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electric charge is a property that some but not all fundamental particles in nature have the most commonly talked about fundamentally charged particles are the electrons which orbit the outside of the atom these are negatively charged there's also the protons which reside inside the nucleus and these are positively charged and the neutrons inside the nucleus don't have any net charge it turns out that all fundamentally charged particles in the universe have charges that come in integer units of the elementary charge so if you find a particle in nature it's going to have a charge of 1 times this number 2 times this number 3 times this number and it could be positive or negative for instance the electron has a charge of negative one point six times 10 to the negative 19 coulombs and the charge of the proton is positive one point six times 10 to the negative 19 coulombs however most atoms in the universe are electrically neutral overall since they'll have just as many negative electrons as they do positive protons but if an atom had too many electrons overall that atom would be negatively charged and if an atom had too few electrons that atom would be overall positively charged and something that's really important to remember is that the electric charge is always conserved for every process in other words the total charge initially is going to equal the total charge finally after any process so it's an example problem involving electric charge look like let's say three identically sized metal spheres start off with the charges seen below positive 5 q positive 3 Q and negative 2 Q if we touch sphere X 2 sphere Y and separate them and then touch sphere Y 2 spheres E and separate them what will be the final charge on each sphere well first when we touch X 2 y the total charge has to be conserved there's a total charge of 8 Q amongst them and since they're identically sized they'll both share the total charge which means after they touch they'll both have positive for Q if one of the spheres were larger it would gain more of the charge but the total charge would still be conserved and now when sphere Y is touched to sphere Z the total charge amongst them at that moment would be positive for Q plus negative 2 Q which is positive 2 Q they would share it equally so sphere Y would have positive cue and spheres II would also have positive q so the answer here is c opposite charges attract and like charges repel and what Coulomb's law does is it gives you a way to find the magnitude of the electric force between two charges the formula for Coulomb's law says that the magnitude of the electric force between two charges Q 1 and Q 2 is going to equal the electric constant K which is nine times 10 to the ninth times the product of the two charges measured in coulombs divided by the center-to-center distance between those two charges squared you can't forget to square this distance and it's got to be in meters if you want to find SI units of Newton's for the force also don't rely on the negative and positive signs of the charges to tell you which way the force points just use the fact that opposite charges attract and like charges repel and use Coulomb's law to get the magnitude of the force so what's an example problem involving Coulomb's law look like let's say two charges exert an electric force of magnitude F on each other what would be the magnitude of the new electric force if the distance between the charges is tripled and the magnitude of one of the charges is doubled well we know the formula for Coulomb's law says that the force between two charges is the electric constant times one of the charges times the other charge divided by the distance between them squared and now once we triple the distance and double a charge the new electric force is going to be the electric constant times one of the charges multiplied by two times one of the charges divided by three times the distance which is squared so I'm going to get a factor of two on top and this three will get squared which gives me a factor of nine on the bottom if I pull out those extra factors I get that the new force is going to be two ninths multiplied by K q1 q2 over d squared but this entire quantity was just the old force F so the new force is going to be two ninths of the old force the electrical current I tells you the amount of coulombs of charge that passes a point in a wire per second so if you watch some point in a wire and you count how many coulombs of charge passed by that point per second that would be the current or an equation form we can see that the current I is going to be the amount of charge that flows past a point in a wire per this gives the units of AI as coulombs per second which we abbreviate as an ampere and since charging time aren't vectors current is not a vector either something that's kind of strange is that the so-called conventional direction of current would be the direction that positive charges flow within a wire however positive charges don't flow within a wire the only charges that actually flow in a wire are negative charges but it turns out that negative charges flowing to the left is physically the same as positive charges flowing to the right so in most physics problems we pretend as if it were the positive charges moving however it's really the electrons which are negative that are moving within the wire so what's an example problem involving electrical current look like let's say three amps flows within a circuit how much charge would pass by a point in that wire during a time interval of five minutes well we know the definition of current is the charge per time that means the charge is going to be the amount of current multiplied by the time so we take our current of three amps and we multiply by the time but we can't multiply by five because that's in units of minutes since amps is coulombs per second we've got to convert five minutes into seconds which would be five minutes multiplied by 60 seconds per minute which would give us a total amount of charge of nine hundred coulombs the resistance of a circuit element measures how much that element will restrict the flow of current the larger the resistance the less current that will be allowed to flow and the definition of a resistance is given by Ohm's law Ohm's law states the amount of current that you'll get through a portion of a circuit is going to be proportional to the voltage across that portion divided by the resistance of that portion of the circuit so between these two points the amount of current that will flow is going to be equal to the voltage between those two points divided by the resistance between those two points so the larger the resistance the less current will flow but the greater the voltage supplied the greater the current will be and this is what Ohm's law says even though Ohm's law gives you a way to define the resistance you can determine the resistance of a circuit element by knowing the size and shape of that circuit element in other words the resistance of a cylindrical resistor is going to be equal to the resistivity which is a measure of an object's natural resistant to current multiplied by the length of that resistor the longer the resistor the greater the resistance and the more it will resist the flow of current and then divided by the cross-sectional area of that resistor which would be this area right here that current is either flowing into or out of if the resistor is cylindrical the area of this circle would be pi times R squared where little R would be the radius of this cross sectional area the units of resistance is ohms and it is not a vector it is always positive or zero so what's an example problem involving Ohm's law or the resistance of a cylindrical resistor look like let's say a battery of voltage V is hooked up to a single cylindrical resistor of length L and radius little R and when that's done a current I is flowing through the battery what is the resistivity Rho of that resistor well we know Ohm's law states that the current that flows through a portion of a circuit will be equal to the voltage across that portion divided by the resistance of that portion and this means the resistance of this resistor is going to be the voltage of the battery divided by the current 2-factor resistivity into this we have to use the formula for the resistance of a cylindrical resistor which is Rho times L over a this gives us the resistance of the resistor which has got to equal V over I and now we can solve for the resistivity Rho we get V times a over IL but since we're given the radius little R we've got to write the area in terms of that radius so there's going to be V times PI R squared divided by I times L which gives us an answer of C when dealing with complicated circuits with many resistors you often have to reduce those resistors into smaller equivalent amounts of resistors and the two ways you do this are by finding two resistors that are in series or in parallel resistors will be in series if the same current that flows through the first resistor flows through the next resistor if the current branched off in between them these resistors would no longer be in series but if they're in series you can find the equivalent resistance of this section of wire by just adding up the two individual resistances so the current for resistors in series must be the same but the voltage might be different since they could have different resistances two resistors will be in parallel if a current comes in splits into two parts goes through one resistor each and then rejoined before hitting anything else in the circuit and if this is the case you can find the equivalent resistance of this portion of the circuit ie between these two points by saying that 1 over the equivalent resistance is going to equal 1 over the resistance of the first resistor plus 1 over the resistance of the second resistor but be careful 1 over r1 plus 1 over r2 just gives you 1 over R equivalent if you want R equivalent you're going to have to take 1 over this entire side in order to get R equivalent so what's an example problem involving resistors in series and parallel look like let's say we have the circuit shown below and we want to know what current flows through the 8 ohm resistor now you might be tempted to say that since Ohm's law says that the current is Delta V over R we can just plug in the voltage of the battery which is 24 volts divided by the resistance of the resistor which is 8 ohms and that would give us 3 amps but that's not right at all when using Ohm's law the current that flows through a resistor R is going to be equal to the voltage across that resistor divided by the resistance of that resistor so if we plug 8 ohms into the denominator we've got to plug in the voltage across that 8 ohm resistor but the voltage across the 8 ohm resistor is not going to be the full 24 volts of the battery it's going to be less than 24 volts in other words the battery provides a voltage between this point and this point of 24 volts but there's going to be voltage drops across the six and 12 ohm resistors which make it so that the voltage across the 8 ohm resistor is not going to be the full 24 volts so we have to reduce these resistors to a single resistance the 6 and the 12 are in parallel so we can say that 1 over 6 plus 1 over 12 would equal 1 over the resistance of that portion of the circuit this is going to equal 3 twelfths which is 1/4 so that means that parallel portion of the circuit has an equivalent resistance of 4 ohms so between this point and this point there are 4 ohms of resistance and that equivalent resistance is in series with this 8 ohm resistor so we can add 4 and 8 and get 12 ohms of total resistance and now I can say that the full 24 volts of the battery is applied across this entire equivalent resistance of 12 ohms so if I come up here and change this 8 ohms to 12 ohms of equivalent resistance for the total circuit I'll get the correct current that flows through the battery of two amps and since that's the current that's flowing through the battery that had to be the current that's flowing through the eight ohm resistor as well since this 8 ohm resistor and the battery are in series elements in a circuit often use electrical power that is to say when current runs through a resistor the electrons moving through that resistor turns some of their electrical potential energy into energies like heat light or sound and the rate at which these electrons are turning their energy into other forms of energy is called the electrical power so the rate at which a resistor is turning electrical potential energy into heat is the electrical power used by that resistor in other words the amount of energy converted into heat divided by the time it took to convert that energy is the definition of the power and there's a way to determine this number of joules per second in terms of quantities like the current the voltage and the resistance the power used by a resistor can be written as the current through that resistor multiplied by the voltage across that resistor or if you substituted Ohm's law into this formula you'd see that this is equivalent to the current through that resistor squared multiplied by the resistance of the resistor or it could rearrange these formulas to get the power used by a resistor would also be the voltage across that resistor squared divided by the resistance of that resistor all three of these if used correctly will give you the same number for the power used by a resistor and if you wanted to determine the number of joules of heat energy converted you could set any one of these equal to the amount of energy per time and solve for that energy the units of electrical power are the same as the regular units of power which is watts ie joules per second an electrical power is not a vector so it's an example problem involving electrical power look like let's say a light bulb of resistance R is hooked up to a source of voltage V and a second light bulb of resistance 2 R is hooked up to a source of voltage 2 V how does the power used by the second light bulb compared to the power used by the first light bulb since we have information about R and V I'll use the version of the power formula that says that the power used by a resistor is going to be Delta V squared over R so in terms of quantities given the power used by the first light bulb is going to be V squared over R and the power used by the second light bulb is going to be equal to the voltage across the second light bulb which is two times the voltage across the first light bulb and we square that divided by the resistance of the second light bulb which is going to be two times the resistance of the first light bulb the two squared on top is going to give me a factor of four and I'll have another factor of two on the bottom so if i factor out this four divided by two I get the power used by the second light bulb is going to be two times V squared over R but V squared over R was the power used by the first light bulb so the power used by the second light bulb is going to be two times the power used by the first light bulb and so the light bulb of resistance to R has twice the power and that means it'll be brighter the quantity that determines the brightness of a light bulb is the electrical power of that light bulb it's not necessarily the resistance or the voltage it's the combination of the two in this formula that will tell you the electrical power and therefore the brightness of the light bulb two of the most useful ideas and circuits are referred to as kirchoff's rules the first rule is called the junction rule and it states that all the current entering a junction must equal all the current exiting that Junction in other words if you add all the current that flows into a junction that has to equal all the current that flows out of that junction because current is just flowing charge and charge is conserved so charge can't be created or destroyed at any point in this circuit no more than water can get created or destroyed within a series of pipes and the second rule is called the loop rule which states that if you add up all the changes in electrical potential ie voltages around any closed loop in a circuit it'll always add up to zero so if you add up all the voltages encountered through a closed loop through a circuit it always adds up to zero and this is just a result of conservation of energy the electrons will gain energy when they flow through the battery and they'll lose energy every time they flow through a resistor but the total amount of energy they gain from the battery has to be equal to the total amount of energy they lose due to the resistors in other words if we consider a complicated circuit that has a battery and three resistors the total current flowing into a junction I 1 would have to be equal to the total current coming out of that Junction I 2 and I 3 since no charge gets created or destroyed and that means when these two currents combine again the total current flowing out of that section is going to again be I won and if we follow a closed loop through this circuit the sum of all the voltages around that loop have to add up to zero ie the voltage of the battery minus the voltage drop across the first resistor minus the voltage drop across the second resistor would have to equal zero so what's an example problem involving Kirchhoff's rules look like let's say we have the circuit below and we wanted to determine the voltage across the six ohm resistor to do this we could use the loop rule I'll start behind the battery and I'll go through the resistor I wanted to determine the voltage across I'll add up all the voltages across that loop and set it equal to zero so the voltage across the battery is going to be positive 24 volts minus the voltage across the six ohm resistor and then minus the voltage across the 8 ohm resistor has to equal zero but we're given this current so we know that 2 amps flows through the 8 ohm resistor and you can always determine the voltage across a resistor using Ohm's law so the voltage across the 8 ohm resistor is going to be 2 amps which is flowing through the 8 ohm resistor multiplied by 8 ohms and we get 16 volts which I can plug in to here and this gives me 24 volts minus the voltage across the 6 ohm resistor minus 16 has to equal zero and if I solve this for the voltage across the six ohm resistor I get 24 volts minus 16 volts which is 8 volts so the voltage across the 6 ohm resistor would be 8 volts note because the 12 ohm resistor and the 6 ohm resistor in parallel the voltage across the 12 ohm resistor would also be 8 volts because the voltage across any two elements in parallel have to be the same volt meters are the device that you use to measure the voltage between two points in a circuit when hooking up a voltmeter you've got to hook it up in parallel between the two points you want to find the voltage across in other words to determine the voltage between this point and this point which would be the voltage across R 3 you would hook up the voltmeter in parallel with r3 a meters are the devices we use to measure the electrical current that passed through a point in a circuit and a meters have to be hooked up in series with the circuit element you want to determine the current through in other words if we wanted to determine the current through r1 we would hook up the ammeter in series with r1 note that for these electrical devices to work well the ammeter should have almost zero internal resistance thereby not affecting the current that flows through the circuit and voltmeter should have near infinite resistance so that it doesn't draw any of the current from the resistor in reality ammeters have a very small but nonzero internal resistance and voltmeters have a very high but not infinite internal resistance so what would an example problem involving voltmeters and ammeters look like let's say we have the circuit shown below and these numbered circles represent possible places we could stick a voltmeter to measure the voltage across the eight ohm resistor which two of these volt meters would correctly give the voltage across the eight ohm resistor and you have to be careful some AP problems are going to require you to select two correct answers for the multiple choice so be sure to read the instructions carefully voltmeter number four is a terrible choice you never hook up your voltmeter in series with the circuit element you're trying to find the voltage across and voltmeter number one is doing nothing really because it's measuring the voltage between two points in a wire with nothing in between that wire so the voltage measured by voltmeter one should just be zero since the voltage across a wire of zero resistance should just give you zero volts so the correct choices would be voltmeter number two which gives you the voltage across the eight ohm resistor and voltmeter number three which also gives you an equivalent measurement of the voltage across the resistor 8 ohms

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