Special Geometries for Electrostatics for AP Physics B & C (page 2)
Practice problems for these concepts can be found at:
There are two situations involving electric fields that are particularly nice because they can be described with some relatively easy formulas. Let's take a look:
If you take two metal plates, charge one positive and one negative, and then put them parallel to each other, you create a uniform electric field in the middle, as shown in Figure 20.3:
The electric field between the plates has a magnitude of
V is the voltage difference between the plates, and d is the distance between the plates. Remember, this equation only works for parallel plates.
Charged parallel plates can be used to make a capacitor, which is a charge-storage device. When a capacitor is made from charged parallel plates, it is called, logically enough, a "parallel-plate capacitor." A schematic of this type of capacitor is shown in Figure 20.4.
The battery in Figure 20.4 provides a voltage across the plates; once you've charged the capacitor, you disconnect the battery. The space between the plates prevents any charges from jumping from one plate to the other while the capacitor is charged. When you want to discharge the capacitor, you just connect the two plates with a wire.
The amount of charge that each plate can hold is described by the following equation:
Q is the charge on each plate, C is called the "capacitance," and V is the voltage across the plates. The capacitance is a property of the capacitor you are working with, and it is determined primarily by the size of the plates and the distance between the plates, as well as by the material that fills the space between the plates. The units of capacitance are farads, abbreviated F ; 1 coulomb/volt = 1 farad.
The only really interesting thing to know about parallel-plate capacitors is that their capacitance can be easily calculated. The equation is:
In this equation, A is the area of each plate (in m3), and d is the distance between the plates (in m). The term ε0 (pronounced "epsilon-naught") is called the "permittivity of free space." This term will show up again soon, when we introduce the constant k. The value of ε0 is 8.84 × 10–12 C/V· m, which is listed on the constants sheet.
Capacitors become important when we work with circuits. So we'll see them again in Chapter 21.
As much as the writers of the AP exam like parallel plates, they love point charges. So you'll probably be using these next equations quite a lot on the test.
But, please don't go nuts … The formulas for force on a charge in an electric field (F = qE ) and a charge's electrical potential energy (PE = qV ) are your first recourse, your fundamental tools of electrostatics. On the AP exam, most electric fields are NOT produced by point charges! Only use the equations in this section when you have convinced yourself that a point charge is creating the electric field or the voltage in question.
First, the value of the electric field at some distance away from a point charge:
Q is the charge of your point charge, k is called the Coulomb's law constant (k = 9 × 109 N · m2/C2), and r is the distance away from the point charge. The field produced by a positive charge points away from the charge; the field produced by a negative charge points toward the charge. When finding an electric field with this equation, do NOT plug in the sign of the charge or use negative signs at all.
Second, the electric potential at some distance away from a point charge:
When using this equation, you must include a + or – sign on the charge creating the potential (see Figure 20.5).
And third, the force that one point charge exerts on another point charge:
In this equation, Q1 is the charge of one of the point charges, and Q2 is the charge on the other one. This equation is known as Coulomb's Law.
Sometimes, the constant k is written as
Writing it this way is very helpful for students in Physics C (we'll discuss why later), but Physics B students shouldn't worry about it too much.
To get comfortable with these three equations, we'll provide you with a rather comprehensive problem.
Yikes! This is a monster problem. But if we take it one part at a time, you'll see that it's really not too bad.
Part 1—Electric field
Electric field is a vector quantity. So we'll first find the electric field at point "P" due to charge "A," then we'll find the electric field due to charge "B," and then we'll add these two vector quantities. One note before we get started: to find r, the distance between points "P" and "A" or between "P" and "B," we'll have to use the Pythagorean theorem. We won't show you our work for that calculation, but you should if you were solving this on the AP exam.
Note that we didn't plug in any negative signs! Rather, we calculated the magnitude of the electric field produced by each charge, and showed the direction on the diagram.
Now, to find the net electric field at point P, we must add the electric field vectors. This is made considerably simpler by the recognition that the y-components of the electric fields cancel … both of these vectors are pointed at the same angle, and both have the same magnitude. So, let's find just the x-component of one of the electric field vectors:
Some quick trigonometry will find cos θ … since cos θ is defined as , inspection of the diagram shows that . So, the horizontal electric field Ex = (510 m) … this gives 140 N/C.
And now finally, there are TWO of these horizontal electric fields adding together to the left —one due to charge "A" and one due to charge "B". The total electric field at point P, then, is
- 280 N/C, to the left.
The work that we put into Part 1 makes this part easy. Once we have an electric field, it doesn't matter what caused the E field—just use the basic equation F = qE to solve for the force on the electron, where q is the charge of the electron. So,
- F = (1.6 × 10–19 C) 280 N/C = 4.5 × 10–17 N.
The direction of this force must be OPPOSITE the E field because the electron carries a negative charge; so,
The nice thing about electric potential is that it is a scalar quantity, so we don't have to concern ourselves with vector components and other such headaches.
The potential at point "P" is just the sum of these two quantities. V = zero!
Notice that when finding the electric potential due to point charges, you must include negative signs … negative potentials can cancel out positive potentials, as in this example.
Practice problems for these concepts can be found at:
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