Phy 202
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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We previously had this conversation:
Find an expression for the electric field strength at distance r from the axis of a capacitor consisting of two co-axial cylinders 3 meters long, with diameters 1.88 cm and 2.56 cm, with the inner cylinder carrying charge - 14.7 `microCoulombs and the outer cylinder carrying charge + 14.7 `microCoulombs. You will find three expressions, depending on whether r is less than 1.88 cm, greater than 2.56 cm or between the two.
• Estimate the average electric field between the cylinders, then determine the approximate work done per unit charge to move a charge from the inner cylinder to the outer. (University Physics students find the exact amount of work, using an integral).
• What is the voltage between the cylinders?
• What is the capacitance of this capacitor?
• What would be the approximate magnitude of the effect on the work, voltage and capacitance if the inner cylinder was not changed, but the outer cylinder shrunk in such a way as to be 10 times closer to the inner?
I honestly have no idea where to even start with this. Can you please just point me in the right direction
The last three problems in Introductory Problem Set 1 address the flux model and how it relates to various situations, including this one.
An imaginary cylinder with radius r between 1.88 cm and 2.56 cm encloses the charge of the inner cylinder. This allows you to determine the electric field at a point between the two cylinders.
An imaginary cylinder with radius r greater than 2.56 cm encloses the charges of both cylinders.
First check out the Introductory Problems, then see if you can apply that information to this situation. You'll probably have questions, which I'll be glad to answer.
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OK so this is where i am after looking at the problem sets questions. we would find field by taking flux/area. the flux would be 4(pi)(k)(charge) while the area would be 2(pi)(r)(h). OK if this is the case then we solve for flux for in the inner and outer cylinder. The middle area will be outer-inner, but what would its charge be?
If h is the 3 meter length of the cylinders, then as long as r lies between the radius of the inner cylinder and the outer cylinder, the charge enclosed by your imaginary cylindrical surface would be the -14.7 microCoulombs residing on the inner cylinder. Your surface would have length h = 3 meters so its area would be 2 pi r * (3 meter) = 6 pi r meters.
The field would therefore be
field = flux / area = 4 pi k * (-14.7 microCoulombs) / (6 pi r meters) = 2/3 k *(-14.7 microCoulombs / meter) / r .
If you work out the units of 2/3 k *(-14.7 microCoulombs / meter) you will find that its units are N * m / Coulomb, i.e., J / C or volts, so that when divided by r in meters you get volts / meter.
You should work this out for yourself for practice, and to verify it, but the expression becomes
2/3 k *(-14.7 microCoulombs / meter) / r = 8.84 * 10^4 volts / r.
If you evaluate this at the two given conductor radii you will find that the fields are very roughly 5 * 10^6 volts/meter and 3.5 * 10^6 volts/meter. The field falls off as 1 / r so the field isn't a linear function of radius, and using the average 4.2 * 10^6 volts/meter of the two fields isn't very accurate; however this estimate will do for the purposes of the General College Physics course.
Multiply this estimated average field by the separation of the two cylinders we find that the voltage difference is
`dV = 4.2 * 10^6 volts / meter * .68 cm = 3 * 10^4 volts, very approximately.
The to find voltage I think i would use V=field*distance. I am not sure what the field is as i stated before.
For capacitance, this would be charge/voltage. Would i find this seperatly for the area with - and the area with +? If not what would I use for charge here?
The stored charge is 14.7 microCoulombs, so the capacitance is
capacitance = charge / voltage = 14.7 microCoulombs / (3 * 10^4 volts) = 5 * 10^-9 Coulombs / volt, or 5 * 10^-9 Farads.
Finally, I think that reducing the distance between the two cylinders will increase voltage which decreases capacitance. I am unsure of the relationship between capacitance, voltage and work. What is this?
Reducing the distance increases the average field. However the average field will not exceed the maximum field between the cylinders, which occurs at the inner cylinder and was roughly estimated as 5 * 10^6 volts / meter. On the other hand the separation can decrease to 0, so voltgage = ave field * separation can decrease to 0.
This doesn't prove that the voltage decreases with separation all the way from the original separation to separation 0 (that turns out to be the case but the proof is most easily done using calculus so we're going to leave it alone for our purposes), but it does at least show that if the cylinders are close enough, the voltage is close to 0. The charge is not assumed to change, so that the capacitance would in fact be unlimited (same charge divided by voltages which approach 0) at least in theory.
In reality there's a limit to how close those two cylinders can get without charge 'jumping' from one to the other, so there is an upper limit on actually achievable capacitance.
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Let me know how my reasoning is or if you see any errors. There are a few questions but I understand the question for the most part now. In addition the help you gave for the other three was great!
Hopefully this will be helpful to you. If not, don't be overly concerned. This is a great problem, but it would be graded somewhat leniently on the General College Physics test.