Return to capacitors in series notes

## Example: Capacitors in Series

Suppose you have the following circuit. Capacitors are labeled 1 and 2 for convenience of reference. You know that the circuit contains a 12-Volt battery, $Q_1=4.5 \mu\text{C}$, and $C_2=0.5 \mu\text{F}$. What is the capacitance of Capacitor 1? What happens to the charge on *both* of the capacitors if we insert a dielectric material with dielectric constant $k = 3$ into Capacitor 1?

#### Facts

- $\Delta V_{\text{bat}} = 12\text{ V}$
- $Q_1 = 4.5 \mu\text{C}$
- $C_2 = 0.5 \mu\text{F}$
- $k = 3$

#### Lacking

- $C_1$
- How the charges change when the dielectric is inserted.

#### Approximations & Assumptions

- The wire has very very small resistance when compared to the other resistors in the circuit.
- We are measuring things like potential differences and charges when the circuit is in a steady state.
- Approximating the battery as a mechanical battery.
- Capacitors are parallel plate capacitors.

#### Representations

- We represent capacitance as $$C=\frac{Q}{\Delta V}$$
- We represent the equivalent capacitance of multiple capacitors arranged in series as

$$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots$$

- We represent the situation with diagram given above.

### Solution

#### Part 1

Let's find $C_1$. In order to use the equation for equivalent capacitance of capacitors in series, as we have here, we first need the equivalent capacitance of the entire circuit. Remember that the charge on the capacitors in series should be the same, so $Q_{\text{equiv}}=Q_1=Q_2$. Now, we can write: $$C_{\text{equiv}}=\frac{Q_{\text{equiv}}}{\Delta V_{\text{bat}}}=0.375 \mu\text{F}$$

Now we can solve for $C_1$ using $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}$$ This gives us $C_1 = 1.5 \mu\text{F}$

#### Part 2

Now we need to consider what happens when we insert a dielectric. It might look something like the circuit to the right. A description of what a dielectric does in a capacitor is here. Its effect on capacitance is: $$C =\frac{k\epsilon_0 A}{d}$$

So when we insert the dielectric, we have a new capacitance for Capacitor 1: $C_{\text{1, new}}=kC_1=4.5 \mu\text{F}$. To find the new value that the capacitors are charged to, we return to the equivalent capacitance of the circuit: $$\frac{1}{C_{\text{equiv, new}}}=\frac{1}{C_{\text{1, new}}}+\frac{1}{C_2}$$ This yields $C_{\text{equiv, new}}=0.45 \mu\text{F}$. Now, we can find the new charge: $$Q_{\text{new}} = C_{\text{equiv, new}}\Delta V_{\text{bat}} = 5.4 \mu\text{C}$$

**This is the charge on both capacitors** since the capacitors are in series. So even if we insert a dielectric in only one of the capacitors, the charge on both will increase.