Electric Field Inside Cylindrical Capacitor

In summary, the conversation discusses the use of the relation between flux, electric field, and charge to calculate the electric field inside a cylindrical capacitor. The integral simplifies to E * A = (q_enclosed)/(ε_naught) when using a cylindrical gaussian surface. The question then addresses the potential impact of negative charges on the outside cylinder and whether they should be considered in the calculation of the electric field. The response explains that the electric flux through the gaussian surface accounts for all field lines, including those produced by the negative charges, and considering them separately would result in double-counting. The conversation ends with a question about the magnitude of the electric field produced by the outer charge inside the cylindrical shell.
  • #1
Idontknowhatimdoing
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Homework Statement
When we want to find the electric field inside of a cylindrical capacitor, we can use Gauss's law and the relation between flux and field to calculate what this field is.
Relevant Equations
Gauss's law
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we know that flux is equal to the area integral of electric field dotted with dA and we can set this equal to charge enclosed divided by epsilon naught. Thus, in this case, the integral simplifies to E * A = (q_enclosed)/(ε_naught) when we choose a cylindrical gaussian surface with radius of r.

My question, then, is why are we allowed to use this relation to find electric field inside of the capacitor. I thought that the electric field that is calculated only describes the electric field that the charges enclosed produce. In this situation, wouldn't the negative charges on the outside cylinder affect the Electric Field too? If it does, the electric field that we find wouldn't be the right electric field between the cylinders. Is there a reason that we can ignore the negative charge on the outside cylinder in this capacitor? Or is my thought process incorrect?
 
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  • #2
Look at the drawing on the right. Whatever field lines leave the inner surface where the positive charges are located terminate at the inner surface of the outer cylinder where the negative charges are. The electric flux through the dotted Gaussian surface essentially counts all the electric field lines. If you were to somehow account for the charges outside the Gaussian surface, you will be double-counting.
 
  • #3
@OP. What do you think is the magnitude of the field produced by the outer charge inside the cylindrical shell?
 

FAQ: Electric Field Inside Cylindrical Capacitor

What is the electric field inside a cylindrical capacitor?

The electric field inside a cylindrical capacitor is radial and depends on the distance from the axis of the cylinder. It can be calculated using Gauss's law, and for a capacitor with an inner radius \(a\) and outer radius \(b\), the electric field \(E(r)\) at a distance \(r\) from the axis is given by \(E(r) = \frac{\lambda}{2 \pi \epsilon_0 r}\), where \(\lambda\) is the charge per unit length and \(\epsilon_0\) is the permittivity of free space.

How does the electric field vary with distance inside a cylindrical capacitor?

The electric field inside a cylindrical capacitor varies inversely with the radial distance \(r\) from the axis of the cylinder. Specifically, it decreases as \(1/r\), indicating that the field strength is stronger closer to the inner conductor and weaker as you move towards the outer conductor.

What assumptions are made when calculating the electric field inside a cylindrical capacitor?

When calculating the electric field inside a cylindrical capacitor, it is typically assumed that the capacitor is infinitely long to ignore edge effects, the charge distribution is uniform along the length of the cylinder, and the dielectric material between the conductors is homogeneous and isotropic.

What role does the dielectric material play in the electric field inside a cylindrical capacitor?

The dielectric material between the conductors in a cylindrical capacitor affects the electric field by reducing it compared to a vacuum. The presence of a dielectric increases the capacitance by a factor equal to the dielectric constant \( \kappa \), and the electric field \(E(r)\) in the presence of a dielectric is given by \(E(r) = \frac{\lambda}{2 \pi \kappa \epsilon_0 r}\).

How can the potential difference between the conductors of a cylindrical capacitor be determined from the electric field?

The potential difference \(V\) between the inner and outer conductors of a cylindrical capacitor can be determined by integrating the electric field over the distance between the conductors. Specifically, \(V = \int_a^b E(r) \, dr\), which for a cylindrical capacitor with vacuum between the conductors, gives \(V = \frac{\lambda}{2 \pi \epsilon_0} \ln(b/a)\).

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