Can a Gaussian Surface Confirm Zero Electric Field Between Two Charges?

AI Thread Summary
The discussion centers on determining the electric field at a point halfway between two positive point charges using a Gaussian surface. It is reasoned that the electric field at this midpoint should be zero due to symmetry and the summation of electric fields from both charges. Participants explore how to apply Gauss's law, emphasizing that external charges do not affect the net electric flux through the Gaussian surface. The concept of electric field lines entering and exiting the surface is highlighted, illustrating that the total flux remains zero despite external influences. Ultimately, the conversation clarifies the relationship between electric fields, Gaussian surfaces, and symmetry in electrostatics.
Gamezwn
Messages
2
Reaction score
0
Warning: Homework Template Must Be Used
1. We have two positive point charges (+q) at a distance from each other
2. Goal is to Find electric field at point A halfway between the point charges
3.By logic and summation of fields we should get ZERO
4. But how would I use a gaussian surface (sphere i assume ) to prove at point A E=0 n/c
5. Personally, I'm attempting a spherical surface around one of the points with radius of :(half distance between charges)
https://drive.google.com/file/d/0B0TaDjU5g_GAVm9hUjliTHhSM1BSbXoyRVp3dXd6MDR4TzE4/view?usp=sharing
 

Attachments

  • 20150320_223627.jpg
    20150320_223627.jpg
    49 KB · Views: 370
Last edited by a moderator:
Physics news on Phys.org
I think you are making it harder than it really is. There is a formula for the electric field created by a point charge. Use that together with symmetry arguments since an electric field is a vector field.
 
Avatrin said:
I think you are making it harder than it really is. There is a formula for the electric field created by a point charge. Use that together with symmetry arguments since an electric field is a vector field.
Yeah I know It is E= kQ/r^2 ...i just want to understand the effect of external charges on a gaussian surface...like in my book they said that external charges don't affect the gaussian surface because all the external field lines come in and come out...no idea what that means tbh
 
Okay, I understand. The idea is that, for a gaussian surface, the same amount of field lines will be pointing into the surface as will be pointing out.

Lets take a very simple example; Think of a perfect sphere in an electric field that is constant everywhere. You are looking directly at the sphere, and let's say the field vectors are pointing towards your right. Using symmetry, it should be easy to see that the net electric flux through the sphere is zero (as much is "coming in" as is "leaving"). Gauss's law says that is true in general; If the electric field is generated by something outside the gaussian surface, the electric flux due to that thing through the surface will be zero.

So, the external charges do affect individual portions of the surface. However, they do not affect the flux through the entire surface. That will remain zero.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top