Find the Electric Field E using Gauss' Law

In summary, the conversation discusses the process of finding the electric field for charges +q and +4q at a distance of r and 3r, respectively. The individual electric fields were calculated using Gauss' Law and then added together using superposition to find the total electric field. The confusion lies in determining the direction of the electric fields, which can be found by using the vector form of Gauss' Law and considering the position of the point where the field is being measured.
  • #1
Nicci
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0
Homework Statement
Two charges +q and +4q are separated a distance of 2r. Use Gauss' Law to find the electric field E at a distance:
a) r around the charge +q
b) 3r around charge +q
Relevant Equations
##A = 4πr^2##
##E.A=\frac {q}{ε_°}##
I tried to work out both a) and b), but I am not sure if I am correct. I drew a picture with a sphere around q first with radius r and then with radius 3r.

For a) ##E.A=\frac {q}{ε_°}## (when using Gauss' Law)
Since ##A=4πr^2##, I substituted this in the equation and solved for E giving me ##E_q= \frac {kq}{r^2}##
My problem is that I am not sure if this is correct and the direction of the field is confusing me. I thought it should be negative...##E_q= -\frac {kq}{r^2}##

I then solved for the electric field using the +4q charge. This I did in the same way as the +q charge. ##E_{4q} = \frac {4q}{4πr^2ε_°}## which gives ##E_{4q}= \frac {4kq}{r^2}## I thought this would be positive.

To get the electric field E: ##E=E_{4q} - E_q## which gives ##E= \frac {3kq}{r^2}##
I am not really sure if I have done this correctly. The directions of the fields are really confusing me.

For b) I basically did the same as for a). The only difference was the distance which increased from r to 3r.

I started by saying ##E_q = \frac {q}{4π(9r^2)ε_°}## and simplifying this gives me ##E_q = \frac {kq}{9r^2}##. I thought the direction would be to the left, making it negative: ##E_q = -\frac {kq}{9r^2}##
For the other charge ##E_{4q} = \frac {4q}{4πr^2ε_°}## which gives ##E_{4q} = \frac {4kq}{r^2}##. I reckon the direction would be to the left, also making it negative: ##E_{4q} =- \frac {4kq}{r^2}##

Getting E:
##E=E_q+E_{4q}## which gives ##E= -\frac {kq}{9r^2} -\frac {4kq}{r^2}##

Upon simplification I get:
##E=-\frac {37kq}{9r^2}##

I am confused about the directions of the electric fields. Can someone maybe help me by explaining how to get the directions?
 
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  • #2
Nicci said:
I am confused about the directions of the electric fields. Can someone maybe help me by explaining how to get the directions?
To get directions, you need to use the vector form of Gauss' Law. Are you familiar with how to use the (2-D) vector form?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html#c1
1565108382784.png
 
  • #3
The electric field depends on the position of the point where you want to know the field not just the distance from one of the charges. Without that information you cannot answer this question. This is a bad problem as stated. Are you sure you are looking for the electric field and not the electric flux through spheres centered at charge +q?
 
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  • #4
The electric field is a consequence of the presence of both charges, therefore you should take ##4q+## into account as well. Superposition theorem.

Untitled.png
 
Last edited:
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1. What is Gauss' Law?

Gauss' Law is a fundamental principle in physics that relates the electric flux through a closed surface to the total charge enclosed by that surface. It is a powerful tool for calculating the electric field in a region where the charge distribution is known.

2. How do I use Gauss' Law to find the electric field E?

To use Gauss' Law, you must first choose a closed surface that encloses the charge distribution of interest. The surface can be any shape, as long as it is closed and the electric field is constant over the surface. Then, you can use the formula E = Q/ε0A, where Q is the total charge enclosed by the surface, ε0 is the permittivity of free space, and A is the area of the surface. This will give you the magnitude of the electric field at any point on the surface.

3. What are the advantages of using Gauss' Law to find the electric field?

Gauss' Law offers several advantages over other methods of calculating the electric field. It is often easier and quicker to use, especially for highly symmetric charge distributions. It also relies on the concept of electric flux, which is a useful and intuitive concept in the study of electromagnetism.

4. Are there any limitations to using Gauss' Law?

While Gauss' Law is a powerful tool, it does have some limitations. It only applies to electric fields that are constant over the surface of the chosen closed surface. Additionally, it can only be used for charge distributions that exhibit a high degree of symmetry. In cases where the charge distribution is not symmetric, other methods such as Coulomb's Law may be more appropriate.

5. Can Gauss' Law be applied to any charge distribution?

No, Gauss' Law can only be applied to charge distributions that exhibit a high degree of symmetry. This means that the charge distribution must have some form of rotational or reflective symmetry. If the charge distribution is asymmetrical, other methods must be used to calculate the electric field.

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