Finding the E due to a non-uniform surface charge distribution in 3D

[Shostakovich]

Homework Statement

Here is the question, which itself is rather confusing.

A nonuniform surface charge lies in the yz plane. At the origin, the surface charge den- sity is 3.5 μC/m^2. Other charged objects are present as well. Just to the right of the origin, the electric field has only an x component of magnitude 4.8 × 10^5 N/C.
What is the x component of the electric field just to the left of the origin? The permittivity of free space is 8.8542 × 10^−12 C2/N · m^2.
Answer in units of N/C

Homework Equations

ø=∫E.dA = Q(enclosed)/ε(naught)

σ = Q/A

The Attempt at a Solution

So i drew the pictures, a 2D surface lying in the yz, with a point x with electric field value 4.8e5 N/C just to the right on the positive axis.
I thought to myself the field will just be negative because its on the opposite side? Nope.
Maybe it's the same? Nope.
I then tried finding the difference, to no avail, and now I'm just stuck.

Is there a better way for me to think about this problem that I'm not seeing, or is it just that obscure?

 

[gneill]

Hmm. Suppose that "just to the right of the origin" implies that the observer there is so close that he "sees" the surface charge in the yz plane as an infinite sheet of charge with the given charge density. What field strength would he expect to see from it? How does that compare to what he actually observes? If there's a difference, what then?

 

[Shostakovich]

Hmm. Suppose that "just to the right of the origin" implies that the observer there is so close that he "sees" the surface charge in the yz plane as an infinite sheet of charge with the given charge density. What field strength would he expect to see from it? How does that compare to what he actually observes? If there's a difference, what then?

As you probably well know, the Electric field due to an infinite sheet is σ/(2ε(naught)). But this gives something about 2.43 times smaller than what was given. Does this proportion have anything to do with how the field distributes itself?

 

[gneill]

As you probably well know, the Electric field due to an infinite sheet is σ/(2ε(naught)). But this gives something about 2.43 times smaller than what was given. Does this proportion have anything to do with how the field distributes itself?

The proportion is not particularly relevant. What is relevant is that the observed field strength is not entirely accounted for by the sheet's field.

If the field due to the sheet of charge doesn't account for the net field observed, what does that tell you about the contribution from the "other charged objects" that were mentioned? What happens if your observer moves (a very small distance indeed) from the right to the left of the sheet?

 

[HalfLostAndSearching]

I would approach this problem by first considering the basic principles of electrostatics. The electric field is a vector quantity, meaning it has both magnitude and direction. It is also a conservative field, meaning the work done by the field in moving a charge from one point to another is independent of the path taken.

In this case, we are given a non-uniform surface charge distribution in the yz plane. This means that the charge density (σ) is not constant throughout the surface. We are also given the electric field at a specific point on the x-axis, just to the right of the origin. This information alone is not enough to determine the electric field at the origin or anywhere else on the surface.

To solve this problem, we need to use the fact that the electric field is a conservative field. This means that we can use Gauss's law to relate the electric field to the charge enclosed by a surface. In this case, we can choose a Gaussian surface in the shape of a cylinder with its axis along the x-axis and passing through the origin. This surface will enclose the non-uniform surface charge distribution as well as the other charged objects present.

Using Gauss's law, we can write:

∫E·dA = Q(enclosed)/ε(naught)

Where Q(enclosed) is the total charge enclosed by the Gaussian surface, E is the electric field, and dA is the differential area element of the Gaussian surface. We can simplify this equation to:

E∫dA = Q(enclosed)/ε(naught)

Since the electric field is constant on the surface of the cylinder, we can pull it out of the integral. We can also rewrite the enclosed charge (Q(enclosed)) in terms of the surface charge density (σ) and the area of the Gaussian surface (A), giving us:

EσA = σA/ε(naught)

Solving for the electric field (E), we get:

E = σ/ε(naught)

Substituting in the given values, we get:

E = (3.5 μC/m^2)/(8.8542 × 10^−12 C^2/N · m^2)

E = 3.95 × 10^16 N/C

Therefore, the x component of the electric field just to the left of the origin is 3.95 × 10^16 N/C. This value is very