Potential inbetween two coaxial cylinders

In summary, the conversation discusses the setup of 2 coaxial cylindrical conductors with different radii and a potential difference between them. Using Laplace's equation in cylindrical coordinates and the given boundary conditions, the potential in the region between the cylinders is found to be V = Vo/[ln(a/b)] * ln(r/b). This result is then confirmed using Gauss' theorem.
  • #1
aftershock
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0

Homework Statement



There are 2 coaxial cylindrical conductors. The inner cylinder has radius a, while the outer cylinder has radius b. There is no charge in the region a < r < b. If the inner cylinder is at potential Vo and the outer cylinder is grounded, we want to find the potential in the region between the cylinders. We assume L >> b > a and neglect end effects.

a) Write Laplace's equation in cylindrical coordinates.

b) Assuming V(r) is a function of the axial distance r alone, integrate the differential equation and use the boundary conditions to find V(r) , a ≤ r ≤ b.



Homework Equations



2V = 0

The Attempt at a Solution



Laplace's equation in cylindrical coordinates would be

(1/r)(d/dr)(r*dV/dr) =0

Since there is no phi or z dependence. Also I know it's a partial and not d but I can't type the symbol.

To solve this

(1/r)(d/dr)(r*dV/dr) =0

(d/dr)(r*dV/dr) =0

r*dV/dr = C

dV/dr = C/r

dV = C/r dr

V = C*ln(r) + K , where C and K are constants.

Using the boundary conditions:

0 = C*ln(b) + K

Vo = C*ln(a) + K

and using that to solve for constants:

V = Vo/[ln(a/b)] * ln(r/b)

Is that right?
 
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  • #2
The result is correct and the other parts look fine, too.

LaTeX code can be used to write partial derivatives: ##\frac{\partial}{\partial r}## --> ##\frac{\partial}{\partial r}##.
 
  • #3
aftershock said:

Homework Statement



V = Vo/[ln(a/b)] * ln(r/b)

Is that right?

Confirmed using Gauss' theorem.
 

Related to Potential inbetween two coaxial cylinders

1. What is the formula for calculating the potential between two coaxial cylinders?

The formula for calculating the potential between two coaxial cylinders is V = (Qln(b/a))/(2πε₀L), where Q is the charge on the inner cylinder, b is the radius of the outer cylinder, a is the radius of the inner cylinder, ε₀ is the permittivity of free space, and L is the length of the cylinders.

2. How does the potential between two coaxial cylinders change with distance?

The potential between two coaxial cylinders decreases with distance. As the distance between the cylinders increases, the potential decreases due to the inverse square law. This means that the potential decreases by a factor of 4 when the distance between the cylinders is doubled.

3. What factors affect the potential between two coaxial cylinders?

The potential between two coaxial cylinders is affected by the charge on the inner cylinder, the radii of the cylinders, the length of the cylinders, and the permittivity of free space. The potential also depends on the distance between the cylinders and follows the inverse square law.

4. How does the potential between two coaxial cylinders change if one of the cylinders is grounded?

If one of the cylinders is grounded, the potential between the two cylinders will decrease. This is because the grounded cylinder acts as a sink for the electric field lines, reducing the potential difference between the cylinders.

5. Can the potential between two coaxial cylinders ever be negative?

Yes, the potential between two coaxial cylinders can be negative. This occurs when the inner cylinder has a negative charge and the outer cylinder has a positive charge. In this case, the potential will be negative for points between the cylinders and positive outside of the cylinders.

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