Electrostatics: cube with 3 charged walls

[Rimantas B.]

Homework Statement

There is a cube with a side length of 0,2m. Its three adjacent walls bear an area charge of -2nC/m².
Find the strength of the electric field at the center of the cube.

Homework Equations

$$E = \frac{\sigma}{2 \varepsilon _{0}}$$
$$dE = \frac{\sigma dS}{4 \pi \varepsilon _{0} r^{2}}$$

The Attempt at a Solution

My problem is that I can think of two methods to solve this and get wildly different results.
First attempt:
The first equation is for an infinite plane. However, the walls of the cube are finite. We could say that each of the walls takes 1/6 of the total field of view, so the field strength of each wall would be $$E _{1} = \frac{\sigma}{12 \varepsilon _{0}}$$.
Then, using vector addition, we can reason that $$E _{total} = \sqrt{3} E_1$$ as they are all perpendicular to each other.
This gives a total field strength of about 65 V/m.

Second attempt:
We could use an area integral to obtain $$E_{1}$$. Because of the symmetry, the components of the field not perpendicular to the walls cancel each other out, therefore we can only care about the perpendicular ones. That gives us
$$dE = \frac{\sigma cos(\alpha) dS}{4 \pi \varepsilon _{0} r^{2}} = \frac{\sigma d dx dy}{4 \pi \varepsilon _{0} r^{3}}$$
herein $$d$$ is the fixed distance in one axis from the wall to the center - that is, $$\frac{a}{2}=0,1m$$
Placing the center of the coordinate system in the center of the wall, we can integrate from x=-0,1 to 0,1 and y=-0,1 to 0,1.
Doing this gives a result more than ten times smaller than the previous method.

Which is correct, if any?

 

[Rick88]

Your first method confuses me.

Only three of the walls bear charges, so why should each wall possesses the same E field?
It depends on the distance from any charge.

Each charge produces an electric force. Try making a diagram of the cube, showing where the charges are placed in what directions the forces point.R.

 

[Rimantas B.]

I meant, all of the charged walls. The distances from each wall to the center are equal. This is a drawing showing the charged walls.

 

[Rick88]

It still was incorrect.

Anyway, if those are the charged walls, just they produce E fields.
So just look closely at where the E fields would point and what their strength would be at the centre of the cube.
Remember E is a vector!

 

[rwisz]

I would recommend using the second method as it takes into account the finite nature of the walls. The first method assumes an infinite plane, which may not accurately reflect the situation. However, it is always good to double check your calculations and make sure all assumptions are accounted for. If possible, try to also calculate the electric field using a different method or formula to verify your results.