Finding electric potential given a cylindrical configuration

In summary, to find the electric potential between two concentric cylinders with radii a and b, use Gauss's Law to find the electric field E between the two cylinders. Then, use the relation E = -ΔV/dr to find the potential difference between the two cylinders. This results in the equation V = constant1 - constant2 ln(r) where r is the distance from the center of the cylinders.
  • #1
carloz
4
0

Homework Statement



You have two concentric cylinders. The inner cylinder has radius a and the external cylinder has radius b. Find the electric potential in the region between the cylinders.

[Hint: The final equation takes the form V(r) = constant1 - constant2 ln(something) ]


Homework Equations



I think:
Gauss's law: [itex]\Phi[/itex]=EA=q/ε
Coulomb's law: E=q/(4πεr^2)

The Attempt at a Solution



I really don't know.
 
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  • #2
Start by finding the electric field E between the two cylinders using Gauss's Law.

How is E related to the potential?
 
  • #3
Okay

So

EA=q/ε

E=q/(εA)

A=2πrL

dA=2πLdr

dE=q/(ε2πLdr)

E = ∫(limits a and b) q/(ε2πLdr) = q/(ε2πL) ∫(limits a and b) 1/dr

Am I on the right track? If so, how to evaluate ∫(1/dr) ?

Thank you.
 
  • #4
Do you know the answer to this? I am just learning this material also, so I don't know if I am thinking through it correctly. But the outer shell should be irrelevant as the electric field is going to be zero for any point inside it assuming it is infinitely long, correct? For the electric field at r distance from the center of cylinder with radius a, would its magnitude be λ/(2πεr) (derived with Gauss's law)? If so, you would end up with Δv(r) = v(r) (taking v_i = 0) = λln(a)/(2πε) - λln(r)/(2πε) where a <= r <= b, λ = q/l. But I have no idea if that is correct.
 
  • #5
carloz said:
Okay

So

EA=q/ε

E=q/(εA)

A=2πrL

dA=2πLdr

dE=q/(ε2πLdr)

E = ∫(limits a and b) q/(ε2πLdr) = q/(ε2πL) ∫(limits a and b) 1/dr

Am I on the right track? If so, how to evaluate ∫(1/dr) ?

Thank you.
No. First off, what's your Gaussian surface and what expression is equal to the amount of charge enclosed in it?
 

Related to Finding electric potential given a cylindrical configuration

1. What is electric potential in a cylindrical configuration?

Electric potential in a cylindrical configuration is a measure of the electrical potential energy per unit charge at a given point in space. It is a scalar quantity that describes the electric potential at any point within the cylindrical configuration.

2. How is electric potential calculated in a cylindrical configuration?

To calculate electric potential in a cylindrical configuration, you will need to know the charge distribution and the distance from the center of the cylindrical configuration to the point where you want to find the electric potential. Then, you can use the formula V = kλ/r, where V is the electric potential, k is the Coulomb constant, λ is the charge per unit length, and r is the distance from the center.

3. What is the relationship between electric potential and electric field in a cylindrical configuration?

The electric field and electric potential in a cylindrical configuration are related by the formula E = -∆V/∆r, where E is the electric field, V is the electric potential, and r is the distance from the center. This means that the electric potential changes with respect to distance, and the electric field is the negative gradient of the electric potential.

4. How does the radius of the cylindrical configuration affect the electric potential?

The radius of the cylindrical configuration does affect the electric potential. As the distance from the center increases, the electric potential decreases. This is because the electric field strength decreases with distance, and the electric potential is directly proportional to the electric field strength.

5. Can the electric potential in a cylindrical configuration be negative?

Yes, the electric potential in a cylindrical configuration can be negative. This indicates that the electric potential energy is negative, which means that work must be done to move a positive charge from infinity to that point. It is important to note that the sign of the electric potential does not affect the direction of the electric field, which always points from positive to negative charges.

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