# Homework Help: Electrostatics:Longitudinal charge density of conductors

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1. Jun 29, 2015

### gruba

1. The problem statement, all variables and given/known data
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.
2. Relevant 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.

3. 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.

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2. Jun 29, 2015

### haruspex

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: Jun 29, 2015
3. Jun 29, 2015

### SammyS

Staff Emeritus
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.

4. Jun 30, 2015

### gruba

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: Jun 30, 2015
5. Jun 30, 2015

### haruspex

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.

6. Jun 30, 2015

### SammyS

Staff Emeritus
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: Jun 30, 2015