EASTMAN LANDRY: Hello, welcome to AP
daily practice sessions for AP Physics C, Electricity
and Magnetism.
My name is Eastman Landry, and I teach at the Kinder High
School for the Performing and Visual Arts in Houston, Texas.
For those of you that jumped over from Dr. Hood's videos
and mechanics, welcome.
In this video, we're solving for the multiple-choice portions
of the electricity and magnetism exam.
As a brief note before we go on, though, in the screen
here or in the description below,
you can find all nine of these upcoming multiple-choices
if you want to try them out before we
start solving them through.
So with that said, I think we're ready to go.
So let's hop to it.
Before we get to deep and actually solving problems,
let's look at a few different tips
to think about when you're doing the multiple choice.
In particular, you have 35 questions to do in 45 minutes,
and that works out to about a minute
and 15 seconds a question.
There's a bunch of different ways you can time it.
As you're going along, I always went for, like,
the halfway method right, so about 22 minutes or so.
You should be at question right around 17 or so,
but really you're going along just
to be thinking like you need to keep moving.
Why does this matter?
Because you should probably do the whole process
of reading the question, glancing at the answers,
categorizing it, and either commit in that moment
or skip to come back later.
You want to play to your strengths.
You want to see all 35 questions.
Some of them, it's not that they go in order
in terms of easy to hard.
It's a complete scramble of all different topics
and all different difficulty levels.
So to give yourself the best representation of yourself,
make sure you see all the questions
and that the best you do is just commit or skip.
If once you commit or you come back to it,
read the question again.
Note the key idea values, make certain sketches,
and then we're going to have to actually solving and smile.
I say smile because each question
that you do should make you like, yes, it's one more done,
and that little bit of endorphin will help kind
of push you through the exam.
So let's look at the breakdown of how this exam looks.
This is all five units for AP Physics, C Electricity
and Magnetism.
And for the multiple-choice section,
this is how the exam is weighted.
If you're wondering, why is it different percentages, because
on any given exam, there might be different questions
on the free-response section, and they're
going to weight it overall in the entire exam.
However, you can kind of see some of the pressures here,
which is why, if you're like, oh, I
hope they have some magnetism questions on there,
they're going to be on there because all the units have
to be represented by definition.
So don't get bogged down, and we can
try to get to all the questions to get to the units
that you feel strongest in, and you can always come back
when you have time at the end.
So let's take a look at our first question.
We have a circular loop of wire that
has some B-field component, and there's an emf
as we read this thing through.
And so this should start making you think of something,
but before you just stop it the question.
Let's also look at some of our answer choices.
Looking at those, I see what looks like math but all
in variables.
This probably feels like a derivation type question
with some calculus in it.
How do I know it's calculus?
Because when I look at electromagnetism,
you should be thinking, wait, whenever I have a changing
flux as a function of time, that gives me
an emf, that gives me a voltage, and where do I find flux?
Flux is just going to be that B field times that area
component.
So if we break that down, our flux
being our area because it said it was a circular loop,
so this pi r squared.
And then we have our term for our B field.
However, we need to make sure we take the derivative of that.
And when we do, this bt squared turns into to 2bt plus
that c, so sure enough, there's our answer choice.
Briefly looking at the other answers,
these first two are looking for you
to think it's the circumference to power, which
might be something like Ampere's law,
and you can kind of break it down that way.
But honestly, as we go through the rest of these slides,
my advice to you is that you actually
try to solve it if you know how to do it.
So you straight to the answer.
You can try to eliminate answer choices,
but sometimes I can come back to get you.
So let's keep moving right along the question 2.
Here we have two small spheres that
have different charts of q and 4q,
and there's an electric force where it's going to be 0.
So I'm looking at all these different points where
it's going to balance out.
Looking at my answer choices, a, b, c, d, e.
OK, well, that's not too bad.
This is going to be in electrostatic conceptual type
question where I just need to figure out conceptually
where this is going to be.
There's no mathematical--
I guess, fully mathematical things where I'm typing things
in to solve for those answers.
So when we look at this setup here,
I have a couple of different regions.
I have region in between the charges and this region
over here to the left.
So starting between the charges.
If it's a positive charge, it's going
to feel a force to the left, and attractive force,
and a repulsive.
There's no way it can be c, d, and e,
so those are automatically out.
That leads me this other side, where
it's going to be attracted to b and a,
so it has to be one of these two answers.
So if you're short on time, pick one.
But if not, take a little bit of time
to ponder how this equation works.
We're trying to balance these electrostatic forces
with Coulomb's law, so the blue ones
can be some kq over r squared.
I'm going to call that one r.
And I want to find some other force
when it's four times the charge, how far it needs to be.
Well, sure enough, if this can be squared,
it needs to be 2 times that radius.
And looking at our picture here, that distance all the way over
and that distance from there to there, that one sure enough is
2 times that distance, so the answer has to be a.
So breaking this one down, we're looking at regions first
to eliminate answers, and then we're
being a little more specific after that.
So this is kind of where thinking about it
first can definitely help you out as you move along.
Let's hop on to question 3.
Ooh, question 3 is actually a question set.
So oftentimes, you'll see these on the exam, where
you'll see one question that says,
use this for the next three.
Whenever you see these, you should immediately say,
here's this part.
But you should immediately think of whenever these things are
in a set that, you have a chunk of time, 3 minutes and 45
seconds.
Because if there's three questions,
I don't have to worry about just sticking to 1 minute and 15.
I can kind of lump that time together.
So any time you invest a think about the problem
or the circuit, can oftentimes pay dividends
on later problems.
So this thing says neglect internal resistance all meters
ideal.
This is our circuits question.
So taking a little bit of time to think about it.
I'm just going to call this current i, why?
Because then I can say that, wait,
these two are each going to contribute half of that current
because they're the same battery that's going to go forward.
And then on the other side, each of these has to split to half
coming back together to i.
So just kind of conceptually seeing the circuit,
because these are equal, I know the current has to split
and it has to combine back together.
And so I haven't actually answered the question yet,
but I've looked at this, I've thought about it a little bit.
And that might help me as I go forward, and it will.
So looking at question number 3, I
want to see what voltmeter would read 6 volts between which two
points?
So I'm looking at these different sets.
OK, this is just feeling like it's
going to be a circuits problem.
It can be conceptual, let me say,
or mathematical because since this is a set,
it's worth maybe investing some more math because one
of the questions along the way is probably
to make me do some math.
There's nothing wrong with investing earlier
on mathematically, but you can also, as you'll see,
get through this one pretty quick just conception.
We're going to solve this one, actually, twice.
Looking at it conceptually.
I know it can't be a and b because they're
right next to each other.
They're going to have 0 volts of potential difference.
The same thing is true for c and d
because they're right next to each other.
And actually, here's the crazy part, a and f and b and f.
If a and b are the same, that means that these two
have the exact same delta b.
You can't have two right answers on the AP multiple-choice.
You only have one.
So just from that, we know that this has to be our answer.
But if you're not pleased with that,
and it feels like a little bit kind of guessy
because you're eliminating answers,
which is absolutely fine, let's actually hop into the math.
We're going to only do the math because we know it's a set
and it'll pay off for us later on.
So if you're 600 resistor in the bottom
plus the resistance between 2 and 3.
Well, these two are in parallel, so when I sum those up,
that's going to be 300 ohms total.
So my total equivalent resistance can be 900 ohms.
Why does that matter?
Because then I can solve for what
this current is going to be.
I know that if I have 9 volts in a total of 900 ohms, then
I'm going to have 0.01 amps.
And sure enough, if there's 0.01 amps running through 600,
that's going to give me 6 volts.
So this that we could actually solve it mathematically.
Let's see if that actually helps us out later on.
It does.
Question 4, I want to see some ammeters that are
inserted at different points.
And I want to see if it's at I, III, I and III,
so and so forth.
And I'm looking for the same current.
Well, this is, once again, our circuit's conceptual bit.
But wait, this is current; didn't we
think about that at the very beginning?
Of course, we did.
We already talked about how point 4 right here
has the exact same current as point 1, 2, and 3.
Therefore, we know our answer for e, and we are done.
If you're concerned about why, you
can go back to that first part.
But ultimately, we know that these two circuits
have to have i over 2 if we're going to call this section i.
So these two also have to split equally because one 600
the other one is also 600.
So now we get to the last question of our set.
What's the total power dissipated
by the three resistors within this given setup?
So we look at our answer choices.
Oh, this one has numbers in it.
Whenever you see numbers, this is
going to be some sort of actually
calculation, mathematical thing, that we have to do.
Well, thankfully, we invested some time earlier in that math.
So we can use that to our advantage.
As a brief note, whenever you see total power,
total power is talking about the battery power
if you took up all the output that the batteries are giving,
so we can use that to say, wait, if I know the current going
through my battery, and I know that power
is my current times my voltage, I can break this down
to say that, oh, one of these is having 0.05 amps times at 9
volts, I get 0.045 watts.
But there are two batteries, so my total answer
is going to be my 0.09 watts.
Is there another way you can do this?
For sure, but because we already did the math before,
unknowingly because we thought that was a way
to solve that problem, it paid off for us earlier.
So think about this when you get to sets.
There's nothing wrong with spending a little bit more time
to break it down because, oftentimes, the more fully you
understand the set, it'll actually help you out,
and later problems maybe save you some of that time.
Well, we're done with this set.
So that's kind of sad.
Let's move on to question number 6.
So for question 6, we have a thin ring
of radius r with this charge as uniformly distribute around
the ring.
And there's some point of charge plus q that's currently
at r under the kinetic energy; when
it reaches this other position here, where it's 2r,
so basically, I let this thing go,
and it's going to go that direction.
So conceptually, I guess, I should look
at the answer choices first.
I have all these fun little equations.
OK, this feels a bit more mathematical
and derivation-like.
So sure enough, the other thing that's slipping in here
is energy.
It's not just electrostatic.
I'm looking for some way to have energy be captured in the set.
You may look at this and think it's, oh, it's a big integral
or something I could do.
You don't, because with energy, remember,
it's not a vector quantity.
So we can just say, wait, I know how far away this is.
It's going to be 2r.
I know how far away this other point is going to be.
And when I look at my energy, I'm
really thinking that the kinetic energy at point r,
I'm calling this r just because it's the one r.
And that electrostatic potential energy at r
is equal to the kinetic energy at 2r and the potential at 2r.
So really, this equation breaks down
to solving for what these potentials are.
So I know potentials take q over r.
And if I look at that whole sum for any given
charge of my loop, it's just going
to be kq over r, where this r is what I already
looked at previously.
That distance and that distance.
So breaking this down a little bit further
to get to this point.
If we get to this equation, where
we have our potential energy at r minus the potential energy
at 2r, that difference is going to give us
our kinetic energy that we have, and that's
what we're looking for.
Tada!
And there's our answer.
You might say, wait, can I just know
that if I'm really clever by saying,
it's the difference of things that
are 1 over square root of 5 and square root of 2?
Sure, you could have jumped straight to this answer
if you're that confident with it,
but just be careful because if you notice,
this one is kind of smiling at you to try
to trick you with bad math.
So avoid your bad math things, and actually,
if you work it out, you're going to see that it stays just
like this, and that's aok.
On the question 7, with a parallel-plate capacitor
is charged and the batteries being removed.
And I want to see what's going to happen with which
the following will decrease.
We'll have all these different options.
This thing just jumps out as being a capacitor type
question that's going to be conceptual in nature.
Maybe this is strength.
If so, you hop in.
If not, you skip and come back later.
A thing to just know right off the bat
is if you see a battery being removed,
the charge in your system is constant.
On the flip side, if the battery hit state attached,
then the voltage would have been constant.
But looking at this, it says constant charge,
which means it can't be that answer because the charge can't
change.
We just talked about that.
It has to be a constant.
We also know resistance of the capacitor just sounds weird.
We've never heard about resistance of capacitors
before, so that one's out.
And we also know the distance between the capacitor plates
also doesn't matter because, when
you put a dielectric, again, in you're not changing
the distance of that circuit.
So we left that this is kind of a and d thing.
If I look at this equation from the equation sheet
with the dielectric constant in it,
then we know the capacitance is going
to increase when I increase my k or my dielectric constant.
Well, that means this can't be the answer.
So our answer has to be a.
Why?
Because if our capacitance increases
for a constant charge, our voltage, by definition,
then has to decrease.
This is just one way to think through it.
There's a couple of different ways.
That's the main one that we are going for.
Looking at 8, this is a classic one,
where we have an electron that's going through a magnetic field.
And we all know which way it's going to be deflected.
We have point a point b.
The key thing with these types of questions
is really setting up your spatial reference frame
to make sure that as you're looking at it,
you're using your proper right-hand rule,
where depending on your particular way of doing it,
you make sure you lined up correctly.
But the key idea is this is an electron, so it's negative.
It's the opposite of what a positive charge will do.
So for me, I would have this thing going towards the screen.
With my right hand, the B-field is going up.
I'm going to think it's going right to c,
but it's an electron.
Therefore, my answer is going to be d, and sure enough,
that is my answer.
So let's hop on the question 9, our last question here.
On this one, we have a total charge of q,
uniformly distributed, has dimensional correct expression
for potential difference between the sphere and its surface.
One of the main things that jumps out
is this is actually the last question on the exam this year.
So a lot of students look at this said, that's crazy,
I want to do that, and skip it.
But there's this key phrase where
it says that this is a dimensionally
correct expression.
So even though it looked mathematical,
this is really a conceptual question.
All this means you have to have proper units.
What is proper units means?
It means that if we're looking for potential difference,
I should have a change in electric potential or units
for electric potential.
Electric potential is that voltage equation.
So I'm really looking for something
that has a q over r because all these 1 over 8 epsilon
knots are the same as the 4 pi epsilon knots.
So breaking that down, the only one
that looks like it matches showing up
is d, and that's our answer.
This is just to point out this is
the last question on this year's exam,
that it's worth skipping ahead.
If you don't know, then see all the questions
before you run out of time and don't even get to look at them.
So with that said, I want to thank you
for stopping by this video, and I look forward
to seeing what the next three.
We're going to be covering the FRQs for electricity
and magnetism.
See you there.