Explanation on the Electric Field in Griffiths' Textbook

In summary, the conversation discusses the discontinuity aspects of the electric field at a surface charge. The use of a pillbox Gaussian surface is mentioned and it is noted that the electric field is pointing in the same direction from both sides of the surface. This is explained by the need to use the same sign convention on both sides of the surface, and examples are given to illustrate this concept. The conversation ends with a thank you to the expert who provided the explanation.
  • #1
Ren Figueroa
Hi. I'm going over the discontinuity aspects of the electric field as we cross the surface charge. A pillbox Gaussian surface was drawn and the electric field for "below" is throwing me off. Can someone explain why the electric field is pointing in the same direction from both sides of the surface? I supplied the image below. Thanks!
C598XwYl61xVUe6uK65kRtgToayN3xKaeS9V4AFctuYMUuBESuUaoJZdjgozDKqg_b6bIs5tNeO3XAs04=w1496-h1058-no.png
 
Physics news on Phys.org
  • #2
I think he does it that way to indicate or emphasize that you need to use the same sign convention on both sides of the surface, which leads to eq. (2.31) having a minus sign, i.e. that you have to use the difference between the two fields. Also, it's actually possible for the field to be in the same direction on both sides of the surface, depending on the situation.

Example #1: if the surface is the only charged object, then the field points upwards above the surface and downwards below it. Using the sign convention with +/- meaning up/down, this might give us something like $$(+5~\rm{N/C}) - (-5~\rm{N/C}) = \frac {\sigma} {\epsilon_0} \\ +10~\rm{N/C} = \frac {\sigma} {\epsilon_0}$$ Example #2: if we place the surface from example #1 between the plates of a large capactor that by itself produces a uniform field of 20 N/C upwards, the net fieid is now upwards on both sides of the surface, and we have $$(+25~\rm{N/C}) - (+15~\rm{N/C}) = \frac {\sigma} {\epsilon_0} \\ +10~\rm{N/C} = \frac {\sigma} {\epsilon_0}$$
 
  • Like
Likes Ren Figueroa
  • #3
jtbell said:
I think he does it that way to indicate or emphasize that you need to use the same sign convention on both sides of the surface, which leads to eq. (2.31) having a minus sign, i.e. that you have to use the difference between the two fields. Also, it's actually possible for the field to be in the same direction on both sides of the surface, depending on the situation.

Example #1: if the surface is the only charged object, then the field points upwards above the surface and downwards below it. Using the sign convention with +/- meaning up/down, this might give us something like $$(+5~\rm{N/C}) - (-5~\rm{N/C}) = \frac {\sigma} {\epsilon_0} \\ +10~\rm{N/C} = \frac {\sigma} {\epsilon_0}$$ Example #2: if we place the surface from example #1 between the plates of a large capactor that by itself produces a uniform field of 20 N/C upwards, the net fieid is now upwards on both sides of the surface, and we have $$(+25~\rm{N/C}) - (+15~\rm{N/C}) = \frac {\sigma} {\epsilon_0} \\ +10~\rm{N/C} = \frac {\sigma} {\epsilon_0}$$

Thanks jtbell! I realized that I forgot to thank you and just put a thumbs up. In my attempt to find my post in the forums, I found a similar post but referencing Jackson book. I thought it was funny how I could have just searched for my question, since you answered that post, but I usually have no luck in searching. Thanks again!
 

Related to Explanation on the Electric Field in Griffiths' Textbook

1. What is the electric field described in Griffiths' textbook?

The electric field is a vector quantity that describes the force experienced by a charged particle at any point in space. It is represented by the symbol E and is measured in units of newtons per coulomb (N/C).

2. How is the electric field calculated?

The electric field is calculated using the equation E = F/q, where E is the electric field, F is the force experienced by the charged particle, and q is the charge of the particle. The direction of the electric field is determined by the direction of the force on a positive test charge.

3. What is the difference between electric field and electric potential?

The electric field is a measure of the force experienced by a charged particle, while electric potential is a measure of the potential energy of the charged particle. Electric potential is related to electric field by the equation V = -∫E·dr, where V is the electric potential, E is the electric field, and dr is the displacement vector.

4. What are some real-world applications of the concept of electric field?

The concept of electric field is used in many areas of science and technology, including electronics, telecommunications, and electricity generation and distribution. It is also important in understanding the behavior of charged particles in fields created by magnets and in the atmosphere during lightning strikes.

5. Is Griffiths' textbook the only source for learning about the electric field?

No, there are many other textbooks and resources available for learning about the electric field. However, Griffiths' textbook is a widely used and highly recommended source for understanding the fundamental concepts and calculations related to the electric field.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
601
  • Electromagnetism
Replies
5
Views
2K
Replies
8
Views
1K
Replies
14
Views
1K
  • Electromagnetism
Replies
4
Views
1K
  • Science and Math Textbooks
Replies
12
Views
2K
  • Electromagnetism
Replies
3
Views
848
  • Introductory Physics Homework Help
Replies
26
Views
696
Back
Top