How Does Gauss's Law Apply to Multiple Charged Sheets?

[Seraph404]

Homework Statement

Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform charge densities $$\sigma$$1, $$\sigma$$2, $$\sigma$$3, & $$\sigma$$4 on their surfaces, as shown in the figure .

These surface charge densities have the values $$\sigma$$1= -6.00E-6 C/m^2, $$\sigma$$2= +5.00E-6 C/m^2,$$\sigma$$3= +2.00E-6 C/m^2, and $$\sigma$$4= +4.00E-6 C/m^2.

A) Use Gauss's law to find the magnitude of the electric field at the point A, 5.00 cm from the left face of the left-hand sheet.

Homework Equations

Gauss's Law

The Attempt at a Solution

Well, at first this confused me because I'm pretty sure I remember that in Gauss's law, electric field depends only on the enclosed charge. But this problem gives me a lot of distances. So I guess the formula

E = $$\sigma$$/(2*$$\epsilon$$₀ )

doesn't work. That's what I tried, anyway, and got the wrong answer. So, how do I approach a problem like this?

 

[Seraph404]

Okay, I changed the problem.

 

[nlightner]

You are right that Gauss's law states that the electric field only depends on the enclosed charge. However, in this problem, we have a situation where there are multiple charges on the surfaces of the two sheets. In order to use Gauss's law, we need to consider the electric field due to each individual charge and then sum up the contributions from all of them.

First, we can calculate the electric field due to each sheet separately using the formula E = \sigma/(2*\epsilon0 ). For the left-hand sheet, the electric field at point A would be:

E1 = \sigma1/(2*\epsilon0 ) = (-6.00E-6 C/m^2)/(2*8.85E-12 C^2/Nm^2) = -3.39E5 N/C

Similarly, for the right-hand sheet, the electric field at point A would be:

E2 = \sigma2/(2*\epsilon0 ) = (5.00E-6 C/m^2)/(2*8.85E-12 C^2/Nm^2) = 2.82E5 N/C

Now, we need to consider the electric field due to the other charges (sigma3 and sigma4) on the surfaces of the two sheets. Since these charges are not directly in front of point A, we need to use the distance formula to calculate the distance between each charge and point A.

For sigma3, the distance between the charge and point A is:

d3 = 10.0 cm + 10.0 cm + 5.00 cm = 25.0 cm = 0.250 m

Similarly, for sigma4, the distance between the charge and point A is:

d4 = 10.0 cm + 10.0 cm + 10.0 cm + 5.00 cm = 35.0 cm = 0.350 m

Now, we can calculate the electric field due to each of these charges at point A using the formula E = \sigma/(2*\epsilon0 *r^2), where r is the distance between the charge and point A.

For sigma3, the electric field at point A would be:

E3 = \sigma3/(2*\epsilon0 *d3^2) = (2.00E-6 C/m^2)/(2*8.85E-12 C^2