## MCAT Physics and Math Review

## Chapter 6: Circuits

### 6.3 Capacitance and Capacitors

Aside from batteries and resistors, the other major circuit element tested on the MCAT is the capacitor. **Capacitors** are characterized by their ability to hold charge at a particular voltage. There are excellent real-world examples of capacitors. Perhaps the most important capacitor you’ll encounter in the clinics is the defibrillator. While a defibrillator is charging, a high-pitched electronic tone sounds as electrons build up on the capacitor. When the defibrillator is fully charged, that charge can be released in one surge of power (after the operator yells *Clear!*). The clouds and the ground during a lightning storm also act as a capacitor, with the charge building up between them eventually **discharging** as a bolt of lightning. The MCAT focuses on a particular type of capacitor called a parallel plate capacitor, and all of our discussion will center on capacitors of this type.

PROPERTIES OF CAPACITORS

When two electrically neutral metal plates are connected to a voltage source, positive charge builds up on the plate connected to the positive (higher potential) terminal, and negative charge builds up on the plate connected to the negative (lower potential) terminal. The two-plate system is a capacitor because it can store a particular amount of charge at a particular voltage. The **capacitance** of a capacitor is defined as the ratio of the magnitude of the charge stored on one plate to the potential difference (voltage) across the capacitor. Therefore, if a voltage *V* is applied across the plates of a capacitor and a charge *Q* collects on it (with +*Q* on the positive plate and –*Q* on the negative plate), then the capacitance is given by

**Equation 6.13**

The SI unit for capacitance is the **farad** . Because one coulomb is such a large quantity of charge, one farad is a very large capacitance. Capacitances are usually given in microfarads (1 *μ*F = 1 × 10^{–6} F) or picofarads (1 pF = 1 × 10^{–12} F). Be careful not to confuse the farad with the Faraday constant, F, which is the amount of charge in one mole of electrons .

The capacitance of a parallel plate capacitor is dependent upon the geometry of the two conduction surfaces. For the simple case of the parallel plate capacitor, the capacitance is given by

**Equation 6.14**

where *ε*_{0} is the permittivity of free space *A* is the area of overlap of the two plates, and *d* is the separation of the two plates. The separation of charges sets up an electric field between the plates of a parallel plate capacitor. The electric field between the plates of a parallel plate capacitor is a **uniform electric field** with parallel field vectors, the magnitude of which can be calculated as

**Equation 6.15**

**KEY CONCEPT**

If you look back to the equations discussed in Chapter 5 of *MCAT Physics and Math Review*, you can see that the equation for *E* here can be derived from the other fundamental electrostatics equations. If and , then *V* = *E* × *r*. *r* in this setup is the distance between the plates, *d*, so we can rewrite this as *V* = *Ed*.

The direction of the electric field at any point between the plates is from the positive plate toward the negative plate. If we imagine placing a positively charged particle between the oppositely charged plates, we would expect the particle to accelerate in that same direction. This should not be surprising, as electric field lines always point in the direction a force would be exerted on a positive charge.

Regardless of the particular geometry of a capacitor (parallel plate or otherwise), the function of a capacitor is to store an amount of energy in the form of charge separation at a particular voltage. This is akin to the function of a dam, the purpose of which is to store gravitational potential energy by holding back a mass of water at a given height. The potential energy stored in a capacitor is

**Equation 6.16**

DIELECTRIC MATERIALS

The term **dielectric material** is just another way of saying insulation. When a dielectric material, such as air, glass, plastic, ceramic, or certain metal oxides, is introduced between the plates of a capacitor, it increases the capacitance by a factor called the **dielectric** **constant** (** κ**). The dielectric constant of a material is a measure of its insulating ability, and a vacuum has a dielectric constant of 1, by definition. For reference, the dielectric constant of air is just slightly above 1, glass is 4.7, and rubber is 7. These numbers need not be memorized; any relevant dielectric constants will be given on Test Day.

The capacitance due to a dielectric material is

*C′* = *κC*

**Equation 6.17**

where *C′* is the new capacitance with the dielectric present and *C* is the original capacitance.

**KEY CONCEPT**

A dielectric material can never decrease the capacitance; thus *κ* can never be less than 1.

*Dielectrics in Isolated Capacitors*

When a dielectric material is placed in an isolated, charged capacitor—that is, a charged capacitor disconnected from any circuit—the voltage across the capacitor decreases. This is the result of the dielectric material shielding the opposite charges from each other. By lowering the voltage across a charged capacitor, the dielectric has increased the capacitance of the capacitor by a factor of the dielectric constant. Thus, when a dielectric material is introduced into an isolated capacitor, the increase in capacitance arises from a decrease in voltage.

*Dielectrics in Circuit Capacitors*

When a dielectric material is placed in a charged capacitor within a circuit—that is, still connected to a voltage source—the charge on the capacitor increases. The voltage must remain constant because it must be equal to that of the voltage source. By increasing the amount of charge stored on the capacitor, the dielectric has increased the capacitance of the capacitor by a factor of the dielectric constant. Thus, when a dielectric material is introduced into a circuit capacitor, the increase in capacitance arises from an increase in stored charge.

The stored energy in a capacitor is only useful if it is allowed to discharge. The charge can be released from the plates either by discharging across the plates or through some conductive material with which the plates are in contact. For example, capacitors can discharge into wires, causing a current to pass through the wires in much the same way that batteries cause current to move through a circuit. The paddles of the defibrillator machine, once charged, are placed on either side of a patient’s heart that has gone into a life-threatening arrhythmia (such as *ventricular fibrillation*). The reason the doctor yells *Clear!* before discharging the paddles is because the current needs to travel through the patient’s heart—not through any other people who might be touching the patient and creating a parallel pathway. On a much larger scale, lightning occurs when a very, very large amount of charge exceeds the capacitance of the Earth’s surface and the underside of the cloud (the two serving, approximately, as a parallel plate capacitor). The large rapid discharge across the plates of a capacitor is considered a failure of the capacitor, while creating a current through the attached wires is the normal function of a capacitor.

**Example:**

The voltage across the terminals of an isolated 3 *µ*F capacitor is 4 V. If a piece of ceramic having dielectric constant *κ* = 2 is placed between the plates, find the new charge, capacitance, and voltage of the capacitor.

**Solution:**

The introduction of a dielectric by itself has no effect on the charge stored on the isolated capacitor. There is no new charge, so the charge is the same as before. The charge stored is therefore given by

*Q′* = *Q* = *CV* = (3 *µ*F)(4 V) = 12 *µ*C

By introducing a dielectric with a dielectric constant of 2, the capacitance of the capacitor is multiplied by 2 (*C′* = *κC*). Hence, the new capacitance is 6 *µ*F.

Now, the new voltage across the capacitor can be determined:

**Example:**

A 3 *µ*F capacitor is connected to a 4 V battery. If a piece of ceramic having dielectric constant *κ* = 2 is placed between the plates, find the new charge, capacitance, and voltage of the capacitor.

**Solution:**

This question is very similar to the previous one, but the voltage is held constant here by a battery. Thus, the new voltage is still 4 V.

By introducing a dielectric with a dielectric constant of 2, the capacitance of the capacitor is multiplied by 2 (*C′* = *κC*). Hence, the new capacitance is 6 *µ*F.

Now, the new charge on the capacitor can be determined:

*Q′* = *C′V′* = (6 *µ*F)(4 V) = 24 *µC*

CAPACITORS IN SERIES AND PARALLEL

Just like resistors, capacitors can be arranged within a circuit either in parallel or in series. They can also be arranged with resistors, although this is beyond the scope of the MCAT in most cases.

*Capacitors in Series*

When capacitors are connected in series, the total capacitance decreases in similar fashion to the decreases in resistance seen in parallel resistors, as shown in Figure 6.3.

**Figure** **6.3.** **Capacitors in Series** C_{s} decreases as more capacitors are added.

This is because the capacitors must share the voltage drop in the loop and therefore cannot store as much charge. Functionally, a group of capacitors in series acts like one equivalent capacitor with a much larger distance between its plates (in fact, with a distance equal to those of each of the series capacitors added together). This increase in distance, as seen earlier, means a smaller capacitance.

Rather than memorizing the following equations independently, understand the conceptual basis for the mathematics of resistors in series and in parallel, and then simply reverse that mathematical approach for capacitors. The equation for calculating the equivalent capacitance for capacitors in series is

**Equation 6.18**

which shows that *C*_{s} decreases as more capacitors are added. Note that for capacitors in series, the total voltage is the sum of the individual voltages, just like resistors in series.

*Capacitors in Parallel*

Capacitors wired in parallel, shown in Figure 6.4, produce a resultant capacitance that is equal to the sum of the individual capacitances.

**Figure** **6.4.** **Capacitors in Parallel** C_{p} increases as more capacitors are added.

Therefore, *C*_{p} increases as more capacitors are added:

*C*_{p} = *C*_{1} + *C*_{2} + *C*_{3} + ⋯ + *Cn*

**Equation 6.19**

Just as we saw with resistors in parallel, the voltage across each parallel capacitor is the same and is equal to the voltage across the entire combination.

**MCAT Concept Check 6.3:**

Before you move on, assess your understanding of the material with these questions.

1. Assuming the plates are attached by a conducting material, how does a capacitor behave after the voltage source has been removed from a circuit?

2. How does a dielectric material impact capacitance? Voltage? Charge?

· Capacitance:

· Voltage:

· Charge:

3. How does adding or removing a capacitor change the total capacitance of a circuit with capacitors in series? In parallel?

· Series:

· Parallel:

4. What physical qualities contribute to the capacitance of a capacitor?