Two capacitors have capacitances 4.0 µF and 2.0 µF…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Two capacitors have capacitances $4.0\ \mu\text{F}$ and $2.0\ \mu\text{F}$ . Each is charged by its own battery, the first to $90\ \text{V}$ and the second to $60\ \text{V}$ , and then both are removed from their batteries. In experiment 1, the two positive plates are joined by a wire and the two negative plates by another wire. In experiment 2, the capacitors are charged again in the same way, but the positive plate of each is joined to the negative plate of the other. In both experiments, the charge is left to settle, and the wires have negligible resistance.

What is the common potential difference in experiment 1?
How much electrical energy is lost, as heat and radiation, in experiment 1?
What is the common potential difference in experiment 2?
How much energy is lost in experiment 2?

After experiment 1, a slab of dielectric constant $3$ is slid in to fill the whole gap of the $2.0\ \mu\text{F}$ capacitor, with the wires still in place and no battery connected.

What is the new potential difference across the pair?

Charge is conserved in each experiment, and only charge that can move is shared. Think about the signs of the charges on the plates that are joined.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $C_1 = 4.0\ \mu\text{F}$ at $90\ \text{V}$ and $C_2 = 2.0\ \mu\text{F}$ at $60\ \text{V}$ . Charges: $Q_1 = 4.0 \times 90 = 360\ \mu\text{C}$ and $Q_2 = 2.0 \times 60 = 120\ \mu\text{C}$ . Energy at the start: $\tfrac12 \times 4.0 \times 90^2 + \tfrac12 \times 2.0 \times 60^2 = 19.80\ \text{mJ}$ .

(a) Like plates joined. The charges on the positive plates add: $360 + 120 = 480\ \mu\text{C}$ . The capacitance is $4.0 + 2.0 = 6.0\ \mu\text{F}$ , so $V = \dfrac{480\ \mu\text{C}}{6.0\ \mu\text{F}} = 80.0\ \text{V}$ .

(b) Energy lost. After joining, $U = \tfrac12 \times 6.0\ \mu\text{F} \times (80\ \text{V})^2 = 19.20\ \text{mJ}$ . The loss is $19.80 - 19.20 = 0.60\ \text{mJ}$ .

(c) Unlike plates joined. The charge on the joined plates is $360 - 120 = 240\ \mu\text{C}$ , since one pair of joined plates carries $+360$ and $-120$ . So $V = \dfrac{240\ \mu\text{C}}{6.0\ \mu\text{F}} = 40.0\ \text{V}$ .

(d) Energy lost. After joining, $U = \tfrac12 \times 6.0\ \mu\text{F} \times (40\ \text{V})^2 = 4.80\ \text{mJ}$ , so the loss is $19.80 - 4.80 = 15.0\ \text{mJ}$ .

(e) The slab. The total charge stays at $480\ \mu\text{C}$ . The $2.0\ \mu\text{F}$ capacitor becomes $3 \times 2.0 = 6.0\ \mu\text{F}$ , so the total capacitance is $4.0 + 6.0 = 10\ \mu\text{F}$ and $V = \dfrac{480\ \mu\text{C}}{10\ \mu\text{F}} = 48.0\ \text{V}$ .

Answer: (a) $80.0\ \text{V}$ . (b) $0.60\ \text{mJ}$ . (c) $40.0\ \text{V}$ . (d) $15.0\ \text{mJ}$ . (e) $48.0\ \text{V}$ .

Check. The loss in an experiment is $\dfrac{C_1C_2}{2(C_1 + C_2)}(V_1 \mp V_2)^2$ . For (b), $\dfrac{4.0 \times 2.0}{2 \times 6.0} \times 30^2\ \mu\text{J} = 0.60\ \text{mJ}$ , and for (d), the same factor times $150^2$ gives $15.0\ \text{mJ}$ .

Watch out. The average of the two potential differences, $75\ \text{V}$ , is wrong because the capacitances differ. And joining unlike plates loses far more energy, since the opposite charges cancel.

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Two capacitors have capacitances 4.0 µF and 2.0 µF…

The worked solution

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