Electric field between inner and outer cylinder

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SUMMARY

The discussion centers on calculating the electric field between two concentric metal cylinders forming a capacitor, with the inner cylinder radius R1 = 1.0 cm and the outer cylinder radius R2 = 1.25 cm. The space between the cylinders is filled with nylon, which has a dielectric constant of 4.0. Using Gauss's law, the electric field E(r) is derived as E(r) = (Q/L)/(2πrε), where Q = 5 nC and ε is the permittivity of the dielectric. The calculated electric fields are 8990 N/C at the inner cylinder and 7190 N/C at the outer cylinder, with the electric field being zero outside the region defined by 1.0 cm < r < 1.25 cm.

PREREQUISITES
  • Understanding of Gauss's law in electrostatics
  • Familiarity with electric fields and capacitance concepts
  • Knowledge of dielectric materials and their properties
  • Ability to graph mathematical functions
NEXT STEPS
  • Learn how to apply Gauss's law to different geometries
  • Explore the effects of dielectric materials on electric fields
  • Study the relationship between charge, electric field, and potential in capacitors
  • Investigate numerical methods for graphing electric fields in cylindrical coordinates
USEFUL FOR

Students studying electromagnetism, electrical engineers, and anyone involved in capacitor design and analysis.

Brennen berkley
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Homework Statement


A capacitor is constructed of two long concentric metal cylinders, each having length of 1.0 meters. The inner cylinder has a radius R1 = 1.0 cm, and the outer cylinder has a radius R2 = 1.25 cm. The hollow space between the two cylinders is filled with nylon having a dielectric constant of 4.0. A charge of Q =5 nC is transferred from the inner cylinder to the outer shell. (a) Find the field between the two using Gauss's law, and use the computer to make a graph of the magnitude of the electric field E(r) from r = 0.5 cm to r = 1.5 cm.

Homework Equations


E = q/εA

The Attempt at a Solution


The field on the inner cylinder would be E = 8990 N/C and the field on the outer cylinder would be 7190 N/C. I am confused though because it is asking for an electric field between the plates. Do they just want an equation for the field based on r? For the graph part I assume they want an equation, but when r < 1 or r > 1.25 the field would be zero right? I just don't totally understand the question, so any input regarding what you think my professor wants would be appreciated.

I am also assuming the cylinders have no top or bottom.
 
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To begin, first work the problem without any dielectric between the cylinders. A charge per unit length Q/L gives a radially outward electric field (using Gauss's law M.K.S. units) obeying ## E(r)*2*\pi*r*L=(Q/L)/\epsilon_o)*L ## ==>> ## E(r)=(Q/L)/(2\pi*r\epsilon_o) ## The dielectric will reduce this E(r) by a factor of the dielectric constant ## \epsilon ##. The electric field is zero outside of the region ## 1.0 cm<r<1.25 cm ## In doing the calculations, be sure to have "r" expressed in meters.
 
Brennen berkley said:
The field on the inner cylinder would be E = 8990 N/C and the field on the outer cylinder would be 7190 N/C. I am confused though because it is asking for an electric field between the plates. Do they just want an equation for the field based on r? For the graph part I assume they want an equation, but when r < 1 or r > 1.25 the field would be zero right? I just don't totally understand the question, so any input regarding what you think my professor wants would be appreciated.
What exactly do you want to know? If, whether the field for r less than 1, or greater than 1.25cm is zero, then yes, it would be.
 

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