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Last-minute concept problems (before exams!)

  • Thread starter Flame
  • Start date

Flame

Hi. This thred will sort of be "my thread" i suspect for various concepts I still do not get after two semesters of college physics (my final is in two days). Without further ado, I'll ask my first question:

Here's how the problem reads:
A long nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m^3) distributed throughout its column. Determine the magnitude of the electric field 5.0 cm from the axis of the cylinder.

If they said to determinte it at a radius larger than the radius of the cylinder itself, I would most likely have no problem solvingit; but how do I solve it when they ask for the electric field 5 cm from the axis (i.e., r<R)?
 

LogicalAtheist

The same way as it is outside. Outside, would you not determin it from the axis, rather than the radius? Thus, this is simply some distance from the axis, regardless of where the radius of said cylinder is?
 

Flame

So it makes no difference, then...

Using Gauss's Law (after being simplified) as follows:

E=(2*k*lambda)/r=(2*(9x10^9)*(5x10^-9))/0.05

I still don't come out with the correct answer (these questions are off of one of our previous tests for which we have the answers, but not how to solve them)

The answer, by the way, is supposed to come out to 14 N/C.

Am I using the wrong approach?
 

ahrkron

Staff Emeritus
Gold Member
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What have you tried?
Hint: Gauss' law, and a cylindrical gaussian surface inside the cilinder.

Think of what should happen with the E field:

1. What would you expect the E field to be at the center of the cylinder? (remember how it is defined... how would you expect a test charge to move if you had it at the center?)
2. You said you can solve it outside the cylinder. What do you get there?
3. How would you connect both results? (1 and 2).

This will probably give you a feeling for the kind of result you are looking for.
 

ahrkron

Staff Emeritus
Gold Member
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Of course it makes a difference!

It does makes a difference if you are inside or outside the cylinder. Say R is the radius of the cylinder and r the point at which you measure the field:

Outside, applying Gauss' law yields something like:

(gaussian integral up to r) = (charge up to R)

Since R is fixed and r is larger than R, when you solve for E you get

E = (coefficient) x (charge density) / r

OTOH, inside the cylinder, you get

(gaussian integral up to r) = (charge up to r)

Notice how the right hand side has now an r on it, so that it combines with the r in the left hand side, and you get a different dependence.
 

Flame

got it!! :)

Thanks. I realized what I was doing wrong. I was using this certain problem as a template for what I was doing, only they were finding the charge of a certain distance from a LINE of charge. Hence, they were using (Lambda)*(L), rather than (ro)*(V), as I was supposed to.

The final equation, for anyone interested is:

E=2*(ro)*(pi)*(r)*(k)

where r is the distance from the axis, andk k=9x10^9

Thanks again. I'm sure I'll have more questions shortly.
 

Flame

Ok. I have a new one.

These circuit problems really get me sometimes. Maybe you can give me some general statements that I haven't caught onto yet to help me think through problems like this. here is goes:

If Va - Vb = 50V, how much energy is stored in the 54-uF capacitor shown?

The image that is shown is roughly this:
Code:
                    ------| 36uF |------
(a)------| 72uF |----                  -----(b)
                    ------| 54uF |------
My idea was to first combine the capacitors to find a single equivalent capacitor (which comes out to 80uF). I was trying to use the relationship:
(Ua)/q0-(Ub)/q0=Va-Vb

Together with this, I found the charge on the supposed 80mF capacitor by using the relationship Q=C(delta-V)

I'm stuck. Any ideas?
 
Last edited by a moderator:

Flame

The diagram

Sorry the diagram didn't come out. it's supposed to be a single line with a 72uF capacitor on it. It then branches into two; one branch has a 36uF capacitor, and the other has a 54uF capacitor. The two then join back together. Hope it's easy enough to understand by just the words.
 

ahrkron

Staff Emeritus
Gold Member
735
1
I haven't done any of these for a while, but I think you're on the right track: with the equivalent capacitance and deltaV, you can get the total charge. Then, you can get the charge and deltaV on each capacitor, and get the energy on the one you want...

you'll need to double check,... I'm a bit rusty on this, and about to fall sleep.

Good luck!
 

LogicalAtheist

Re: Ok. I have a new one.

I don't see any 54uF capacitor on your picture.

Problem just doesn't look posted good enough for an answer.

Try it again, perhaps I can help

Make a .BMP and post it!
Originally posted by Flame
These circuit problems really get me sometimes. Maybe you can give me some general statements that I haven't caught onto yet to help me think through problems like this. here is goes:

If Va - Vb = 50V, how much energy is stored in the 54-uF capacitor shown?

The image that is shown is roughly this:

------| 36uF |------
(a)------| 72uF |---- -----(b)
------| 64uF |------

My idea was to first combine the capacitors to find a single equivalent capacitor (which comes out to 80uF). I was trying to use the relationship:
(Ua)/q0-(Ub)/q0=Va-Vb

Together with this, I found the charge on the supposed 80mF capacitor by using the relationship Q=C(delta-V)

I'm stuck. Any ideas?
 

ahrkron

Staff Emeritus
Gold Member
735
1
Re: Re: Ok. I have a new one.

Originally posted by LogicalAtheist
I don't see any 54uF capacitor on your picture.
It is the one marked "64".
 

ahrkron

Staff Emeritus
Gold Member
735
1
Re: Ok. I have a new one.

Here:

Code:
                            ------| 36uF |------
(a)------| 72uF |----                         -----(b)
                            ------| 64uF |------
Flame: you just need to put your ascii drawing between {CODE}{/CODE}
(in square instead of curly brackets)
 

LogicalAtheist

Re: Re: Ok. I have a new one.

AHA! And now it makes sense!

I would have to consult my book to get the answer, as this is one area of physics that I didn't remember to remember.

If you seriously need help, like if you don't get it in 2 days, I can dit it up. But there's probably some people tommorrow that will help ya.

But I do know the basics. From the change in voltage use forumula to get some value. Consolidate the capacitors. Use said value, plus total capicatance and work backwords. Consolidate first the last two only to get the first one. the opposite.

Originally posted by ahrkron
Here:

Code:
                            ------| 36uF |------
(a)------| 72uF |----                         -----(b)
                            ------| 64uF |------
Flame: you just need to put your ascii drawing between {CODE}{/CODE}
(in square instead of curly brackets)
 

Flame

Don't know why...

I attach the file, tried both jpg and bmp. Neither one is showing up...
 

ahrkron

Staff Emeritus
Gold Member
735
1
Ok,...

I think you need to:

1. get the total equivalent capacitance, C
2. Obtain the total charge in the circuit (from C and V),
3. "Unfold" the big C in two capacitances: one is the original C1 (72 uF), and the second one is the equivalent to the remaining two (call it C23).
4. From there, you can get the voltage difference on each of C1 and C23. IIRC, it goes like V23 = V(C1/(C1+C23)), but you'll need to double check this.
5. Once with V23 (which is the v. diff. in the system formed by C2 and C3), you know that the voltage across C3 is also V23 (so you can get the charge Q3).
6. Finally, you can get the energy from Q3 and C3.

Makes sense?
 

schwarzchildradius

capacitance = C = q/V
U = (1/2)CV2
capacitors sum in series (I always think of "caparallel, plus plus plus")
Ceq=&sum;C}par
Ceq=1/&sum; (1/C)}ser
you should solve it with Kirchoff's rules:
input to every junction = output from every junction
and
sum of voltages around a closed loop = 0
 

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