Three identical capacitors C1, C2 and C3 have a capacitance of 1.0 uF each and they are uncharged in

Q: Why is $C_3$ the only capacitor charged initially?

When $S_2$ is open, the middle branch containing $C_1$ and $C_2$ is disconnected from the circuit. Therefore, no current can flow through them, and only $C_3$ gets charged by the battery $V_0$.

Q: How do we know $C_1$ and $C_2$ are in series?

In the final state, $S_1$ is open (disconnecting the battery) and $S_2$ is closed. The charge leaving $C_3$ has only one path to flow, which goes through both $C_1$ and $C_2$ sequentially. Thus, they are in series and must have the same charge.

Q: Why does the voltage across $C_3$ equal the sum of voltages across $C_1$ and $C_2$?

In the final closed loop formed by the middle and right branches, $C_3$ is connected in parallel with the series combination of $C_1$ and $C_2$. In a parallel connection, the potential difference across the branches is equal.

Q: What happens to the total energy of the circuit?

The total electrostatic energy decreases. When charge redistributes among the capacitors, some energy is dissipated as heat in the connecting wires due to their resistance, even if it's negligibly small.

Q: Does the polarity of the capacitors matter when applying the loop rule?

Yes. Since $C_1$ and $C_2$ are charged by the discharge of $C_3$, their positive plates will face the positive plate of $C_3$. Traversing the loop, you will experience a potential drop across $C_1$ and $C_2$ and a potential gain across $C_3$, leading to $V_3 - V_1 - V_2 = 0$.

Question

Three identical capacitors $C_1, C_2$ and $C_3$ have a capacitance of $1.0\mu	ext{F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_r$. The cell electromotive force (emf) $V_0 = 8	ext{V}$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $C_3$ is found to be $5\mu	ext{C}$. The value of $\varepsilon_r = \dots$

Sigma Percentile · Accepted Answer

### The Initial State: Charging Up

When we first look at the circuit, switch $S_1$ is closed and $S_2$ is open. This specific configuration isolates the middle branch, meaning no current can flow through capacitors $C_1$ and $C_2$.

The battery $V_0$ is directly connected across the rightmost branch, which contains only capacitor $C_3$.

Since $C_3$ is fully charged by the $8	ext{V}$ battery, we can easily calculate its initial charge using the fundamental relation $Q = CV$.

$$Q_0 = C_3 V_0 = (1.0 \mu	ext{F})(8 	ext{V}) = 8 \mu	ext{C}$$

This $8 \mu	ext{C}$ is the total charge available in our system for the next phase of the experiment.

### The Switch: A New Loop Emerges

The physical setup changes dramatically when $S_1$ is opened and $S_2$ is closed. Opening $S_1$ completely removes the battery from the active circuit, ensuring that the total charge trapped in the capacitors remains conserved.

Closing $S_2$ connects the middle branch to the right branch, forming a new closed loop. In this loop, the charge stored in $C_3$ acts as a source, redistributing itself to the uncharged capacitors $C_1$ and $C_2$.

Because the charge leaving $C_3$ has only one path to follow, it must flow through both $C_1$ and $C_2$ sequentially. This tells us that **$C_1$ and $C_2$ are in series**, and they will acquire the exact same amount of charge.

We are given that the final charge on $C_3$ is $5 \mu	ext{C}$. By the principle of conservation of charge, the remaining charge must have flowed to the middle branch.

$$q = Q_0 - q_3 = 8 \mu	ext{C} - 5 \mu	ext{C} = 3 \mu	ext{C}$$

Thus, both $C_1$ and $C_2$ now hold a charge of $3 \mu	ext{C}$.

### The Math: Balancing the Voltages

In our final closed loop, the series combination of $C_1$ and $C_2$ is connected in parallel with $C_3$. In any parallel circuit, the potential difference across the branches must be equal.

This gives us our master equation: the voltage across $C_3$ equals the sum of the voltages across $C_1$ and $C_2$.

$$V_3 = V_1 + V_2$$

Substituting $V = \frac{Q}{C}$ for each capacitor, we get:

$$\frac{q_3}{C_3} = \frac{q_1}{C_1} + \frac{q_2}{C_2}$$

Remember that $C_1$ is filled with a dielectric of relative permittivity $\varepsilon_r$, so its capacitance is $\varepsilon_r 	imes 1.0 \mu	ext{F}$. Plugging in our known values:

$$\frac{5}{1} = \frac{3}{\varepsilon_r} + \frac{3}{1}$$

$$5 = \frac{3}{\varepsilon_r} + 3$$

$$2 = \frac{3}{\varepsilon_r}$$

Solving for the unknown dielectric constant, we find:

$$\varepsilon_r = \frac{3}{2} = 1.50$$

This elegant result showcases how charge conservation and Kirchhoff's loop rules seamlessly work together to reveal hidden properties of materials!

Enter Numerical Value:

Visualized Solution

$S_{1} Closed, S_{2} Open$

$Q = C V$

$Q_{0} = C_{3} V_{0}$

$Q_{0} = 8 μ C$

$S_{1} Open, S_{2} Closed$

$V_{3} = V_{1} + V_{2}$

$q = Q_{0} - q_{3}$

$\frac{q _{3}}{C _{3}} = \frac{q _{1}}{C _{1}} + \frac{q _{2}}{C _{2}}$

$\frac{5}{1} = \frac{3}{ε _{r}} + \frac{3}{1}$

$2 = \frac{3}{ε _{r}}$

$ε_{r} = 1.50$

\text{Symmetry in Series}

The Sigma Insight: Combination of Capacitors

The Initial State

The Switch

The Math

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​ Closed, S2​ Open

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q=CV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q0​=C3​V0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q0​=8μC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​ Open, S2​ Closed

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V3​=V1​+V2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q=Q0​−q3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C3​q3​​=C1​q1​​+C2​q2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 15​=εr​3​+13​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2=εr​3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } εr​=1.50