” and no more current flows. A typical RC circuit is shown below; when the switch is closed, the capacitor discharges with an exponentially decreasing current.
 \begin{align*}Q\end{align*} refers to the amount of positive charge stored on the high voltage side of the capacitor; an equal and opposite amount, \begin{align*}Q\end{align*}, of negative charge is stored on the low voltage side of the capacitor.
 Many homeelectronic circuits include capacitors; for this reason, it can be dangerous to mess around with old electronic components, as the capacitors may be charged even if the unit is unplugged. For example, old computer monitors (not flat screens) and TVs have capacitors that hold dangerous amounts of charge hours after the power is turned off.
 When solving capacitor circuits remember that connected points have the same potential. Also remember that charge must be conserved. So if the capacitor(s) are disconnected from a power supply, the total charge must remain the same.
Chapter 15: Electric Circuits – Capacitors Problems
 Design a parallel plate capacitor with a capacitance of \begin{align*}100\;\mathrm{mF}\end{align*}. You can select any area, plate separation, and dielectric substance that you wish.
 You have two \begin{align*}42 \mu \mathrm{F}\end{align*} and one \begin{align*}39 \mu \mathrm{F}\end{align*} all wired in parallel. Draw the schematic and calculate the total capacitance of the system .
 You have two \begin{align*}42 \mu \mathrm{F}\end{align*} and one \begin{align*}39 \mu \mathrm{F}\end{align*} all wired in series. Draw the schematic and calculate the total capacitance of the system .
 You have a \begin{align*}5 \mu \mathrm{F}\end{align*} capacitor.
 How much voltage would you have to apply to charge the capacitor with \begin{align*}200\;\mathrm{C}\end{align*} of charge?
 Once you have finished, how much potential energy are you storing here?
 If all this energy could be harnessed to lift you up into the air, how high would you be lifted?
 Show, by means of a sketch illustrating the charge distribution, that two identical parallelplate capacitors wired in parallel act exactly the same as a single capacitor with twice the area.
 A certain capacitor can store \begin{align*}5\;\mathrm{C}\end{align*} of charge if you apply a voltage of \begin{align*}10\;\mathrm{V}\end{align*}.
 How many volts would you have to apply to store \begin{align*}50\;\mathrm{C}\end{align*} of charge in the same capacitor?
 Why is it harder to store more charge?
 A certain capacitor can store \begin{align*}500\;\mathrm{J}\end{align*} of energy (by storing charge) if you apply a voltage of \begin{align*}15\;\mathrm{V}\end{align*}. How many volts would you have to apply to store \begin{align*}1000\;\mathrm{J}\end{align*} of energy in the same capacitor? (Important: why isn’t the answer to this just \begin{align*}30\;\mathrm{V}\end{align*}?)
 Marciel, a bicycling physicist, wishes to harvest some of the energy he puts into turning the pedals of his bike and store this energy in a capacitor. Then, when he stops at a stop light, the charge from this capacitor can flow out and run his bicycle headlight. He is able to generate \begin{align*}18\;\mathrm{V}\end{align*} of electric potential, on average, by pedaling (and using magnetic induction).
 If Mars wants to provide \begin{align*}0.5\end{align*} A of current for 60 seconds at a stop light, how big a \begin{align*}18\;\mathrm{V}\end{align*} capacitor should he buy (i.e. how many farads)?
 How big a resistor should he pass the current through so the RC time is three minutes?
 Given a capacitor with \begin{align*}1\;\mathrm{cm}\end{align*} between the plates a field of \begin{align*}20,000\;\mathrm{N/C}\end{align*} is established between the plates. (a) What is the voltage across the capacitor? (b) If the charge on the plates is \begin{align*}1 \mu \mathrm{C}\end{align*}, what is the capacitance of the capacitor? (c) If two identical capacitors of this capacitance are connected in series what it the total capacitance? (d) Consider the capacitor connected in the following circuit at point \begin{align*}B\end{align*} with two switches \begin{align*}S\end{align*} and \begin{align*}T\end{align*}, a \begin{align*}20 \Omega\end{align*} resistor and a \begin{align*}120\;\mathrm{V}\end{align*} power source: (i) Calculate the current through and the voltage across the resistor if \begin{align*}S\end{align*} is open and \begin{align*}T\end{align*} is closed (ii) Repeat if \begin{align*}S\end{align*} is closed and \begin{align*}T\end{align*} is open
Figure for Problems 1012:
 Consider the figure above with switch, \begin{align*}S\end{align*}, initially open:
 What is the voltage drop across the \begin{align*}20 \Omega \end{align*} resistor?
 What current flows thru the \begin{align*}60 \Omega\end{align*} resistor?
 What is the voltage drop across the \begin{align*}20\end{align*} microfarad capacitor?
 What is the charge on the capacitor?
 How much energy is stored in that capacitor?
 Find the capacitance of capacitors \begin{align*}B\end{align*}, \begin{align*}C\end{align*}, and \begin{align*}D\end{align*} if compared to the \begin{align*}20 \mu \mathrm{F}\end{align*} capacitor where...
 \begin{align*}B\end{align*} has twice the plate area and half the plate separation
 \begin{align*}C\end{align*} has twice the plate area and the same plate separation
 \begin{align*}D\end{align*} has three times the plate area and half the plate separation
 Now the switch in the previous problem is closed.
 What is the total capacitance of branch II?
 What is the total capacitance of branches I, II, and III taken together?
 What is the voltage drop across capacitor \begin{align*}B\end{align*}?
 Reopen the switch in the previous problem and look at the \begin{align*}20 \mu \mathrm{F}\end{align*} capacitor. It has a plate separation of \begin{align*}2.0 mm\end{align*}.
 What is the magnitude and direction of the electric field?
 If an electron is released in the center to traverse the capacitor and given a speed \begin{align*}2/3\end{align*} the speed of light parallel to the plates , what is the magnitude of the force on that electron?
 What would be its acceleration in the direction perpendicular to its motion?
 If the plates are \begin{align*}1.0\;\mathrm{cm}\end{align*} long, how much time would it take to traverse the plate?
 What displacement toward the plates would the electron undergo?
 With what angle with respect to the direction of motion does the electron leave the plate?
 Design a circuit that uses capacitors, switches, voltage sources, and light bulbs that will allow the interior lights of your car to dim slowly once you get out.
 Design a circuit that would allow you to determine the capacitance of an unknown capacitor.
 The voltage source in the circuit below provides \begin{align*}10\;\mathrm{V}\end{align*}. The resistor is \begin{align*}200 \Omega\end{align*} and the capacitor has a value of \begin{align*}50 \mu \mathrm{F}\end{align*}. What is the voltage across the capacitor after the circuit has been hooked up for a long time?
 A simple circuit consisting of a \begin{align*}39 \mu \mathrm{F}\end{align*} and a \begin{align*}10 \text{k} \Omega \end{align*} resistor. A switch is flipped connecting the circuit to a 12 V battery.
 How long until the capacitor has 2/3 of the total charge across it?
 How long until the capacitor has 99% of the total charge across it?
 What is the total charge possible on the capacitor?
 Will it ever reach the full charge in part c.?
 Derive the formula for V(t) across the capacitor.
 Draw the graph of V vs. t for the capacitor.
 Draw the graph of V vs. t for the resistor.
 If you have a \begin{align*}39 \mu \mathrm{F}\end{align*} capacitor and want a time constant of 5 seconds, what resistor value is needed?
Answers to Selected Problems
 .
 \begin{align*}123 \mu \mathrm{F}\end{align*}

\begin{align*}0.073 \mu \mathrm{F}\end{align*}
 \begin{align*}4 \times 10^7 \;\mathrm{V}\end{align*}
 \begin{align*}4 \times 10^9 \;\mathrm{J}\end{align*}
 .
 \begin{align*}100 \;\mathrm{V}\end{align*}
 A greater voltage created a stronger electronic field, or because as charges build up they repel each other from the plate.

\begin{align*}21 \;\mathrm{V}, \;\mathrm{V}\end{align*} is squared so it doesn’t act like problem \begin{align*}4\end{align*}
 \begin{align*}3.3 \;\mathrm{F}\end{align*}
 \begin{align*}54 \ \Omega\end{align*}
 \begin{align*}200 \;\mathrm{V}\end{align*}
 \begin{align*}5 \times 10^{9} \;\mathrm{F}\end{align*}
 \begin{align*}2.5 \times 10^{9} \;\mathrm{F}\end{align*}
 \begin{align*}6\mathrm{V}\end{align*}
 \begin{align*}0.3\mathrm{A}\end{align*}
 \begin{align*}18\mathrm{V}\end{align*}
 \begin{align*}3.6 \times 10^{4}\mathrm{C}\end{align*}

\begin{align*}3.2 \times 10^{3}\mathrm{J}\end{align*}
 \begin{align*}80 \mu \mathrm{F}\end{align*}
 \begin{align*}40\mu \mathrm{F}\end{align*}
 \begin{align*}120 \mu \mathrm{F}\end{align*}
 \begin{align*}26.7\mu \mathrm{F}\end{align*}
 \begin{align*}166.7\mu \mathrm{F}\end{align*}
 \begin{align*}19.0 \times 10^3 \;\mathrm{N/C}\end{align*}
 \begin{align*}1.4 \times 10^{15} \;\mathrm{N}\end{align*}
 \begin{align*}1.6 \times 10^{15} \;\mathrm{m/s}^2\end{align*}
 \begin{align*}3.3 \times 10^{11} \;\mathrm{s}\end{align*}
 \begin{align*}8.9 \times 10^{7} \;\mathrm{m}\end{align*}
 \begin{align*}5.1 \times 10^{30}\end{align*}
 .
 .
 10 V
 0.43 seconds
 1.8 seconds
 \begin{align*} 4.7 \times 10^{4} \text{C} \end{align*}
 No, it will asymptotically approach it.
 The graph is same shape as the Q(t) graph. It will rise rapidly and then tail off asymptotically towards 12 V.
 The voltage across the resistor is 12 V minus the voltage across the capacitor. Thus, it exponentially decreases approaching the asymptote of 0 V.
 about \begin{align*}128 \text{k} \Omega \end{align*}