# Electrostatics:Longitudinal charge density of conductors

• gruba
In summary: And any charge on the central cylinder must reside on its outer wall.Could you show the expression for longitudinal charge density of grounded conductor and explain if it is not equal to zero? Also, did I derived correctly longitudinal charge densities for first two conductors?The expression for the longitudinal charge density of the grounded conductor is:Q' = (2πε0Va)/(ln(b/a)+ln(c/b))This is derived using the equation for electric potential (V=∫Edl) and the integral for Va, which is from (a to b)+(b to c). The grounded conductor has a potential of zero, so the integral for it is just zero. However, in order for the electric field to be zero at points outside of
gruba

## Homework Statement

Three very long (theoretically infinite long) hollow cylindrical conductors, with radius a,b,c (c>b>a) are in vacuum. Inner and central conductor are charged, and outer conductor is grounded. Potentials of inner and central conductors with reference point relative to outer conductor are Va,Vb. Find longitudinal charge density of all three conductors.

## Homework Equations

Electric field of cylindrical conductor can be derived using Gauss law for vacuum: E=Q'/(2πrε0), where Q' is longitudinal charge density.
Electric potential of cylindrical conductor is given by: V=∫Edl, where dl represents integration by radius.

## The Attempt at a Solution

If outer conductor is grounded, and it is a referent point to potentials Va and Vb, integration for Va will be from (a to b)+(b to c),
Va=(Q'/(2πε0))*(ln(b/a)+ln(c/b))

Integration for Vb will be from (b to c),
Vb=(Q'/(2πε0))*ln(c/b)

We need to find longitudinal charge density for each capacitor, so for first we derive it from Va:
Q'=(2πε0Va)/(ln(b/a)+ln(c/b))

For second conductor, we derive it from Vb:
Q'=(2πε0Vb)/(ln(c/b))

Third conductor is grounded, so the potential of the third conductor is equal to zero, thus the longitudinal charge density of the third conductor is equal to zero.

Could someone check this, and help if something is not correct?
Thanks for replies.

#### Attachments

• image.PNG
5.9 KB · Views: 443
Each shell generates a field. At each shell, the potential is the sum of the potentials generated by the three shells at that radius. Start by creating unknowns for the charge densities on each shell.
The grounded shell has a potential of zero, but it must have a charge in order to neutralise the fields generated by the other two.

Last edited:
gruba said:

## Homework Statement

Three very long (theoretically infinite long) hollow cylindrical conductors, with radius a,b,c (c>b>a) are in vacuum. Inner and central conductor are charged, and outer conductor is grounded. Potentials of inner and central conductors with reference point relative to outer conductor are Va,Vb. Find longitudinal charge density of all three conductors.

## Homework Equations

Electric field of cylindrical conductor can be derived using Gauss law for vacuum: E=Q'/(2πrε0), where Q' is longitudinal charge density.
Electric potential of cylindrical conductor is given by: V=∫Edl, where dl represents integration by radius.

## The Attempt at a Solution

If outer conductor is grounded, and it is a referent point to potentials Va and Vb, integration for Va will be from (a to b)+(b to c),
Va=(Q'/(2πε0))*(ln(b/a)+ln(c/b))

Integration for Vb will be from (b to c),
Vb=(Q'/(2πε0))*ln(c/b)

We need to find longitudinal charge density for each capacitor, so for first we derive it from Va:
Q'=(2πε0Va)/(ln(b/a)+ln(c/b))

For second conductor, we derive it from Vb:
Q'=(2πε0Vb)/(ln(c/b))

Third conductor is grounded, so the potential of the third conductor is equal to zero, thus the longitudinal charge density of the third conductor is equal to zero.

Could someone check this, and help if something is not correct?
Thanks for replies.
Your basic approach seems fine. You need to be careful of details in solving this problem.

The commonly used symbol for charge per unit length is λ (Greek 'lambda'), which is called longitudinal charge density in your problem statement.

Let λa, λb, and λc be linear charge densities on cylinders of radii a, b, and c respectively.

Your expression for the electric field (coming from Gauss's Law) looks good. ##\displaystyle \ E_r=\frac{\lambda_\text{in}}{2\pi\epsilon_0 r} \ ##

Then use ##\displaystyle \ V_{r_2}-V_{r_1}=-\int_{r_1}^{r_2}E_r\,dr \ .##

I would start with ##\displaystyle \ V_{b}-V_{c} \,,\ ## and proceed on from there.

haruspex said:
Each shell generates a field. At each shell, the potential is the sum of the potentials generated by the three shells at that radius. Start by creating unknowns for the charge densities on each shell.
The grounded shell has a potential of zero, but it must have a charge in order to neutralise the fields generated by the other two.

Could you show the expression for longitudinal charge density of grounded conductor and explain if it is not equal to zero? Also, did I derived correctly longitudinal charge densities for first two conductors?

Last edited:
gruba said:
Could you show the expression for longitudinal charge density of grounded conductor and explain if it is not equal to zero? Also, did I derived correctly longitudinal charge densities for first two conductors?
Just assign unknown linear charge densities to each shell, as SammyS and I already suggested. Write out the equations for the resulting potentials and solve.

gruba said:
Could you show the expression for longitudinal charge density of grounded conductor and explain if it is not equal to zero? Also, did I derived correctly longitudinal charge densities for first two conductors?
The longitudinal charge density of grounded conductor depends upon the particular values one might be given for Va and Vb. In general, the longitudinal charge density of grounded conductor is not zero. To see this, for the electric field to be zero, beyond the grounded conductor (that is for r > c ) the sum of all of the longitudinal charge densities on all of three conductors must be ______ . (Use Gauss's Law to fill in the blank.)

By the way:
In this problem it appears that the thickness of the conductors is small enough to be ignored. However, you should know that any charge on the outer cylinder must reside on its inner wall.

Last edited:

## 1. What is the longitudinal charge density of conductors?

The longitudinal charge density of conductors refers to the amount of electric charge per unit length along the direction of current flow. It is usually denoted by the symbol λ and is measured in coulombs per meter (C/m).

## 2. How is the longitudinal charge density of conductors related to electric current?

The longitudinal charge density is directly proportional to the electric current flowing through a conductor. This means that as the current increases, the charge density also increases, and vice versa.

## 3. Are there any factors that affect the longitudinal charge density of conductors?

Yes, there are several factors that can affect the longitudinal charge density of conductors. These include the material of the conductor, its length and cross-sectional area, and the applied voltage or potential difference across the conductor.

## 4. How does the longitudinal charge density of conductors affect the electric field within the conductor?

The longitudinal charge density is directly related to the electric field within a conductor. As the charge density increases, the electric field also increases, and vice versa. This is because the electric field is created by the movement of charges, and a higher charge density means more charges are present in a given area.

## 5. Can the longitudinal charge density of conductors be changed?

Yes, the longitudinal charge density of conductors can be changed by altering the factors that affect it, such as the current, material, and dimensions of the conductor. It can also be changed by applying an external electric field to the conductor.

• Introductory Physics Homework Help
Replies
19
Views
293
• Introductory Physics Homework Help
Replies
10
Views
313
• Introductory Physics Homework Help
Replies
23
Views
921
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
26
Views
917
• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
748
• Introductory Physics Homework Help
Replies
2
Views
299
• Introductory Physics Homework Help
Replies
15
Views
4K