A resistor (20 ohm) is made of a very thin piece of metal wire, length = 3mm and diameter = .1mm. Given that it has a potential of 8 volts and .4 amps of current running through it, what is the electric field inside the wire?
I know there's no electrostatic equilibrium here, so I can't just...
So (using my limited knowledge of matrixes) my equations should be solved so they look like this:
a = d+e
c= a+b
c = b+d+e
c-b-d-e = 0
e = ((200*d + 40)/80)
e - 5/2*d = 1/2
d = ((20*b+300)/200)
d - b/10 = 3/2
b = ((200*d-300)/20)
b - 10d = 15
c=((-200*d-80)/70)
c...
I'm going to assume I didn't make any mistakes in modeling the circuit. I've done two other problems similar to this and I'm confident that I've ironed out my misconceptions about modeling the circuits, if not my difficulty with the algebra involved in solving said systems of equations.
Jacob Chestnut:
The constants? I have said equations (see above) should my input be of the form:
c=((-200*d-80)/70)
-200, 80, 70
Or:
c + 20/7d = -8/7
1, 20/7, -8/7
(I had some additional help that counseled me to solve for the variables then input the constants into my...
Ok, I took your advice cookie, I tried to enter the equations in the solver (as I'm rusty on my matrixes), here is what I entered:
(for convenience I assigned my difficult to enter variables to single letters easily accessible on the calculator keypad:
I_1 = a
I_2 = b
I_3 = c
I_5 = d...
(see attached for circuit diagram)
E1 = 40V
E2 = 300V
E3 = 80V
r1 = 200 ohm
r2 = 80 ohm
r3 = 20 ohm
r4 = 70 ohm
I've come up with some very ugly equations to express the current flow across various parts of the circuit, as follows:
I_1 = I_5 + I_6
I_3 = I_1 + I_2
I_3 =...
A question surprisingly similar to this:
https://www.physicsforums.com/showthread.php?t=17219
I = 1000amp
V = 700,000volt
distance = 100miles
Resistence of the wire = .5 ohm / mile
The resistence of my line is 50 ohms, original power is 7e5 * 1e3 = 7e8 watts. Final power, due to P =...
Edit:
Moved from statistics -
https://www.physicsforums.com/showthread.php?s=&postid=164120#post164120
I'm doing some homework over the break (!) so I don't have access to my usual lines of help. I've hit a wall:
I don't really know how to solve a problem like this, from the...
I know there is charge present on each plate of a capacitor, as they store charge (by definition). The charges are equal and opposite and it is the electric forces generated by that charge that allows said charge to be stored. Taken as a system and preserving the signs, yes there is no net...
Quanta? I'll assume it means amount. So my intuition was correct about the maximum available charge is spread between the cylinders.
Calculating the electric field at the outer cylinder:
So you are correct, the electric field does extend beyond the outer cylinder and some distance...
First I set about to calculate the electric field:
Solve for Q with the capacitance of two coaxial cylinders:
substituting:
Where 'r' is a distance measured from the center. Trying several distances (r = 5cm -> E = 69.2 Volts / meter, r = 2cm [on the internal cylinder] -> 432.8 Volts...
Yes, it seems to me that this should be the answer. Maybe it is a misprint, all the evidence seems to point that way.
Calculating the charges on the two capacitors:
Q_1 = C1 * V_1 = 5600*2 = 11200
Q_2 = C2 * V_1 = 1400*8 = 11200
It seems that the charges on the capacitors are equal...
A group of physics students designed the following experiment to test the model of conservation of charge.
a) They first charged a capacitor C1 = 5600 micro F by applying a voltage V_0 = 10 volts (as shown in attachment Exam 2 - Problem 2 - Part a.JPG)
b) Then the connected a second...
A group of physics students designed the following experiment to test the model of conservation of charge.
a) They first charged a capacitor C1 = 5600 \mu\f
cookiemonster -
I'm sorry, but those explanations do nothing for me. We haven't learned T-tests as far as I know.
Damned charming :) -
I don't have power point, thanks though.
I can't get those pictures (?) to paste, but you get the idea. How did you do that anyway? That's kool...
Huh? I don't have a formula for this ditribution nor do I know how to make one.
Haven't learned any tests that look like that.
Okay? I have formulas? Huh?
I'm going to email my teacher obviously I'm missing something.
I'm doing some homework over the break (!) so I don't have access to my usual lines of help. I've hit a wall:
I assume the "N(12.2, .04)" notation refers to that the distribution of the boxes has mean 12.2 and variance .04. I think 'a' has something to do with the normal distribution, how...
Wow Holly, good advice. I never even thought about asking another professor. Of course the shape I was in when I wrote that I probably couldn't done anything creative much less constructive. I'll try some of those things tomorrow. Heh your adivce may have stopped other from responding...
I need help. I'm studying physics, electricity and magnetism. I'm completely lost. My teacher does nothing for me, I have learned absolutely nothing from her since the course began. What I have learned, I have learned from doing homework problems and inferring from those who help me here...
So the if I'm interpreting this correctly, the charge on the inner surface of the outer sphere should be -3q and the surface charge on the outer sphere is 2q. The "internal" electric field (inside the material of the outer sphere) should be equal and opposite the field due to the inner sphere?
Looking at that sentence, I'm not even sure what I meant, ^_^. I'll try to reconstruct my thought patterns as best I can. i suppose I was saying that the charges create an equal and opposite relationship, the charge on the inner sphere is balanced by the negative net charge on the inside of...
gnome:
So charge effects charge internal to the outer sphere but the electric fields extend beyond either.
This is a almost a direct contradiction of what my teacher told me about a previous problem.
Why isn't there an electric field produced internally to the out sphere? There's an...
paul11273:
I was thinking in straight lines, excuse me. From my understanding the negative plate would not put forth an electric field, it only receives electric field "lines." But from what you said I think I have an inkling of what you mean. You're saying that the electric field from...
Excerpt from:
Beicher and Serway, "Physics for Scientists and Engineers with
Modern Physics, 5th edition"
Chapter 24, Problem 53, part e:
A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed throughout its volume. Concentric with...
Excerpt from:
Beicher and Serway, "Physics for Scientists and Engineers with
Modern Physics, 5th edition"
Chapter 24, problem 58, part c:
Two infinite, non conducting sheets of charge are parallel to each other, as show in the figure. The sheet on the left has a uniform...
So due the symmetry of the sphere any point inside it experiences no net electric field because any for exterted on any point in that sphere is countered by an equal but opposite field, produced by every other point on the sphere.
I think that's what I mean. These ideas don't really lend...
Thank you that's much clearer.
I am unclear on Gauss's law apparently:
Why is this? My teacher is unclear at best, so you'll have to bear with me as I stumble through this. Thanks for your help.
I was able to copy the image from the homework problem. It has to be in zip format, I don't know why the forum code won't allow me to attach a 630x432 image when its well within the maximum attachment size. But whatever there it is.
I cannot post the diagram for some reason, but I will do my best to explain. The rod is lying horizontally, the balloons are lying 3cm away from the center of the rod (from the edge of the rod to the centers of the balloons) and are 2cm apart (from edge to edge).
Part 'A' says:
"Find the...
I'm currentlt taking physics 4b (electricity and magnetism) and I'm having problems deciphering a homework problem.
I'm just mystified at what the question is asking, could someone restate the question in simpler terms? I have copied the question verbatim, I will attempt to supply...