1. ELECTRIC CURRENT AND CIRCUITS TOPIC 1 – ELECTRICAL ENERGY AND VOLTAGE TOPIC 2 – CAPACITANCE TOPIC 3 – CURRENT AND RESISTANCE TOPIC 4 – ELECTRIC POWER TOPIC 5 – EQUIVALENT RESISTANCE AND CIRCUITS

2. TOPIC 1 – ELECTRICAL ENERGY AND VOLTAGE Learning Goal: You will understand how to determine the potential energy of a charged object or a system of charged objects and how electrical potential energy relates to the concept of voltage. Success Criteria: You will know you have met the learning goal when you can truthfully say: 1.I can calculate the potential energy of a charge or a system of two charges. 2.I can relate voltage to potential energy and calculate the voltage for a given situation. Image(s) from Bing Images

3. Success Criteria 1: I can calculate the potential energy of a charge or a system of two charges. Remember when we learned about the potential energy of an object raised up in a gravitational field: U = mgh This equation told us how many joules of work an object could do if it were lowered from that position to the ground. The capacity to do work came from the objects mass and the force of gravity. The electrical potential energy of a charged object can be calculated as well, which is the amount of work a charged object can do due to its location in an electric field: U electric = -qEd (for a uniform electric field) U electric = k c q 1 q 2 /r (for a non-uniform electric field created by point charges) Since the electric fields inside of wires are uniform, and the main point of this unit is to learn about electric current, we’ll mainly (but not entirely) work with the uniform electric field equation.

4. Success Criteria 1: I can calculate the potential energy of a charge or a system of two charges. U electric = -qEd (for a uniform electric field) U electric = k c q 1 q 2 /r (for a non-uniform electric field created by point charges) The equation for a uniform electric field is very similar to the equation for gravitational potential energy that we have used previously. This is because near the Earth’s surface, the strength of the Earth’s gravity doesn’t vary much with height, so we could treat it like a uniform gravitational field. U gravity = mgh …is analogous to… U electric = -qEd U gravity  U electric Mass (m)  Charge (q) The acceleration due to gravity: 9.81m/s 2 (g)  The electric field strength (E) Height (h)  Displacement (d)

5. Success Criteria 1: I can calculate the potential energy of a charge or a system of two charges. Since potential energy is defined relative to a reference point, such as the ground or a charged plate, we can also talk about the change in potential energy using the same concepts, such as when a 5.0 kg object is raised from 11 m off the ground to 15 m off the ground: Change in U gravity = Δ U gravity = mg Δ h = 5 kg x 9.81m/s 2 x (15 m -11 m) = 20 J This is sometimes a more useful concept to understand, as we’ll see when we get to our discussion of voltage. Thus, the equation we learned for the potential energy of a charge in a uniform electric field can also apply to changes in potential energy: U electric = -qEd can also be: Δ U electric = -qE Δ d

6. Success Criteria 1: I can calculate the potential energy of a charge or a system of two charges. So why is there a negative in front of the q in Δ U electric = -qE Δ d? It is because it takes an input of energy to move an object against a force. Take this example (see picture): Consider in this case up to be positive and down to be negative. Let’s say the charges are both 1 C, the E field is 1 N/C, and the arrows indicate displacements of 1 m, again with up being positive and down being negative. Since an object moving against a force should gain potential energy ( Δ U would be positive) and an object moving with a force should lose potential energy, Δ U electric should be negative for these charges. Ex: For the negative charge: Δ U electric = - q x E x d Δ U electric = - (-1 C) x (1 N/C) x (-1 m) Δ U electric = -1 J Image(s) from Bing Images

7. Success Criteria 1: I can calculate the potential energy of a charge or a system of two charges. Task 7.1.1 (8 points): Answer the questions. a)What is the change in potential energy of a +3.0 C charge that gets 1.5 m closer to a positively charged plate a emitting a 220 N/C uniform electric field? b)How many joules of energy would you need to move a -0.18 C charge 0.52 m closer to the negatively charged plate through a uniform 314 N/C electric field? c)If a +2.4x10 -4 C charge moved 1.62 m through a uniform 1168 N/C electric field in the same direction as the field, what would its change in potential energy be? d)Imagine a block of Styrofoam has a charge of +2.42 µC. If 0.00346 J of potential energy were removed from the block by bringing it closer to a negatively charged plate emitting a uniform 3840 N/C field, how much closer would it get to the plate?

8. Success Criteria 1: I can calculate the potential energy of a charge or a system of two charges. Task 7.1.1 (8 points): Answer the questions e)How much potential energy is there between two point charges, +2.4x10 -6 C and -1.6x10 -6 C, if they are separated by 0.50 m? f)If the charges described in part e) were brought from 0.50 m apart to 0.10 m apart, what would the change in potential energy be? g)From your answer to part f), does a system composed of two oppositely changed particles gain or lose potential energy when they are brought closer together? h)Given that work = force x displacement, would it be correct to say “It takes work to separate charges of opposite signs.” or “It takes work to bring charges of opposite signs together.” Explain.

9. Success Criteria 1: I can calculate the potential energy of a charge or a system of two charges. Task 7.1.2 (5 points): Consider these data tables for a charge moved to different locations in an electric field. a)Is the electric field uniform or non- uniform? How do you know? b)If the charge is +16 µC, and is moved 25 cm further away from the source, what is the change in potential energy? Distance from + source of E field (cm)20.040.060.080.0100.0 E Field strength (N/C)552 Distance from + source of E field (cm)20.040.060.080.0100.0 E Field strength (N/C)13703421528555 c)Is the electric field uniform or non- uniform? How do you know? d)If the charge is +16 µC, what is the potential energy of the system when the charges are 40 cm apart? e)How many Coulombs is the charge producing this electric field?

10. Success Criteria 2: I can relate voltage to potential energy and calculate the voltage for a given situation. Voltage is a term we often hear, but rarely consider the actual meaning of. Voltage is another term for electric potential, and is defined as the potential energy of a charged object divided by the charge of the object. V = U electric /q Since the change in potential energy is usually the more useful concept to consider, “voltage” generally refers to the potential difference ( ΔV) : Δ V = Δ U electric /q One volt is thus when a one coulomb charge is placed at a point in an electric field where it has one joule of potential energy. For a uniform electric field, which is what is normally relevant when dealing with electric current, the equation that relates E field to voltage is: Δ V = -EΔ d Image(s) from Bing Images

11. Success Criteria 2: I can relate voltage to potential energy and calculate the voltage for a given situation. More conceptually, voltage is kind of like the pressure pushing water through a pipe. A higher voltage can cause electricity to flow through a material with more resistance. A common misconception is that high voltage is itself dangerous. A static electric shock you get on a dry day is around 6000 volts. Think of it like a squirt gun vs a river. Water is shot out of a squirt gun at a higher pressure that the water flowing down a river, but you aren’t in any danger of getting swept away by the squirt gun because there isn’t enough water to provide the necessary force. Amperage is what is dangerous, which, as we’ll learn more about in topic 7.3, is the amount of electric current. Amperage is analogous to the amount of water flowing down a river. Image(s) from Bing Images

12. Success Criteria 2: I can relate voltage to potential energy and calculate the voltage for a given situation. Task 7.1.3 (6 points): Answer the questions. a)Describe the difference between voltage and amperage. b)A lot of water under low pressure would be analogous to a high ________ and a low ________. c)If a +0.15 C charge gains 30 J of potential energy, what is the potential difference? d)Suppose a charge moes 42 cm in a 832 N/C uniform electric field. What is the voltage? e)If a 1.5 V potential difference is detected 0.28 m across a uniform electric field what is the strength of the electric field? f)If a charge moves perpendicularly through an electric field, does the voltage charge? Why or why not?

13. Success Criteria 2: I can relate voltage to potential energy and calculate the voltage for a given situation. A battery or an electrical outlet has a voltage associated with it. Most batteries range from around 1 V to around 12 V. Outlets in the U.S. have been standardized to 120 V. When connected to a battery or an outlet, a uniform electric field is created inside the wire. Image(s) from Bing Images Task 7.1.4 (2 points): How does a battery provide the force that it takes to move electrons through a wire?

14. Task 7.1.5 (4 points): Write at least 8 things you learned in this topic (1/2 point each). If you do this in your notebook, please do it in list form, rather than paragraph form.