Q: What would happen to the capacitance value, C, for a capacitor if the charge, q, were doubled?

A: The MCAT loves proportions, but this is a trick question. Possible answer choices could be:

a. double
b. triple
c. no change
d. cannot be determined

Without a good understanding of capacitance, many might choose a, but that is the incorrect answer. Although q and C are directly proportional, capacitance is dependent on the geometry of the capacitor. Thus, the correct answer would be c.

Q: Suppose you have two capacitors C1 = 1F and C2 = 2F. C1 is charged to a voltage of 5 V by a battery. The battery is disconnected from C1 and C1 is then connected directly to C2. What will be the potential across each capacitor?

A: This question is a bit more involved. When C1 is connected to C2, C1 will begin to discharge. The total charge on C1 will begin to disperse between the two capacitors. After all, where else can the charge on C1 go? At equilibrium, the potential differences across each capacitor will be the same. Knowing this you can solve the problem.

qtot = q1 + q2

C1Vinitial = C1V + C2V

V = VinitialC1/C1 + C2

V = 5/3 volts or about 1.67 volts

It is important to note that charge cannot be destroyed—thus the initial charge should equal the final charge.

Q: Suppose you have a capacitor connected to a battery and introduce a dielectric during charging. What happens to the potential difference and capacitance?

A: The capacitance always increases if a dielectric is introduced regardless of whether the battery is still connected. But what happens to the potential difference? It stays the same. How can it increase or decrease if it is still connected to the battery? It will be at the same potential as the battery.

There are many different ways to store potential energy. You can stretch a spring, compress a gas or lift an object over your head. You can also use an electric field to store potential energy and a capacitor does just that. Capacitors have many uses in electronic devices. Examples include computers and the defibrillator. Capacitors also come in different shapes—from the parallel plate capacitor to even spherical and cylindrical. For the MCAT, the parallel plate capacitor is most important.

The standard parallel plate capacitor can be described by two parallel plates separated by some distance. The plates have area A and the distance between them is d. When an external agent, like a battery, charges a capacitor the plates acquire equal but opposite charge. To calculate the charge (q) acquired on any plate:

Where C is the capacitance and V is the voltage. The voltage is the potential difference between the plates. The MCAT likes proportions so try the following question:

Q: The charge is doubled but the potential difference between the plates is kept constant. What happens to the capacitance?

A: Since this is a proportion, and q = C, then the capacitance will double. WRONG!! The capacitance stays the same!

The value of the capacitance is dependent on the geometry of the capacitor and is independent of the charge and potential difference (voltage). For a parallel plate capacitor the capacitance (C) is equal to:

C = єA/d

Where є is the permittivity constant, A is the area of the plate and d is the distance between the plates. Gauss’ law is needed to derive this equation (surprisingly Gauss’ law is a topic on the AAMC list of topics for the MCAT). Don’t worry about the value of the permittivity constant because it will be provided on the MCAT if it necessary—which is very unlikely because you are more likely to encounter a proportion question. From the equation if the area of the plates is increased the capacitance will increase but if the distance between the plates is increased the capacitance will decrease. A larger area means more room for charge to be stored—more charge means more energy. But, what about the distance—why does an increase mean a decrease in capacitance? That will be addressed in a bit!

What if you put capacitors in parallel or series?

Case 1: Capacitors in parallel arrangement.

When capacitors are connected in a parallel fashion to a battery source, all the capacitors are at the same potential. This potential is the voltage of the battery source. How do you calculate the equivalent capacitance for two capacitors, C1 and C2, in parallel? For resistors in parallel, the equivalent resistance is 1/Req = 1/R1+…..1/Rn. That is not true for capacitors. For capacitors in parallel, the potential difference is the same across each. Therefore:

The total charge is then:

q = q1 + q2 = (C1 + C2)V

C1 + C2 represents the equivalent capacitance, Ceq, so:

q = CeqV ——– Ceq = q/V = C1 + C2

Thus, for capacitors arranged in a parallel fashion, the equivalent capacitance is found by:

Case 2: Capacitors in series arrangement.

When capacitors are arranged in series to a battery, the total potential difference across all the capacitors is equal to the voltage of the battery. This means that all capacitors end up with the same charge, q. How is this? Consider the case where you have two capacitors, C1 and C2, connected in series to a battery source. The negative terminal of the battery is connected to C2 and the positive terminal to C1. This means that the bottom plate of C2 will acquire a negative charge. As the bottom plate acquires negative charge, this will repel negative charge on the top plate of C2. Due to repulsion, the top plate of C2 will acquire a positive charge. The repelled negative charge will move to the bottom plate of C1. The charge repelled will be the magnitude of the charge on C2. Thus, both capacitors end up having the same charge. How do you find the equivalent capacitance? The potential difference across each capacitor:

V = V1 + V2 = q ( 1/C1 + 1/C2) ——– (1/C1 + 1/C2) = Ceq

Ceq = q/V = 1/(1/C1 + 1/C2)

Thus,

1/Ceq = 1/C1 + 1/C2

For capacitors in series, the equivalent capacitance is then found by:

So, how exactly is energy stored in a capacitor? As charge builds up on the plates of the capacitor, equal but opposite charge, work is being done by an external agent to charge the capacitor. As equal but opposite charge builds up, an electric field builds up between the plates of the capacitor. Remember, positive and negative charges want to attract one another, but in order to separate the two opposite charges external work is needed. For capacitors, this is most often done by a power source such as a battery. Since opposite charges are being separated by one another, potential energy is being stored within the electric field. Work is done by the battery over the distance of the plates in order to store this potential energy in the electric field. This is why as the distance between the plates of a parallel plate capacitor is increased, the capacitance, the ability to store charge, decreases. More work needs to be done by the battery as charge builds up on the plates and over time it becomes more difficult to charge the capacitor. Over time, the potential energy stored in the capacitor is:

What if you put some material between the plates of the capacitor? What happens to its ability to store charge? This is the case of a dielectric material. Will the capacitance increase as a result of a dielectric? The answer is YES! For the MCAT you don’t need to understand why but it is the result of the electric field between the plates decreasing. Since the electric field is less between the plates due to a dielectric, more charge can be stored and the capacitance increases. The dielectric, depending on the material, is some constant denoted by: қ. For a parallel plate capacitor with a dielectric, the capacitance is given by:

C = є қ A/d——the capacitance increases by the dielectric constant!!

What is electric potential? By definition, electric potential is the potential energy per unit charge. The unit for electric potential is the volt (V). According to the definition of electric potential, the unit can also be expressed as (J/C). This is very important because the MCAT may use both interchangeably. So, what exactly does electric potential mean? The best way to explain electric potential is to use the concepts that govern the macroscopic world and to apply them to the microscopic world of charges. Objects possess some charge that can be positive, negative or neutral. The charge which constitutes the objects sets up an electric field around the object that acts as a “detector.” This electric field is what allows the object to detect neighboring objects or fields. If the field points outward than the object has a positive net charge and if the field points inward the object has a negative net charge. If another charged object were to enter the field, an electric force would act on the object accelerating it in some direction. Which direction? The direction of the net force acting on the charged object—this is what Newton’s second law tells us. But, how do you determine direction? Imagine that you have a positive source charge. The direction of the electric field for the source charge will be outward. Now, you place a positive test charge in the electric field. What happens? The direction of the net force acting on the test charge is outward since it will repel the source charge. Thus, it will accelerate in the direction of the field. The opposite will happen with a negative test charge—it will accelerate in the direction opposite of the electric field. You can calculate the magnitude of the force using the following equation:

F=qE

Where q is the test charge and E is the magnitude of the electric field of the source charge.

The electric field is a vector approach to explain the behavior of charged objects (similar to velocity and moving objects). Electric potential is a scalar approach to explain the behavior of charged objects (similar to kinetic energy and moving objects). From Newton’s second law, if an object moves in the direction of the net force acting on it, the velocity of the object will increase. This means the kinetic energy will increase. Conversely, if the object moves opposite the direction of the net force acting on it, the kinetic energy will decrease. For charged objects, the electric potential explains the energy of an object with respect to its position in some electric field. So, how do you explain the potential and kinetic energy of a charged object? Consider the macroscopic world. If you were to lift a book from the ground over your head, what kind of energy does the book gain? It gains potential energy. If you were to then drop the book from above your head, what happens? The potential energy translates into kinetic energy. When you lift the book over your head, you are doing something the book would not normally do. When you drop the book from the top of your head, the book is doing what it wants to do—to move in the direction of gravity and to the floor. This means that when an object does something it “naturally” will not do, it will gain potential energy. When an object does something it “naturally” wants to do, it will gain kinetic energy (This makes sense according to Newton’s first law). This same idea can be applied to the microscopic world. When a positive test charges moves in the direction of the electric field due to a positive source charge, the test charge is doing something it naturally wants to do, potential energy is lost and kinetic energy is gained. This means for the system to gain potential energy some external agent needs to act on the system to push the two “like” charges near one another. Only then will potential energy be gained.

In a nutshell:

• When two like charges repel, this is a natural tendency. Thus potential energy is lost and kinetic energy is gained.

• When opposite charges attract, this is a natural tendency. Thus potential energy is lost and kinetic energy is gained.

• In order to move two like charges near one another, this will not naturally happen because they will repel, external work needs to be done on the system. This means potential energy is gained and kinetic energy is lost.

So now that the concepts are explained, let me introduce some equations. The mathematical representation for electric potential is:

V = U/q

Where V is the electric potential, U is the potential energy and q is the charge.

The potential energy of a test charge is:

U = qV.
This will be negative when potential energy is lost and positive when potential energy is gained. This is a very important point for the MCAT because when asked to calculate the potential energy of the system, you can knock out choices you know are wrong based on what you know of the tendency of the system. This saves time on the test and prevents unnecessary calculations. Work is defined as:

W = -ΔU

Thus negative work means a positive change in potential energy. This is another important point for the MCAT.

I think I will tackle the question on electric fields, electric potential and work done in an electric field in parts. First, I will discuss electric fields.

What is an electric field? Imagine you have a positive point charge and place a test positive charge in its vicinity. Coulomb’s law tells us that there is repulsion between the two charges. How exactly do the two charges feel one another? Then answer is the positive point charge sets up an electric field in the space around it. At any point within the space, the electric field has both a magnitude and a direction. The magnitude is dependent on the value of the charge and direction depends on the electrical charge. The electric field is a vector field and is defined for each point in space. By definition, the electrostatic force a test charge qo will feel in the vicinity of a point charge is:

E = F/ qo

The unit for electric field is N/C.

How do we draw electric field lines? Electric field lines give us a representation of the electric field. Say we have a sphere of uniform negative electric charge. If a positive test charge were placed in the field, the test charge would feel an electrostatic force pointed toward the sphere. This tells us that for negative charges, electric field lines point in. What if we have a sphere of uniform positive electric charge? If a positive test charge were placed in the field, the test charge would feel an electrostatic force pointed away from the sphere. Therefore, for positive charges, electric field lines point outward. How do electric field lines relate to electric field vectors? Simple! The electric field vector is tangent to the electric field line. For straight field lines, this means the electric field is that direction!

What is the electric field due to a point charge? Instead of a uniform sphere of charge, suppose we have a point charge and place a test charge qo in the vicinity. How would we calculate the electric field? From above, we know E = F/ qo. F is the electrostatic force which is given by Kq qo/r2, where r is the distance between the charges. Using the equation above, we see this reduces to: Kq/r2. Now, for some important points:
1. The above equation gives us the value for the electric field for all points in space for a point charge
2. The electric field is a vector—VERY IMPORTANT!!!
3. If there is more than one point charge, the electric fields are additive.

What would happen if we were to place a test charge in an electric field produced by a moving charge? Well, the electrostatic force that the test charge would feel is given by:

F=qE

Where q is the electric charge of the particle, and E is the electric field produced by the moving charge. Examining this equation, we see that if a negative charge is placed in the electric field produced by a positive charge, there is a negative electrostatic force due to attraction (This attraction can be verified using Coulomb’s law). Therefore, the direction of the electrostatic force has the direction of the electric field if the test charge is positive and will have the opposite direction if the test charge is negative. This is important! If on the MCAT, you are given a diagram of a electric field and given a test charge, and are asked the direction of the electrostatic force, you can use the procedure outlined above. Given all this information, you can also calculate the acceleration of the test charge. If the only force acting on the test charge is the electrostatic force, than you can set that value equal to F=ma, where F is the sum of the net forces. This can all be related to Newton’s second law!!

Question: (imagine this was a MCAT question) What if you are given a diagram of a negative test charge in a electric field pointing outward and you asked to determine whether the acceleration would increase if the test charge were moving in the direction of the electric field? First, if the electric field is pointing outward, than the point charge is positive. From Coulomb’s law, we know there is attraction between the test charge and point charge. Therefore, the electrostatic force, the net force, should be pointing inward. However, the test charge is moving away from the point charge. Since the electric field decreases with distance (Kq/r2), this means if qE=ma, than the acceleration will decrease. That’s it! On the MCAT, you would circle the answer choice with decreases!

Now, for a real world application of electric fields and charge. Believe it or not, you can use this analysis for ink-jet printers. You don’t need to know this for the MCAT!!!

In the printer, when the print command is received, a generator generates drops. These drops then pass through a charge unit and receive a charge q. These drops then pass through a pair of conducting plates where the electric field points down from the top plate to the bottom plate. Once these drops pass through the plates, they are deflected land on the paper. By varying the magnitude of the electric field and charge of the drop, you get drops that land at different points on the paper. When this is done thousands of times, you get a printed copy of your work!!! (FYI: This is how older versions of ink-jet printers operate. There is much more to how they operate, but this is the basics!!) Who knows, a passage such as this might be on the MCAT!!!