No, Vab is not equal to Vbatt. So Vab = 5V + Iac*R, but I'm not sure how to calculate that voltage (more specifically, what the current is). My guys would be 0.1 amps, but I'm assuming Ohm's Law

Okay, I can see that figuring out the current and using that requires a leap that you aren't quite ready to make -- besides, that leap actually gives you the answer without calculating the current, so it's probably not the best way to go anyway.

But it let's me make another really valuable point -- namely that you can choose ANY path between two point and sum up the voltage along that path to find the voltage difference between the two points.

So at this point we have that

Vab = Vac + Vca

which gave us

Vab = Vbatt + Iac*R

But we can go the other way, too, namely through the capacitor and over to the battery. If we define Vcap to the the voltage across the capacitor with the positive side on top, the we have that

Vcap = Vb - Va = Vba

Since Vxy = Vx - Vy and since Vyx = Vy - Vx, it stands that Vxy = -Vyx. Hence, in our case,

Vab = -Vab = -Vcap

Like the other equation, this is a general result (with the switch closed) and so is valid at all times (that the switch is closed).

We've previously established that, at t=0, the voltage across the capacitor is 0V. So what is Vab at t=0?

Notice that this let's you find the current very easily, as well. Since both of these equations are valid for all time t>0, we can set them equal:

Vab = Vbatt + Iac*R = -Vcap

Notice that, at t=0, we know Vcap, Vbatt, and R. What does that mean we can do?

I'm kind of confused, was I wrong in my answers then? Or the logic

The part you were responding to here was my comment regarding your answers to part C, in which you indicated that the current would be 0.1A (but without indicating the direction -- please, please, please always be sure to indicated direction/polarity for an current or any voltage).

If there is ANY current in a capacitor, can the voltage across that capacitor be constant?

If you answer no, then what does that tell you is required for the case when we are in DC steady state?

 

Notice that this let's you find the current very easily, as well. Since both of these equations are valid for all time t>0, we can set them equal:

Vab = Vbatt + Iac*R = -Vcap

Notice that, at t=0, we know Vcap, Vbatt, and R. What does that mean we can do?

So Vab = 0 ==> Iac = -0.5 Amps. But, I have to define the positive end of the capacitor to make sense of this (negative) value correct?

The part you were responding to here was my comment regarding your answers to part C, in which you indicated that the current would be 0.1A (but without indicating the direction -- please, please, please always be sure to indicated direction/polarity for an current or any voltage).

If there is ANY current in a capacitor, can the voltage across that capacitor be constant?

No

If you answer no, then what does that tell you is required for the case when we are in DC steady state?

0 current is provided by the battery

 

So Vab = 0 ==> Iac = -0.5 Amps. But, I have to define the positive end of the capacitor to make sense of this (negative) value correct?

How did you get -0.5A?

You need to show your work. Remember, the grader can give you partial credit for partially correct answers. But an answer with no work is all or nothing (or may only get partial points as a penalty for not showing work).

The positive end of the capacitor has already beed defined. Iac is defined as the current flowing from Point A to Point C. So a negative value simple means that current is actually flowing from Point C to point A.

But let me ask you something. If it had been 0.5A and not -0.5A, would you have still said that you needed to define the positive end of the capacitor to make sense of it? If not, then how could a positive value make sense when a negative value didn't? Either you needed to define the positive end of the capacitor to make sense of Iac or not.

The rest of your answers here are correct.

 

How did you get -0.5A?

Sorry, just a mathematical error on my part

 

Okay so let me take another, hopefully final, shot:

a) What is the voltage difference, stored charge, and current at T < 0?

Voltage diff = Vbatt = 5V

Stored charge = 0, assuming this capacitor has never been charged before, or if it has, the charge has bled out over the "long time"

Current = 0, since it is an open circuit

b) What is the voltage difference, stored charge, and current at T = 0?

Voltage diff between A and B =0V

Stored charge on capacitor = 0, since it can be modeled as a short

Current provided by battery = 0.1 Amps

c) What is the voltage difference, stored charge, and current at T --> ∞?

Voltage diff = 5V

Stored charge on capacitor = 50 micro coulombs (or 0.00005 C) (Q = VC)

Current provided by battery = 0, since it is open-circuit model

 

Okay so let me take another, hopefully final, shot:

a) What is the voltage difference, stored charge, and current at T < 0?

Voltage diff = Vbatt = 5V

Stored charge = 0, assuming this capacitor has never been charged before, or if it has, the charge has bled out over the "long time"

Current = 0, since it is an open circuit

b) What is the voltage difference, stored charge, and current at T = 0?

Voltage diff between A and B =0 V

Stored charge on capacitor = 0, since it can be modeled as a short

Current provided by battery = 0.1 Amps

c) What is the voltage difference, stored charge, and current at T --> ∞?

Voltage diff = 5V

Stored charge on capacitor = 50 micro coulombs (or 0.00005 C) (Q = VC)

Current provided by battery = 0, since it is open-circuit model

 

Sorry, just a mathematical error on my part

I figured that was the case. We all make them. My point was that if you have

I = -5V/500Ω = -0.5A

Then the grader can see that you had the right data and were using it correctly and you just screwed up the math. Thus they can give you most of the points for getting the concepts and the techniques correct and take of a little for the math blunder. But when you just give the result, they can't do that.

Also, even if there's no grading involved. If you put -5V/500Ω you are far less likely to follow that with a -0.5A because even if you make the math blunder, your brain is likely to catch the discrepancy between the two when they are right next to each other.

Furthermore, if you don't catch it, you are (hopefully) going to have to track it down eventually. So you can either start over from scratch and redo everything, or have enough work shown so that you can find the error and track it on through from that point on.

 

Correction:

b) What is the voltage difference, stored charge, and current at T = 0?

Voltage diff between A and B =0 V

Stored charge on capacitor = 0, since it can be modeled as a short

Current provided by battery = -0.1 Amps

 

You still haven't indicated a polarity for your battery current.

By convention, the current in a source is the current leaving the positive terminal of the source. So unless you explicitly define it otherwise, that is the interpretation that will be used.