(nz045.jpg)

Announcements  Physics 2135 spreadsheets for all sections, with Exam 1 scores, will be posted today on the Physics 2135 web site. You need your PIN to find your grade.  Physics 2135 Exam 1 will be returned in recitation Thursday. When you get the exam back, please check that points were added correctly. Review the course handbook and be sure to follow proper procedures before requesting a regrade. Get your regrade requests in on time! (They are due by next Thursday’s recitation.) On a separate sheet of paper, briefly explain the reason for your regrade request. This should be based on the work actually shown on paper, not what was in your head. Attach to the exam and turn it in by the end of your next Thursday’s recitation.  Preliminary exam average is about 69.6% (10 sections out of 12 reporting). Could be better. Scores ranged from a low of 15 to a high of 200 (3 students). I will fill in the ??’s during the “live” lecture and in its “.ppt” file.

Today’s agenda: Energy Storage in Capacitors. You must be able to calculate the energy stored in a capacitor, and apply the energy storage equations to situations where capacitor configurations are altered. Dielectrics. You must understand why dielectrics are used, and be able include dielectric constants in capacitor calculations.

Energy Storage in Capacitors Let’s calculate how much work it takes to charge a capacitor. The work required for an external force to move a charge dq through a potential difference  V is dW = dq  V. VV +- +q-q + dq From Q=C  V (   V = q/C): q is the amount of charge on the capacitor at the time the charge dq is being moved. We start with zero charge on the capacitor, and end up with Q, so

The work required to charge the capacitor is the amount of energy you get back when you discharge the capacitor (because the electric force is conservative). Thus, the work required to charge the capacitor is equal to the potential energy stored in the capacitor. Because C, Q, and V are related through Q=CV, there are three equivalent ways to write the potential energy.

All three equations are valid; use the one most convenient for the problem at hand. It is no accident that we use the symbol U for the energy stored in a capacitor. It is just another “version” of electrical potential energy. You can use it in your energy conservation equations just like any other form of potential energy! There are now four parameters you can determine for a capacitor: C, Q, V, and U. If you know any two of them, you can calculate the other two.

Example: a camera flash unit stores energy in a 150  F capacitor at 200 V. How much electric energy can be stored? If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Example: compare the amount of energy stored in a capacitor with the amount of energy stored in a battery. A 12 V car battery rated at 100 ampere-hours stores 3.6x10 5 C of charge and can deliver at least 4.3x10 6 joules of energy (we’ll learn how to calculate that later in the course). If a battery stores so much more energy, why use capacitors? Application #1: short pulse magnets at the National Magnet Laboratory, plus a little movie of a short pulse magnet at work.short pulse magnets 10 6 joules of energy are stored at high voltage in capacitor banks, and released during a period of a few milliseconds. The enormous current produces incredibly high magnetic fields.high magnetic fields A 100  F capacitor that operates at 12 V can deliver an amount of energy U=CV 2 /2=7.2x10 -3 joules. If you want your capacitor to store lots of energy, store it at a high voltage. Truth in advertising: I set up the comparison in a way that makes the capacitor look bad.

Application #2: quarter shrinker.quarter shrinker Application #3: can crusher.can crusher Some links: shrinking, shrinking (can you spot the physics mistake), can crusher,. Don’t do this at home. Or this.shrinking can crusherthis

Energy Stored in Electric Fields VV +- +Q-Q E d area A Energy is stored in the capacitor: The “volume of the capacitor” is Volume=Ad

Energy stored per unit volume (u): VV +- +Q-Q E d area A The energy is “stored” in the electric field!

VV +- +Q-Q E f area A This is not a new “kind” of energy. It’s the electric potential energy resulting from the coulomb force between charged particles. Or you can think of it as the electric energy due to the field created by the charges. Same thing. “The energy in electromagnetic phenomena is the same as mechanical energy. The only question is, ‘Where does it reside?’ In the old theories, it resides in electrified bodies. In our theory, it resides in the electromagnetic field, in the space surrounding the electrified bodies.”—James Maxwell This is on page 2 of your OSE sheet. Do not use until later! (Unless you really know what you are doing.)

Today’s agenda: Energy Storage in Capacitors. You must be able to calculate the energy stored in a capacitor, and apply the energy storage equations to situations where capacitor configurations are altered. Dielectrics. You must understand why dielectrics are used, and be able include dielectric constants in capacitor calculations.

If an insulating sheet (“dielectric”) is placed between the plates of a capacitor, the capacitance increases by a factor , which depends on the material in the sheet.  is the dielectric constant of the material. dielectric In general, C =  0 A / d.  is 1 for a vacuum, and  1 for air. (You can also define  =  0 and write C =  A / d). Dielectrics

The dielectric is the thin insulating sheet in between the plates of a capacitor. dielectric Any reasons to use a dielectric in a capacitor?  Lets you apply higher voltages (so more charge).  Lets you place the plates closer together (make d smaller).  Increases the value of C because  >1.  Makes your life as a physics student more complicated. Gives you a bigger kick when you discharge the capacitor through your tongue! A lot of interesting physics happens in the dielectric, but we’ll skip that section.

dielectric This is equivalent to two capacitors in parallel. Each of the two has half the plate area. The two share the total charge, and have the same potential difference Homework hint: what if the dielectric fills only half the space between the plates? C1C1 C2C2 Q2Q2 Q1Q1 CQ

If you charge a capacitor and then remove the battery and manipulate the capacitor, Q must stay the same but C, V, and U may change. (What about E?) If you charge a capacitor, keep the battery connected, and manipulate the capacitor, V must stay the same but C, Q, and U may change. (What about E?) If exactly two capacitors are connected such that they have the same voltage across them, they are probably in parallel (but check the circuit diagram). Some things for you to ponder… If you charge a capacitor and then remove the battery and manipulate the capacitor, Q must stay the same but C, V, and U may change. (What about E?) If you charge a capacitor, keep the battery connected, and manipulate the capacitor, V must stay the same but C, Q, and U may change. (What about E?) If exactly two capacitors are connected such that they have the same voltage across them, they are probably in parallel (but check the circuit diagram).

If you charge two capacitors, then remove the battery and reconnect the capacitors with oppositely-charged plates connected together… draw a circuit diagram before and after, and use conservation of charge to determine the total charge on each plate before and after. If you charge two capacitors, then remove the battery and reconnect the capacitors with oppositely-charged plates connected together… draw a circuit diagram before and after, and use conservation of charge to determine the total charge on each plate before and after.

Example: a parallel plate capacitor has an area of 10 cm 2 and plate separation 5 mm. 300 V is applied between its plates. If neoprene is inserted between its plates, how much charge does the capacitor hold.  V=300 V d=5 mm A=10 cm 2  = V

Example: how much charge would the capacitor on the previous slide hold if the dielectric were air? The calculation is the same, except replace 6.7 by 1. Or just divide the charge on the previous page by 6.7 to get…  V=300 V d=5 mm A=10 cm 2  =1 300 V

Visit howstuffworks to read about capacitors and learn their advantages/disadvantages compared to batteries!howstuffworks A capacitor connected as shown acquires a charge Q. V While the capacitor is still connected to the battery, a dielectric material is inserted. Will Q increase, decrease, or stay the same? Why? V V=0 Conceptual Example

Back to our Example: find the energy stored in the capacitor.  V=300 V d=5 mm A=10 cm 2  = V

Example: the battery is now disconnected. What are the charge, capacitance, and energy stored in the capacitor?  V=300 V d=5 mm A=10 cm 2  =6.7 The charge and capacitance are unchanged, so the voltage drop and energy stored are unchanged. 300 V

Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor?  V=300 V d=5 mm A=10 cm 2  =6.7  V=? d=5 mm

 V=? d=5 mm A=10 cm 2 Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor? The charge remains unchanged, because there is nowhere for it to go.

 V=? d=5 mm A=10 cm 2 Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor? Knowing C and Q we can calculate the new potential difference.

 V=2020 V d=5 mm A=10 cm 2 Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor?

Huh?? The energy stored increases by a factor of  ?? Sure. It took work to remove the dielectric. The stored energy increased by the amount of work done.

A “toy” to play with… (You might even learn something.)