This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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Preparing for the AP Physics 1 Exam

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This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Introduction to Circuits

Topics include elective charge, electric force, and DC Circuits. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

In this module, we will talk about electric charge, current, and

Â electric circuits.

Â When discussing electric charges, you may have heard the phrase like charges repel,

Â and opposite charges attract.

Â This electric repulsion or attraction is a new type of fundamental force.

Â Up until this point we've only discussed one other type of fundamental force,

Â gravitation.

Â A gravitation force is different,

Â however, in that masses have always been observed to attract and never repel.

Â Generally speaking, the electrostatic force is a stronger force than gravity.

Â We measure charge with the SI unit coulombs.

Â An electron carries the smallest unit of charge,

Â often represented with the symbol little e.

Â The charge of an electron is negative e, or -1.6 x 10 to the -19 C.

Â This means that 1 coulomb of negative charge

Â is made up of 6.25 x 10 to the 18 electrons.

Â Let's take a look at how objects with a net charge behave.

Â It is possible to give a balloon a net charge,

Â due to friction, by rubbing it on your hair.

Â Even with this small amount of rubbing, electrons,

Â which are negatively charged, will transfer fro your hair to the balloon.

Â This leaves your hair positively charged and the balloon negatively charged.

Â This results in the attraction that makes your hair stand up

Â when next to the balloon.

Â This method of charging through contact is called conduction.

Â It is also possible to charge an object by induction.

Â When the objects do not touch and therefore do not transfer charge.

Â For instance, if you now take the negatively charged balloon next to some

Â small neutral pieces of paper, you will see these pieces attract to the balloon.

Â But if these pieces of paper are neutral, why do they attract at all?

Â The pieces of paper undergo a process that is called polarization.

Â The nearby net negative charge on the balloon repels the electrons found in

Â the paper, just a little bit, resulting inn a charge separation.

Â The side closer to the balloon is now positive

Â while the far side is now negative.

Â Just like with gravitation,

Â the force of electricity is stronger the closer the charges are together.

Â This means that the attraction between the opposite charges on the paper and

Â the balloon is stronger than the repulsion

Â between the like charges on the paper and the balloon.

Â Notice that in this example of charging by induction.

Â No charge was ever transferred to or from the paper.

Â It is still neutral overall.

Â It is also worth pointing out, at this point,

Â that electric charge is a conserved quantity.

Â Charge can be transferred from an object to an object during the process, but

Â it is never created or destroyed.

Â Let's look at an example that involves charging by conduction.

Â Okay so this problem tells us that we have two identical metal spheres and

Â they're at rest near one another but not touching.

Â So I'm going to go ahead and draw that.

Â Here's one sphere and here's the other, the two spheres are insulated from

Â their environment, so they're not gaining or losing charge from their environment.

Â The one on the left has an excess charge of -6 Coulombs.

Â That's a lot of electrons, not just 6.

Â That's many, many electrons.

Â The two spheres then touch, and are again separated.

Â What's the net charge on each sphere now?

Â So this would be a charging by conduction question.

Â If we would have touched this by wire or physically take them and

Â put them in contact and them move them away from one another,

Â what you will notice is that the charge sets here at the same sides equalizes.

Â What I mean by that is imagine that this

Â is made up of billions of excess electrons.

Â That doesn't mean it doesn't have protons.

Â That just means it has a lot more negative charges electrons

Â than it does protons in the first place so it's positive charges.

Â Well, what excess charge wants to do is those are all like charges.

Â They repel one another.

Â Now that it's connected and it has more space over here that it can expand to,

Â that's exactly what's going to happen.

Â You're going to see a lot of these electrons space out

Â evenly distributed along the surface as much as possible.

Â So,really what we're having is this negative six is going to bounce off

Â the two.

Â Magnetically I would show it something like this.

Â There's a negative six coulombs on the left sphere.

Â Zero charge on the one on the right 0C because it's neutral.

Â And we're dividing it by 2, literally averaging it.

Â I'm left with -3C on each sphere afterward.

Â That would not be the same if they were different sizes.

Â If one was larger than the other,

Â the larger one would actually end up with more charge,

Â because again they're trying to space out those electrons as much as possible.

Â Now, let's look at a slightly different variation on this question.

Â Supposing that we have the same two spheres as before,

Â one still with a charge of a negative 6 coulombs, one neutral.

Â It says now what's different this time Is the one on the right

Â does not touch the sphere, instead you connect the wire to the ground.

Â And that's our symbol for ground down here.

Â Well the ground is so large, it will either give or

Â take as many electrons as the problem requires as is necessary.

Â So in this scenario, notice what happens.

Â This sphere on the right has positive and negative charges, but in equal amounts.

Â But there's this negatively, excess negatively charged sphere on the left.

Â And what it's going to do is the negative charges that are in the sphere on

Â the right will be repelled some.

Â And what you will see is electrons start to leave this metal sphere and

Â go to the ground, because, like we said, the ground is very large.

Â It can take or give as many electrons as necessary.

Â It wants to know what kind of charge, if any,

Â now exists on this previously neutral sphere?

Â Well, notice what happens is that these electrons leave, then they cut that line.

Â That means this sphere on the right is actually positively charged now.

Â We can't be certain exactly what charge because it depends on how far away we are,

Â how strong the repulsion is.

Â There's other factors in play, but

Â we can be sure now it becomes positively charged, and

Â this is charging by induction because it never actually touches the other sphere.

Â >> We try to mathematically describe physical phenomenon whenever possible.

Â We can calculate the electrostatic force between two charges using an equation

Â called Coulomb's law seen here.

Â Notice that you must have two charges to have a force.

Â This is a consequence of Newton's third law.

Â In the diagram seen here for instance,

Â the positive charge on the right would feel a repulsion force to the right.

Â The positive charge on the left would feel an equal force of repulsion to the left.

Â Recall that force is a vector and

Â the direction is important when describing these interactions.

Â Suppose that two small charged spheres are separated by a distance

Â d between their centers and they exert a force on each other.

Â If the distance between these charges is tripled and the charge on one of

Â the spheres is half by what fraction has the original force changed?

Â So for this question let's look at our equation for Coulomb's law first.

Â We know that force is equivalent to K times Q1Q2.

Â Over R which is the distance separating the two charges squared.

Â Now in this case you have to read the question really carefully.

Â They're telling you that the distance between the two charges is tripled and

Â that the charge of one of the spheres is half.

Â So here's what that really means.

Â This means that my r is triple, so it's three times the original value,

Â and, say, one of the charges, I could pick either one, say I go with q1, is half.

Â It's half of whatever its original value was.

Â Notice, I don't need those numerical values in order to solve for

Â this question.

Â So, if I look at the proportionality over here,

Â I'm only considering the values that have changed.

Â In other words, everything else remains the same, so I can keep that as 1.

Â So if I have in my equation Kq1q2 over r squared, K could be equivalent to 1,

Â q2 could be equivalent to 1, in that they're not changing.

Â So I can write this as q1, which is halved, over r,

Â which is tripled, but remember, that's squared,

Â I can't just put 3, I have to put 3 squared.

Â So one-half divided by 3 squared, well,

Â that gives me 1 over 18.

Â Once again, what the question wanted,

Â they wanted us to get the fraction by which the original force has changed.

Â So I can say over here that my new force

Â is going to be one-eighteenth of my original force,

Â and that's the answer to that one.

Â Since electric force is a vector, we may need to break up forces into components in

Â the x and y direction to solve efficiently.

Â Let's look at an example where the net force on a charge is at an angle.

Â Okay, so in order to solve this question, if you've noticed,

Â I've just gone ahead and sketched the points on a graph to help me with this.

Â The first one, I've got charge one, q1,

Â which is positive 3 microcoulombs, that micro is going to be important,

Â I'm going to have to keep that in mind when I'm doing my calculations.

Â Next I've got q2,

Â which is negative 2 microcoulombs.

Â And then I have the charge that's placed at the origin, and

Â that is positive 4 microcoulombs, that's the one that I'm solving for.

Â The question specifically asks for

Â the net force that's experienced by this charge that's at the origin.

Â So in order for me to solve for this, I'm going to have to look at my equation for

Â Coulomb's Law, which is F = Kq1q2 over r squared,

Â where r is the distance that's separating the two charges.

Â My strategy here is going to be to solve for force 1 exerted by charge 1,

Â force 2 exerted by charge 2, solve for those forces first, and then break down my

Â horizontal and vertical components to get to the net force, so let's look at that.

Â So I'm going to solve for F1 first,

Â F1 is the force exerted by charge 1.

Â I use the value for k, which is 9 times

Â 10 to the 9th, then I use the value for

Â q1, 3 times 10 to the negative 6th.

Â Now since I'm using times 10 to the negative 6th, I am going from

Â microcoulombs to coulombs, and then I've got 4 times 10 to the negative 6th.

Â The distance separating them is that vertical 4 meters.

Â However, keep in mind, it would be 4 squared because it is r squared.

Â The answer I get when I solve for this is 0.00675 Newtons,

Â force is a vector, the direction in this case would be down,

Â that's the direction that the force is exerted by this q1 over here.

Â It's pointing downwards, exerting a force on our charge at the origin.

Â Now I want to solve for F2.

Â Using my same equation, plug in my value for K.

Â I've got the charge, charge 2.

Â Notice how I'm not plugging in the negative,

Â you do not plug in negative values for the charges here.

Â I did, however, put a direction for force, and I'll continue to do that

Â Okay, so now I need to figure out what my distance is that's separating these two.

Â So if you notice up here, we'll use a different color here,

Â I need to figure out what this distance over here is.

Â Well, I know that this horizontal distance is 2 meters.

Â I know that this vertical distance is 3 meters.

Â So I can go ahead and use my trig, and solve for

Â what this side over here would be.

Â And when I do that, using a squared plus b squared equals c squared,

Â I get 3.6 meters for that side.

Â So, I've got 3.6 squared, it is r squared.

Â Okay, so when I solve for this,

Â I end up with 0.00556 Newtons, and

Â keep in mind that I'm going to break this F2 down

Â into its horizontal and vertical components.

Â Now, in order for me to do that, it's important for

Â me to know what this angle over here is.

Â So if I'm going to solve for this angle, I can go ahead and

Â use, I can use theta equals tan inverse,

Â opposite over adjacent, which would be 3 over 2,

Â which gives me the angle of 56.3.

Â Degrees, make sure you're in degree mode.

Â All right, so now I'm ready to start looking at the components for this.

Â I look at the sum of my vertical forces, and I've got F1, which is is completely

Â vertical, it's pointing towards the charge, so it's downwards.

Â I've got F2 sine theta, keep in mind charge 2 was negative, so

Â it's opposite, it's pointing away, so it's directly vertically upwards.

Â I've got negative 0.00675,

Â this was the one that we had solved for

Â earlier, plus my F2,

Â which is 0.00556 sine of 56.3.

Â When I solve for this, I get negative, or

Â downward, 0.0021 Newtons.

Â So that's my vertical, sum of my vertical forces.

Â Now I can go ahead and look at the sum of my horizontal forces.

Â In this case, I only have one, since charge 1 was completely vertical,

Â I didn't have a horizontal component for it.

Â So what I have is this F2 Cosine theta caused by charge 2, which was negative.

Â So I've got my F2, which was 0.00556.

Â The force in this case is pointed towards the right.

Â And I'm solving for this horizontal component, I'm using cosine of 56.3.

Â And my sum of horizontal forces is 0.003 Newtons,

Â and it's positive so it's directed towards the right.

Â Here's what I can go ahead and do from here.

Â I can go ahead and solve for my net force now and for the angle because remember,

Â force is a vector, so I need direction.

Â My net force.

Â Square root of your sum of your vertical forces squared

Â plus sum of your horizontal forces squared.

Â Add that 0.0021 squared + 0.003 squared.

Â And I take the square root of that, and

Â that gives me my net force of 0.0037.

Â Had they asked for the magnitude of the net force, we'd be done right now.

Â But because they just said net force, that means that they're asking for

Â the direction as well.

Â So I can go ahead.

Â And solve for it, my horizontal component was pointed towards the right.

Â It was positive, and that was 0.003, the vertical was downwards, negative, 0.0021.

Â I need to solve for the angle over here.

Â Then I use theta equals inverse tangent,

Â opposite over adjacent,

Â 0.0021 over 0.003.

Â This gives me 34.99, or 35 degrees.

Â So, I can go ahead and say that I have a net

Â force of 0.0037 Newtons that's

Â acting at 35 Degrees below the positive x-axis.

Â And you can see that by my diagram over here,

Â the angle is below the positive x-axis.

Â And that is our final answer for this question.

Â >> Up until this point, we've only discussed charge and

Â rest often called electrostatics but even simple circuits like a battery connected

Â to a light bulb have moving charge.

Â To help us analyze moving charge we use a concept called current.

Â We define current as the amount of charge moving through the cross section of a wire

Â at any point per unit time.

Â Or i equals q divided by t.

Â The units for current are coulombs per second also called amps.

Â At the atomic level electrons move from atom to atom

Â as they travel through a DC circuit.

Â For our purposes and other historical reasons,

Â we usually analyze what is called conventional current.

Â This is the direction that a pretend positive charge would flow.

Â It is exactly equivalent to the negative charge flowing in the opposite direction.

Â If a question wants us to discuss the negative current, it will ask for

Â electron current flow instead.

Â Let's look at an example, in this problem a steady current

Â of 200 milliamp exists in a wire for 2.5 minutes.

Â It wants us to calculate both the total charge that

Â passes through a point in the wire during that time and

Â how many electrons that corresponds to I also notice that.

Â I have some non-SI units here that I'm going to have to convert.

Â I want my milliamps to be in amps, and my minutes

Â to be in seconds before we actually calculate anything in our problem.

Â So to start with, I want to find out the total charge, and

Â I notice it gives me current.

Â I'm going to actually use our definition of current, which is the total amount of

Â charge passing through a cross-sectional area of a piece of wire per unit time.

Â It gives me the total amount of current, and,

Â again, I want to convert this into amps.

Â So milliamps means * 10 to the -3, that's my current.

Â The charge is what we're solving for.

Â The time frame again converted to seconds.

Â There are 2.5 minutes and 60 seconds in every minute.

Â So I'm going to multiply that by 60 to put it in terms of seconds.

Â Now I'm solving here for Q,

Â the total amount of charge passing that cross sectional area.

Â I get 30 coulombs for the first part of this problem.

Â So, that's the first answer it wanted.

Â The next thing it wants to know is how many electrons this corresponds to, and

Â remember, the electrons have a very small charge, so

Â we should end up with a very large answer as a check.

Â Again, I'm going to convert, so, 30 coulombs,

Â I know that there are 1.6x10 to the -19 coulombs for every 1 electron.

Â So what I'm doing is I'm dividing 30 by 1.6x to the -19 as my conversion.

Â That way, my Cs, my coulombs cancel out, and I'm left with the number of electrons.

Â When I solve for that in my calculator, again I'm checking to make sure

Â I end up with a very large answer, 1.88 x 10 to the 20 electrons.

Â Which is again, what I would expect.

Â You have probably heard of a six volt battery or a nine volt battery.

Â Voltage, also called electric potential,

Â is a measure of how much energy a charge has based on its location.

Â And will be discussed in more detail in AP Physics II.

Â For now it is important to know that a charge can gain or

Â lose energy and therefore gain and lose voltage.

Â This will occur as a charge travels through a circuit.

Â While the charge might gain voltage as it travels through a battery.

Â It might lose voltage as it travels across five volts.

Â >> The relationship that helps to tie together voltage and

Â current is called Ohm's Law and is seen here.

Â The changing electric potential is measured in volts and is scalar.

Â I is the current in the circuit element and is measured in amps.

Â R is the resistance of the circuit element and is measured in ohms.

Â Notice that the Current is directly related to the changed in Voltage.

Â This relationship Ohms law is how we define Resistance.

Â For example, suppose we were to measure the Current in a wire

Â as we change the Voltage across it by touching it to two various batteries.

Â As we increase the Voltage across the wire,

Â we will observe more Current problems with the wire.

Â This linear relationship is what we expect from ohmic resistors.

Â The slope of this graph is equal to V over I, which is the resistance of that wire.

Â The resistance of that wire doesn't change as the voltage and current change.

Â The slope stays constant.

Â You can only change the resistance of an object by changing its construction.

Â To calculate the resistance of the wire, we use the equation seen here.

Â The variable L represents the length of the wire in meters, and the variable

Â A represents the cross-sectional area of the wire if m is in meters.

Â The constant symbol rho is called resistivity of the material and

Â is a high value for insulators and a low value for conductors.

Â Let's look at an example that brings this all together.

Â So this question actually has two parts to it.

Â First, it tells me find the resistance of the entire length of the wire,

Â so I'm going to start with that.

Â My equation for resistance is resistance equals the resistivity rho multiplied

Â by L, the length of the wire, divided by A, the cross-sectional area of the wire.

Â So I can go ahead and

Â say that the resistivity constant which they give me over here for

Â copper is 1.68 times 10 to the -8 ohms times meters.

Â And then you have the length of the wire, which is 100 meters.

Â And you've gotta be careful with your conversions because there's two things to

Â look out for here.

Â They're giving you the diameter of your wire,

Â the cross sectional area is pi r squared.

Â So, divide the diameter by 2 to get the radius,

Â and then I change it from millimeters to meters.

Â That's also something you need to be careful about with questions like these.

Â And really all questions that you come across.

Â So then from here you solve for your resistance and you get 0.53 ohms.

Â That's part one of this question, the next part says

Â what is the maximum allowed voltage difference across this wire?

Â Voltage difference or potential difference, so if I were to solve for

Â this, I would use my Ohm's Law V = IR.

Â I know what R is, they're asking me to solve for V and

Â in this question they had given me the current in the wire which is 15A.

Â So at this point, I plugged in

Â the 15 amps for current, the 0.53 ohms for resistance.

Â And I end up with a maximum voltage difference of 8.02 Volts.

Â So those would be my two answers for this question.

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