Kvant Physics Problem 318

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Problem

Fig. 3

At what $C_1$ is the capacitance of the capacitor system shown in Fig. 3 equal to:

$C$;
$kC$ ($k \ne 1$);
$C_1$?

B. B. Bukhovtsev

Setup and Assumptions

A system of ideal capacitors is connected as shown in Fig. 3, containing a parameter capacitor $C_1$ and identical capacitors of capacitance $C$. The task is to determine the value of $C_1$ such that the equivalent capacitance $C_{\text{eq}}$ of the entire network equals $C$, equals $kC$ for $k \ne 1$, and equals $C_1$.

All capacitors are assumed ideal, with no leakage, no parasitic inductance, and no dielectric losses. The connections are treated as perfect conductors so that all nodes connected by wires are equipotential. The system is analyzed in the static regime where capacitance is defined via $Q = C_{\text{eq}}U$.

Physical Principles

The defining relation for capacitance is $Q = CU$, where $Q$ is the charge stored on a conductor and $U$ is the potential difference across the capacitor.

For networks of capacitors, the following relations are used. For capacitors in parallel, capacitances add as $C_{\parallel} = \sum_i C_i$. For capacitors in series, reciprocals add as $\frac{1}{C_{\text{series}}} = \sum_i \frac{1}{C_i}$. More generally, the equivalent capacitance is obtained by enforcing charge conservation at floating nodes and equality of potential differences at nodes connected by ideal conductors.

For symmetric networks, equipotential nodes can be identified, reducing the circuit to combinations of series and parallel blocks.

Derivation

The circuit in Fig. 3 has a mirror symmetry about the central vertical axis passing through $C_1$. The two identical branches on the left and right therefore carry equal charges when a potential difference is applied between the external terminals. This symmetry implies that corresponding midpoints of the left and right branches are at equal potential, so no charge flows through the symmetry axis. The network reduces to two identical subcircuits connected in parallel, each containing a reduced series combination of $C$ and $C_1$.

Each branch consists of a capacitor $C_1$ in series with an effective capacitance formed by two identical capacitors $C$ in parallel, giving $2C$. The series combination of $C_1$ and $2C$ yields the branch capacitance

$$C_{\text{branch}} = \frac{C_1 \cdot 2C}{C_1 + 2C}.$$

Since the two branches are identical and connected in parallel, the total capacitance is

$$C_{\text{eq}} = 2C_{\text{branch}} = \frac{2C_1 \cdot 2C}{C_1 + 2C} = \frac{4CC_1}{C_1 + 2C}.$$

For case 1, the condition $C_{\text{eq}} = C$ gives

$$\frac{4CC_1}{C_1 + 2C} = C.$$

Dividing by $C$ yields

$$\frac{4C_1}{C_1 + 2C} = 1,$$

$$4C_1 = C_1 + 2C,$$

$$3C_1 = 2C,$$

$$C_1 = \frac{2}{3}C.$$

For case 2, the condition $C_{\text{eq}} = kC$ gives

$$\frac{4CC_1}{C_1 + 2C} = kC.$$

Dividing by $C$ yields

$$\frac{4C_1}{C_1 + 2C} = k,$$

$$4C_1 = k(C_1 + 2C),$$

$$4C_1 = kC_1 + 2kC,$$

$$(4 - k)C_1 = 2kC,$$

$$C_1 = \frac{2k}{4 - k}C.$$

For case 3, the condition $C_{\text{eq}} = C_1$ gives

$$\frac{4CC_1}{C_1 + 2C} = C_1.$$

Multiplying both sides by $C_1 + 2C$ yields

$$4CC_1 = C_1^2 + 2CC_1.$$

Rearranging,

$$0 = C_1^2 - 2CC_1,$$

$$C_1(C_1 - 2C) = 0.$$

The nontrivial solution is

$$C_1 = 2C.$$

Result

For $C_{\text{eq}} = C$,

$$C_1 = \frac{2}{3}C.$$

For $C_{\text{eq}} = kC$ with $k \ne 1$,

$$C_1 = \frac{2k}{4 - k}C.$$

For $C_{\text{eq}} = C_1$,

$$C_1 = 2C.$$

Sanity Checks

The expressions have correct dimensional consistency since all results scale linearly with $C$, so $C_1$ has units of farads.

For $k \to 1$, the second expression gives $C_1 \to \frac{2}{3}C$, matching the first case.

For $k \to 4$, the formula $C_1 = \frac{2k}{4-k}C$ diverges, which corresponds to the network approaching a configuration where the equivalent capacitance cannot reach $4C$ for any finite $C_1$ within this topology.

For $C_1 = 2C$, substitution into $C_{\text{eq}} = \frac{4CC_1}{C_1 + 2C}$ yields $C_{\text{eq}} = 2C$, confirming self-consistency of case 3.