# Electric field of a cylindrical capacitor

• Eigenvaluable
In summary, the electric field of a cylindrical capacitor can be calculated using the formula $$E = \frac{Q}{2 \pi r L \epsilon_0}$$ but this formula assumes that the length of the cylinder is much larger than the radius of the larger cylinder. This assumption is similar to the one made in analyzing parallel plate capacitors, where the plate dimensions are taken to be much larger than the separation between them. These assumptions may not hold true everywhere in the field.
Eigenvaluable

## Homework Statement

Calculate the electric field of a cylindrical capacitor comprised of a smaller cylindrical conductor of radius ##a## enclosed within a larger cylindrical conductor of radius ##b## where ##b>a##. The smaller cylinder has charge ##+Q## and the larger cylinder has charge ##-Q##.

## Homework Equations

$$\oint \vec{\mathbf{E}} \cdot \vec{\mathbf{dA}} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$

## The Attempt at a Solution

I've already solved the problem, as follows:

Define a Gaussian surface in which the cylinder completely encloses the smaller cylinder and is completely within the larger cylinder, i.e. a Gaussian cylinder with radius ##r## where ##a<r<b##.

Furthermore, let ##L## denote the length of the cylinder.

$$(2 \pi r L) E = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$
$$E = \frac{Q}{2 \pi r L \epsilon_0}$$

While this is trivially easy, this reference sheet assumes that ##L >> b## prior to calculating anything.

Why must this be done?

Thank you,
Eigenvaluable

Hello Eigenvaluable, Welcome to PF.

Eigenvaluable said:

## Homework Statement

Calculate the electric field of a cylindrical capacitor comprised of a smaller cylindrical conductor of radius ##a## enclosed within a larger cylindrical conductor of radius ##b## where ##b>a##. The smaller cylinder has charge ##+Q## and the larger cylinder has charge ##-Q##.

## Homework Equations

$$\oint \vec{\mathbf{E}} \cdot \vec{\mathbf{dA}} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$

## The Attempt at a Solution

I've already solved the problem, as follows:

Define a Gaussian surface in which the cylinder completely encloses the smaller cylinder and is completely within the larger cylinder, i.e. a Gaussian cylinder with radius ##r## where ##a<r<b##.

Furthermore, let ##L## denote the length of the cylinder.

$$(2 \pi r L) E = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$
$$E = \frac{Q}{2 \pi r L \epsilon_0}$$

While this is trivially easy, this reference sheet assumes that ##L >> b## prior to calculating anything.

Why must this be done?
A similar stipulation is made in the analysis of parallel plate capacitors where the plate dimensions are taken to be much larger than the plate separation. Consider the assumptions being made about the field between the capacitor plates. Do they hold true everywhere?

## 1. What is the formula for calculating the electric field of a cylindrical capacitor?

The formula for calculating the electric field of a cylindrical capacitor is E = Q / (2πε0lr), where E is the electric field, Q is the charge on the capacitor, ε0 is the permittivity of free space, l is the length of the cylinder, and r is the radius of the cylinder.

## 2. How does the electric field vary with distance from the center of a cylindrical capacitor?

The electric field of a cylindrical capacitor is directly proportional to the distance from the center of the capacitor. This means that as the distance increases, the electric field decreases. This relationship is described by the formula E ∝ 1/r, where r is the distance from the center of the capacitor.

## 3. Can the electric field of a cylindrical capacitor be negative?

Yes, the electric field of a cylindrical capacitor can be negative. This occurs when the charge on the capacitor is negative and the electric field points in the opposite direction of the positive charge. However, the magnitude of the electric field is always positive.

## 4. How does the electric field of a cylindrical capacitor compare to that of a parallel plate capacitor?

The electric field of a cylindrical capacitor is similar to that of a parallel plate capacitor in that both have a constant electric field between their plates. However, the electric field of a cylindrical capacitor is stronger near the edges of the cylinder, while the electric field of a parallel plate capacitor is more uniform between the plates.

## 5. What factors affect the strength of the electric field in a cylindrical capacitor?

The strength of the electric field in a cylindrical capacitor is affected by the charge on the capacitor, the length and radius of the cylinder, and the permittivity of free space. Additionally, the material between the plates of the capacitor can also affect the strength of the electric field.

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