Charge flows until the voltages match
Charge a \(2\,\mu\text{F}\) capacitor to \(100\,\text{V}\) and take away the battery. It holds \(Q = 200\,\mu\text{C}\). Now join its plates to those of an uncharged \(3\,\mu\text{F}\) capacitor, top plate to top plate and bottom to bottom. At the first instant the top plates differ in potential by \(100\,\text{V}\). So charge flows through the wire from the full capacitor to the empty one.
The flow stops when there is no potential difference left along the wire. Then the two capacitors have one common voltage \(V\), and they are simply in parallel. No charge can leave the pair of joined plates, so the total stays \(200\,\mu\text{C}\). With \(C_1 + C_2 = 5\,\mu\text{F}\), the common potential is \(V\) \(= 200/5\) \(= 40\,\text{V}\).
The \(2\,\mu\text{F}\) keeps \(2 \times 40\) \(= 80\,\mu\text{C}\) and the \(3\,\mu\text{F}\) takes \(3 \times 40\) \(= 120\,\mu\text{C}\). Together they still hold \(200\,\mu\text{C}\). The larger capacitor ends with the larger share, in the ratio of the capacitances.
Before the switch closes, the \(2\,\mu\text{F}\) at \(100\,\text{V}\) stores \(10\,\text{mJ}\). After the charge has moved and settled, how much energy do the two capacitors store together?
Play the figure and watch the signs move from the \(2\,\mu\text{F}\) to the \(3\,\mu\text{F}\). The count of signs never changes, but the energy bar of the pair shrinks while the heat bar grows. The next topic finds the loss exactly.
A \(6\,\mu\text{F}\) capacitor charged to \(50\,\text{V}\) is joined plate to plate with an uncharged \(3\,\mu\text{F}\) capacitor. What is the common voltage, in volts?
Where the energy goes
Count the energy before and after. Before, only the \(2\,\mu\text{F}\) is charged: \(U_i\) \(= \tfrac12 \times 2 \times 10^{-6} \times 100^2\) \(= 0.01\,\text{J}\). After, the pair of \(5\,\mu\text{F}\) is at \(40\,\text{V}\): \(U_f\) \(= \tfrac12 \times 5 \times 10^{-6} \times 40^2\) \(= 4 \times 10^{-3}\,\text{J}\). So \(6 \times 10^{-3}\,\text{J}\) has left the capacitors.
It became heat in the connecting wire, while the current flowed. Some also leaves as light and sound in the spark at the switch, and as radio waves. Charge is conserved when capacitors are joined, but stored energy is not. The loss has a general formula.
\[ V = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2} \]
Why. the charge \(C_1V_1 + C_2V_2\) is conserved and ends on the parallel pair \(C_1 + C_2\).
\[ U_i = \tfrac12 C_1V_1^2 + \tfrac12 C_2V_2^2, \qquad U_f = \tfrac12(C_1 + C_2)V^2 = \dfrac{(C_1V_1 + C_2V_2)^2}{2(C_1 + C_2)} \]
Why. the energy of each capacitor before, and of the joined pair after.
\[ H = U_i - U_f = \dfrac{(C_1V_1^2 + C_2V_2^2)(C_1 + C_2) - (C_1V_1 + C_2V_2)^2}{2(C_1 + C_2)} \]
Why. put both over the common denominator \(2(C_1 + C_2)\).
\[ H = \dfrac{C_1C_2(V_1^2 - 2V_1V_2 + V_2^2)}{2(C_1 + C_2)} = \dfrac{C_1C_2(V_1 - V_2)^2}{2(C_1 + C_2)} \]
Read it. on expanding, the terms \(C_1^2V_1^2\) and \(C_2^2V_2^2\) cancel.
(a) Check the \(6 \times 10^{-3}\,\text{J}\) of the \(2\) and \(3\,\mu\text{F}\) with the formula. (b) A \(4\,\mu\text{F}\) capacitor at \(200\,\text{V}\) is joined to an uncharged \(4\,\mu\text{F}\). Find the final charges, the common voltage and the heat.
(a) \(H\) \(= (2 \times 3 \times 10^{-12} \times 100^2)/(2 \times 5 \times 10^{-6})\) \(= 6 \times 10^{-3}\,\text{J}\), as counted.
(b) The charge \(800\,\mu\text{C}\) is shared equally: \(400\,\mu\text{C}\) each, at \(100\,\text{V}\).
Before, \(U_i\) \(= \tfrac12 \times 4 \times 10^{-6} \times 200^2\) \(= 0.08\,\text{J}\). After, \(U_f\) \(= \tfrac12 \times 8 \times 10^{-6} \times 100^2\) \(= 0.04\,\text{J}\). The heat is \(0.04\,\text{J}\), exactly half.
Values used: none of the constants; only \(C\), \(Q\) and \(V\).
For two equal capacitors, one charged to \(V\) and one empty, the common voltage is \(V/2\). The stored energy falls from \(\tfrac12CV^2\) to \(\tfrac12(2C)(V/2)^2\) \(= \tfrac14CV^2\). Half the energy is always lost, whatever the size of \(C\) or \(V\).
Does a thick copper wire lose less than a thin one? No. The formula holds no resistance at all. A thin wire lets a small current flow for a long time; a thick one lets a large current flow for a short time. The heat comes out the same, fixed by the charges and the capacitances alone. Even a wire of almost zero resistance loses it, mainly as radiation from the spark.
Think of two water tanks side by side, one full and one empty, joined by a pipe at the bottom. Open the valve and the water rushes across and sloshes to and fro until both levels match. The friction of that sloshing turns part of the stored energy into heat. A wide pipe or a narrow one changes how fast the water settles, not how much energy is lost.
A \(6\,\mu\text{F}\) capacitor charged to \(50\,\text{V}\) is joined to an uncharged \(3\,\mu\text{F}\). How much energy is lost as heat, in millijoules?
A student says that two capacitors joined by a wire of zero resistance make no heat. So, the student says, no energy is lost. Explain what is wrong with this.
The final state is fixed by conservation of charge alone: the pair ends at the common voltage whatever the wire. So the stored energy after is the same, and the loss \(C_1C_2(V_1 - V_2)^2/2(C_1 + C_2)\) is the same. Resistance only sets how fast the charge flows and how the loss is shared between heat and radiation. With almost no resistance, a large current surges and oscillates, and the energy leaves as electromagnetic radiation and the light and sound of the spark.
Two charged capacitors, either way round
Now let both capacitors start charged: \(2\,\mu\text{F}\) at \(100\,\text{V}\), holding \(200\,\mu\text{C}\), and \(3\,\mu\text{F}\) at \(50\,\text{V}\), holding \(150\,\mu\text{C}\). Join them plus to plus and minus to minus. The joined positive plates hold \(200 + 150\) \(= 350\,\mu\text{C}\), shared over \(5\,\mu\text{F}\). The common voltage is \(70\,\text{V}\), between the two starting values.
Join them the other way, the positive plate of one to the negative plate of the other. Now the joined plates hold \(+200\) and \(-150\,\mu\text{C}\), and these partly cancel. Only \(50\,\mu\text{C}\) is left to share, so the common voltage is \(50/5\) \(= 10\,\text{V}\). In the formulas this is the same as giving \(V_2\) a minus sign.
The \(2\,\mu\text{F}\) at \(100\,\text{V}\) and the \(3\,\mu\text{F}\) at \(50\,\text{V}\) are joined (a) plus to plus, (b) plus to minus. Find the heat produced in each case.
(a) \(H\) \(= 2 \times 3 \times 10^{-12} \times (100 - 50)^2/(2 \times 5 \times 10^{-6})\) \(= 1.5 \times 10^{-3}\,\text{J}\). The voltages were close, so little is lost.
(b) Now \(V_2 = -50\,\text{V}\) and \(V_1 - V_2 = 150\,\text{V}\): \(H\) \(= 6 \times 10^{-12} \times 150^2/(10^{-5})\) \(= 0.0135\,\text{J}\).
Check (b): before, \(0.01 + 3.75 \times 10^{-3}\) \(= 0.01375\,\text{J}\); after, \(\tfrac12 \times 5 \times 10^{-6} \times 10^2\) \(= 2.5 \times 10^{-4}\,\text{J}\). Nearly all the energy is lost.
Choose the chips of the figure to see both cases. Plus to plus, the charge moves only a little and the heat bar stays short. Plus to minus, the signs on the \(3\,\mu\text{F}\) turn over, and almost the whole store becomes heat.
A last case joins this lesson to the one before. Charge the \(2\) and \(3\,\mu\text{F}\) in series across \(100\,\text{V}\), as in Lesson 3: each holds \(120\,\mu\text{C}\), with \(60\,\text{V}\) and \(40\,\text{V}\). Take them out and join them in parallel, plus to plus. They share \(240\,\mu\text{C}\) over \(5\,\mu\text{F}\), at \(48\,\text{V}\). The heat is \(6 \times 10^{-12} \times 20^2/10^{-5}\) \(= 2.4 \times 10^{-4}\,\text{J}\).
Two identical capacitors are each charged to \(100\,\text{V}\). They are joined plus to minus and minus to plus. What happens?
Two spheres joined by a wire
The same happens with any two conductors, each acting as a capacitor with its other plate at infinity. The chapter Conductors in Electrostatics found how the charge shares out. Take a sphere of radius \(10\,\text{cm}\) with \(6\,\mu\text{C}\), far from an uncharged sphere of radius \(20\,\text{cm}\), and join them by a long thin wire. Fig 4.1 shows the two states.
The spheres have \(C_A = R_A/k\) and \(C_B = 2C_A\). Before, all the energy is on the small one: \(U_i = Q^2/2C_A\) \(= (6 \times 10^{-6})^2 \times 9 \times 10^9/0.2\) \(= 1.62\,\text{J}\). After, the \(6\,\mu\text{C}\) spreads over three times the capacitance, so \(U_f\) is a third: \(0.54\,\text{J}\). The small sphere keeps \(2\,\mu\text{C}\) and the large one takes \(4\,\mu\text{C}\), both at \(1.8 \times 10^5\,\text{V}\). So \(1.08\,\text{J}\) is lost.
Some books say the energy is lost "because of the resistance of the wire". The resistance is where the heat appears, not what decides its amount. The loss, \(1.08\,\text{J}\) here, is fixed by conservation of charge and the two capacitances. It would be the same through any wire.
Two charged metal spheres, far apart, are joined by a thin wire. Which statement is true once the charge stops moving?
A capacitor keeps its charge after the power is switched off. The fan capacitor of Lesson 1, or the flash capacitor of Lesson 2, can still give a shock hours later. Electricians discharge a large capacitor through a resistor before touching it. A screwdriver across the terminals also empties it, but all the stored energy then goes into one bright spark. The spark can pit the terminals and throw hot metal.
- Taking the stored energy to be conserved when capacitors are joined.Charge is conserved; stored energy falls by \(C_1C_2(V_1 - V_2)^2/2(C_1 + C_2)\).
- Forgetting the minus sign when opposite plates are joined.Plus to minus, write \(V_2\) as negative: the charges on the joined plates partly cancel.
- Blaming the loss on the resistance of the wire.The resistance sets how fast the loss happens, not how much.
- Sharing the charge equally between two different capacitors.They end at one voltage, so each takes \(Q_i = C_iV\).