Electric field between inner and outer cylinder

In summary, the problem involves a capacitor made of two concentric metal cylinders with a dielectric material in between. A charge of 5 nC is transferred from the inner cylinder to the outer shell. Using Gauss's law, the electric field between the two cylinders can be calculated and graphed from r = 0.5 cm to r = 1.5 cm. The field is zero outside of the region between the two cylinders and is reduced by a factor of the dielectric constant within the region. The field on the inner cylinder is 8990 N/C and the field on the outer cylinder is 7190 N/C. The question is asking for an equation for the field based on r and if the field is zero for
  • #1
Brennen berkley
19
0

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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  • #2
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.
 
  • #3
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.
 

What is the concept of electric field between inner and outer cylinder?

The electric field between inner and outer cylinder refers to the region of space between two concentric cylinders where an electric field is present due to the presence of charges on the inner and outer surfaces of the cylinders.

How is the electric field calculated between inner and outer cylinder?

The electric field between inner and outer cylinder can be calculated using the formula E = Q/(2πεR), where E is the electric field, Q is the charge on the cylinder, ε is the permittivity of the medium between the cylinders, and R is the distance between the cylinders.

What factors affect the strength of the electric field between inner and outer cylinder?

The strength of the electric field between inner and outer cylinder is affected by the magnitude of the charges on the cylinders, the distance between the cylinders, and the permittivity of the medium between the cylinders.

Is the electric field between inner and outer cylinder uniform?

No, the electric field between inner and outer cylinder is not always uniform. It depends on the distribution of charges on the cylinders and the distance between them. If the charges are evenly distributed and the distance between the cylinders is constant, then the electric field can be considered uniform.

How is the direction of the electric field determined between inner and outer cylinder?

The direction of the electric field between inner and outer cylinder is determined by the direction of the force that a positive test charge would experience if placed in the electric field. The direction of the electric field is always from positive charges to negative charges.

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