Parallel Plate Capacitor: Potential Difference Vs. Spacing

LivePhoto Physics Activity 29 Name: ______Page 1 of 4 Parallel Plate Capacitor: Potential Difference vs. Spacing

In this assignment you will consider how a charged capacitor constructed from a fairly large pair of parallel metal plates behaves when the spacing between the plates increases. In particular we want you to explore how the voltage (a.k.a. potential difference) across the plates varies as the distance between the charged plates increases. Theoretical Review: [1] A capacitor is defined as any two conductors, separated by an insulator d where each conductor carries a net excess charge that is equal in magnitude and opposite in sign. Its capacitance, C, is defined as Q C  [Eq. 1] V where Q is the magnitude of the excess charge on each conductor and V is the voltage (or potential difference) across the plates. [2] We can use Gauss’ Law to analyze a parallel plate capacitor if we assume that most of the electric field lines is perpendicular to the plates. According to Gauss, if air is the insulator, the capacitance, C, is related to Figure 1: The spacing, d, between the surfaces of two parallel metal plates is changed while the the area of the plates, A, and the spacing between voltage between the plates is measured. them, d, by the equation  A C  0 . [Eq. 2] d

 0 is known as the electric constant (or permittivity).

1. Preliminary Questions   Note: You will receive full credit for each prediction made in this preliminary section whether or not it matches conclusions you reach in the next section. As part of the learning process it is important to compare your predictions with your results. Do not change your predictions!

As you proceed with this assignment, you’ll be working with a short video clip entitled . It shows the spacing between a pair of parallel plates increasing by a known amount each time a knob is twisted by a half turn. The potential difference between the plates is measured with both an electrometer and voltage probes connected to a computer (see Fig. 1) so data can be recorded with the Logger software. Before proceeding, you should view the video clip. (a) Use Equations 1 and 2 to derive an equation that describes how the voltage across a parallel plate capacitor depends on the plate spacing, d, and area, A. Show your work.

(b) If the plate spacing increases by 1.25 mm for each full turn of a knob, by what amount does the spacing change in millimeters (mm) when the knob goes through half a turn?

Physics with Video Analysis 29 - 1 (c) You should have noticed in frame 5 of the movie that only one plate of the capacitor is being charged positively. The other plate is “grounded.” The right-hand plate and the left-hand plate are separated by only one-half of a turn of the dial. How can we claim this set up is actually a capacitor and thus has equal and opposite excess charges on both plates? Hint: What happens immediately after the first plate is charged? Is induction possible?

(d) What do you expect will happen to the capacitor system as the spacing between the plates increases? Hint: Refer to the equation you derived in section 1(a). The potential difference will: Increase Decrease Stay the same

The excess charge on each plate will: Increase Decrease Stay the same

(e) When the potential difference is large the capacitor system is storing more energy than when it is small. Where does the additional energy “come from” as the plate spacing increases?

(f) If the two plates behave like an ideal capacitor, sketch the shape of a graph of voltage vs. spacing you might expect to measure and explain your reasoning.

29 - 2 Physics with Video Analysis 2. Activity-Based Questions In this section, you will start by working with the Logger Pro file that has the movie inserted in it. The electrometer shown in the background has numbers that are too small to read, but you can easily see the deflection of the needle. However, separate Logger Pro voltage probe readings were recorded as the movie was made. In the < SensorDataSet.cmbl> file the movie and the voltage readings have been synchronized. Although the capacitor plates look as if they are in contact at first, they are actually separated by 0.625 mm or a “half turn.” After a few seconds the left capacitor plate is charged to about 13 volts relative to the right plate. Then the spacing between the two plates is increased a half turn at a time. You can see this process by opening the Logger Pro file and clicking on the Start button in the Replay window.

(a) Enter data into Logger Pro for voltage vs. plate spacing: In use the Examine tool to find a typical value for the voltage between the plates at each spacing. Enter the results in the Manual Entries table below. Also sketch the resulting graph of V vs. d. Note: We’ve assumed for you that when the plates are in contact at d = 0 mm the charges neutralize, so V = 0.0 v.

Table 1–Manual Entries

Rotations D V_readings (# of turns) (mm) (volts)

(b) Is this an ideal capacitor? Close your original Logger Pro file and then open a second file entitled . Enter the data from Table 1 into the designated columns manually. A graph of V_readings vs. d will appear. Use the Curve Fit tool in the Logger Pro Analyze menu to determine whether the graph seems to show a proportional relationship between V and d. Hint: Do not perform a linear fit. Instead select Curve Fit from the Analyze menu and select a Proportional Fit (denoted y = Ad). Does the capacitor act like an ideal capacitor? Explain.

Physics with Video Analysis 29 - 3 (c) Find the area of the plates in square millimeters. The faces of the capacitor plates are disks. Use the video analysis tools to find the data needed to calculate the area, A, of each plate. Then calculate A in mm2 to three significant figures. Show your calculations. Hints: (1) The yellow ruler at the bottom of each frame is 20.0 cm long, and (2) The Enable Video Analysis ( ), Set Scale ( ), and Photo Distance ( ) tools will come in handy!

(d) Convert the area of the plates from mm2 to m2.

(e) Determine excess charge on the capacitor. When d = 1.25 mm, V is about 25 V. Rearrange the equation you derived in Section 1(a) and calculate the magnitude of the excess charge on each plate in Coulombs. Represent the 2 12 2 2 plate area in m , the plate spacing in meters and note that 0  8.8510 C /N m . Show your calculations with detailed units so you can perform a units check as you do your calculations. Express your answer using two significant figures. Hints: It can be shown that 1 volt = 1 [N.m/C]. Your answer should be between 7 and 10 nC. 

3. Reflections on Your Findings (a) Consider your analysis of the graph of V vs. d. The assignment forced you to look at the relationship as it exists when there is a very small separation distance between the plates of a parallel plate capacitor and to categorize it as being proportional. Do you see any evidence of deviations from ideal capacitor behavior in the graph of these data as the spacing was increased more and more? Explain

29 - 4 Physics with Video Analysis