(a) Check that the results of Exs. 2.4 and 2.5, and Prob....

(a) Check that the results of Exs. 2.4 and 2.5, and Prob. 2.11, are consistent with Eq. 2.33. (b) Use Gauss's law to find the field inside and outside a long hollow cylindrical tube, which carries a uniform surface charge $\sigma .$ Check that your result is consistent with Eq. 2.33. (c) Check that the result of Ex. 2.7 is consistent with boundary conditions 2.34 and 2.36.

(a) Check that the results of Exs. 2.4 and 2.5, and Prob. 2.11, are consistent with Eq. 2.33.
(b) Use Gauss's law to find the field inside and outside a long hollow cylindrical tube, which carries a uniform surface charge $\sigma .$ Check that your result is consistent with Eq. 2.33.
(c) Check that the result of Ex. 2.7 is consistent with boundary conditions 2.34 and 2.36.

Key Concepts

Boundary Conditions in Electrostatics

Boundary conditions in electrostatics involve the constraints on electric fields at interfaces between different media or at surfaces bearing charge. These conditions, such as the continuity of the tangential component of the electric field and the discontinuity in the normal component related to the surface charge density, are essential for ensuring consistent solutions across different regions and are often used to verify the correctness of computed fields.

Surface Charge Distributions and Their Electric Fields

Surface charge distribution pertains to charges spread over a two-dimensional surface, which creates an electric field that can be computed using Gauss's law. The evaluation of the electric field inside and outside such distributions is crucial in understanding how charges influence their surroundings, particularly when the configuration adheres to high symmetry like that seen in hollow cylindrical tubes.

Gauss's Law

Gauss's law is one of the fundamental laws of electromagnetism, stating that the total electric flux out of a closed surface is proportional to the enclosed charge. This law is especially powerful in solving problems with high symmetry, allowing for the evaluation of electric fields in systems such as spheres, cylinders, or planes by simplifying the integral form of Maxwell's equations.

Cylindrical Symmetry in Electrostatics

Cylindrical symmetry refers to the spatial invariance of a system under rotations about an axis and translations along it. In the context of long hollow cylinders or tubes with uniformly distributed surface charges, this symmetry simplifies the calculation of electric fields by ensuring that the field depends only on the radial distance from the axis, rather than on angular or axial coordinates.

Transcript

00:01 For this problem on the topic of electrostatics, we want to check that the results of 4, 5 and 11 are consistent with the given equation.

00:08 We then want to use galser's law to find the field inside and outside a long hollow cylindrical tube that carries a uniform surface charge of sigma.

00:18 And we then want to check the result of example 7, or exercise 7 rather, is consistent with the given boundary conditions.

00:25 So firstly for part a and exercise 2 .4, we have the electric field above equal to sigma over 2 epsilon 0 times n hat.

00:46 And the field below is minus sigma over 2 epsilon 0 n hat with n hat always pointing upward.

00:57 So the difference in fields that is e above minus e below is equal to simply sigma over epsilon n hat as required.

01:21 For exercise 2 .5 at each surface for one side we have e equal to zero and on the other side we have the electric field magnitude e equal to sigma over epsilon not, which therefore means that the difference delta e is equal to sigma over epsilon not again as required.

02:01 For problem 2 .11, we have the outward electric field to be sigma r squared divided by epsilon not little r squared r hat and the inside electric field e in to be zero which means that delta e and if we simplify the first equation this is simply sigma over epsilon not r hat so delta to e is again sigma over epsilon knot r hat agreeing with galus law.

02:56 Now for part b of the problem, if we look at the outside first, the integral of the field e.

03:10 Dot the area vector da is equal to e times 2 pi s times l by galza's law.

03:21 This is 1 over epsilon not times the enclosed charge...

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