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Five capacitors, all uncharged at first, are joined to a 60 V battery as in the figure…

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

Five capacitors, all uncharged at first, are joined to a $60\ \text{V}$ battery as in the figure. The battery's positive terminal is at the corner $X$ and its negative terminal at the corner $Y$ . Between $X$ and $Y$ there are two paths. On the upper path, $C_1 = 2.0\ \mu\text{F}$ joins $X$ to the point $M$ , and $C_2 = 4.0\ \mu\text{F}$ joins $M$ to $Y$ . On the lower path, $C_3 = 4.0\ \mu\text{F}$ joins $X$ to the point $N$ , and $C_4 = 2.0\ \mu\text{F}$ joins $N$ to $Y$ . A fifth capacitor $C_5 = 3.0\ \mu\text{F}$ joins $M$ to $N$ . Each capacitor is a parallel-plate capacitor of its own, and the wires have no capacitance worth counting. Take the potential of $Y$ as zero. The charges have settled.

A network of five capacitors in a diamond. The left corner is X and the right corner is Y. The upper path goes from X through C1 of 2.0 microfarads to the top corner M, then through C2 of 4.0 microfarads to Y. The lower path goes from X through C3 of 4.0 microfarads to the bottom corner N, then through C4 of 2.0 microfarads to Y. C5 of 3.0 microfarads joins M and N. A 60 V battery below the diamond joins X to Y, positive terminal at X.

What is the potential of point $M$ ?
What is the potential of point $N$ ?
What is the charge on $C_5$ ?
What is the capacitance of the whole network between $X$ and $Y$ ?
What is the total energy stored in the network?
With the battery still connected, $C_4$ is filled with a slab of dielectric. The slab fills the gap of $C_4$ completely. For what dielectric constant does $C_5$ end up with no charge at all?

The worked solution

Have you tried it?

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.

Plan. The network is not a simple series and parallel one, because $C_5$ links the two paths. Use the fact that the conductors at $M$ and $N$ stay neutral.

Step 1. Node equations (a, b). At $M$ : $C_1(V_M - 60) + C_2V_M + C_5(V_M - V_N) = 0$ , so $9V_M - 3V_N = 120$ . At $N$ : $C_3(V_N - 60) + C_4V_N + C_5(V_N - V_M) = 0$ , so $9V_N - 3V_M = 240$ . Solving, $V_M = 25.0\ \text{V}$ and $V_N = 35.0\ \text{V}$ .

Step 2. $C_5$ (c). It has $10.0\ \text{V}$ across it, $N$ higher. Its charge is $3.0\ \mu\text{F} \times 10.0\ \text{V} = 30.0\ \mu\text{C}$ .

Step 3. The network (d). The plates at $X$ hold $C_1(60 - V_M) + C_3(60 - V_N) = 70 + 100 = 170\ \mu\text{C}$ . So $C_{\text{eq}} = \dfrac{170\ \mu\text{C}}{60\ \text{V}} = 2.83\ \mu\text{F}$ .

Step 4. Energy (e). $U = \tfrac12 C_{\text{eq}}E^2 = \tfrac12 \times 170\ \mu\text{C} \times 60\ \text{V} = 5.10\ \text{mJ}$ . The five capacitors' own $\tfrac12 CV^2$ add to the same.

Step 5. Balance (f). $V_M = V_N$ needs $\dfrac{C_1}{C_2} = \dfrac{C_3}{C_4'}$ , so $C_4' = \dfrac{C_2C_3}{C_1} = 8.0\ \mu\text{F}$ . The slab must multiply $C_4$ by $K = 4.0$ .

Answer: (a) $25.0\ \text{V}$ . (b) $35.0\ \text{V}$ . (c) $30.0\ \mu\text{C}$ . (d) $2.83\ \mu\text{F}$ . (e) $5.10\ \text{mJ}$ . (f) $4.0$ .

Watch out. Leaving out $C_5$ and taking two series pairs in parallel gives $2.67\ \mu\text{F}$ . That is right only when the bridge is balanced.

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