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...