Capacitors and Capacitance
4.1 Capacitors and Capacitance
Learning Objectives
By the end of this section, you will be able to:
- Explain the concepts of a capacitor and its capacitance
- Describe how to evaluate the capacitance of a system of conductors
Capacitance
We just spent a significant amount of time discussing the electric field and potential difference between two parallel, oppositely-charged plates. This is because it the basis for how a capacitor works, which is what this part of the module is all about. In the introduction, you saw a video on defibrillators and how they can store and release electric energy. Now we will discuss how they work. But before we do, we need to define capacitance.
A capacitor is a device that stores electrical energy. To charge a capacitor, you can connect one to a battery. The potential difference across the battery terminals (what you typically call voltage) will do work to pull electrons off of one plate of the capacitor and push them onto the other plate. This results in an equal amount of positive and negative charge on the capacitor. The battery will continue to do this work moving electrons from one plate to the other until the potential difference across the capacitor plates is equal to the voltage of the battery. At that point the battery does not have enough energy to move any more electrons.
If you measured the potential difference across the capacitor as it was being charged by the battery, you would find that ΔVC is proportional to the amount of charge (the number of electrons) that has been transferred from one plate to the other. So if you plotted q versus ΔVC you’d get:
where the slope of this line is defined to be the capacitance. So, by definition, capacitance is:
where q is the amount of charge transferred from one plate to the other and ΔV is the potential difference (or voltage) across the capacitor plates. The capacitance is a property of the capacitor, not of the battery that charged it. The units of capacitance are called farads, symbol: F, which is named after (my favorite) physicist Michael Faraday. One farad (1 F) is a HUGE amount of capacitance and you will likely not see any problems with that large a value. Most capacitors have a capacitance more like 1 µF (10-6 F) or even 1 pF (1 picofarad, 10-12 F).
| Pause & Predict 4.1.1 |
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| What is ΔVC? |
| Pause & Predict 4.1.2 |
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| What is the energy stored in the capacitor after you pull the plates apart? |
Practice!
| Practice 4.1.1 |
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| A parallel-plate capacitor has a capacitance of 1.97 pF and a plate area of 5.86 cm2. What is the separation distance between the plates? |
| Practice 4.1.2 |
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| A 1.97-pF capacitor is connected to a 9.0-V battery and fully charged. How many electrons did the battery transfer from one capacitor plate to the other? |
| Practice 4.1.3 |
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| A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery and fully charged. Then the separation between the plates is adjusted so that the energy stored in the capacitor is increased by a factor of 3.5. What is the new plate separation distance? |
Discuss!
Reflect on these questions and take notes on how you would answer them. Then we will share these thoughts together in a class discussion.
In the simulation above, click on the Capacitance option (not the light bulb just yet) and then slide the slider on the battery upward to 1.5 V. Adjust the area of the plates and see what happens to the capacitance. Then adjust the spacing between the plates and see what happens to the capacitance.
You should notice that a larger area results in a higher capacitance (think of capacitance as the capacity to hold charge). A larger area means the excess charges can spread out more, and since these plates are conductors, the charges easily spread out. With more room, more charge can be transferred from one plate to the other.
You should also notice that a smaller separation distance increases the capacitance. The attractive coulomb forces between the opposite charges on each plate would counteract some of the repulsive forces between the like charges on the plates. This sort of leaves more room for more charges to collect on the plates.
Go ahead and play around with the simulation.
Parallel-Plate Capacitor
The type of capacitor will we focus on a lot in this course is called a Parallel-Plate Capacitor. Here is a diagram of one:
This parallel-plate capacitor is connected to a battery and has been fully charged. The +Q and -Q represent the amount of charge that the battery moved from one plate to the other. Both plates start out neutral before being connected to the battery. The +Q on the left plate arises from removing -Q and the -Q on the right plate comes from adding the -Q that was removed from the left plate. This means the plates will ALWAYS have the same amount of charge, just opposite in sign. We describe the geometry of the parallel-plate capacitor in terms of the size of each plate and the spacing: A is the area of a plate (the plates are almost always the same size as each other) and d is the separation distance.
Practice!
| Practice 4.1.4 |
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A parallel plate capacitor is connected to a 12 V battery and given enough time to come to a full charge. If the capacitor stays connected to the battery and the plate spacing is doubled, what happens? |
| Practice 4.1.5 |
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| A parallel plate capacitor is connected to a 12 V battery and given enough time to come to a full charge. If the capacitor is disconnected from the battery and the plate spacing is now doubled, what is the new value of the voltage? |
Discuss!
Reflect on these questions and take notes on how you would answer them. Then we will share these thoughts together in a class discussion.
A parallel-plate capacitor is connected to a 90 V power supply. Given are plots of the electric field and the voltage between the plates. (The plates are 2 cm apart.)
Select the correct plots of E and V versus distance between two parallel plates of a capacitor.
Example problem: Work through this example on your own. When you’re done, watch my video solution to check your work.
A parallel plate capacitor has circular plates of 8.2 cm radius separated by 1.3 mm of air. They are connected to a 240 V power supply and allowed to charge up before being disconnected.
- Calculate the capacitance of this capacitor.
- What charge will appear on the plates?
- If the plates are pulled apart to a separation of 2.6 mm without affecting the charge distribution, what is the potential difference between the plates with the new separation?
- What happens to the electric field between the plates?
Capacitance of a Spherical Capacitor
I have a mistake in the video that I need to correct. It is a careless algebra mistake, but the physics is correct.
When solving for C, it should be:
Then, after plugging in the values for a and b:
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