Two Infinite Non-Conducting Planes

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The electric field values for the regions around two infinite non-conducting planes are given as 19j V/m above plane 'r', -31j V/m between the planes, and -19j V/m below plane 's'. To find the surface charge density (σr) for plane 'r', the correct relationship to use is E = σ/(2ε0), taking into account the direction of the electric field. The total electric field in each region is the sum of the contributions from both planes, requiring careful attention to the signs of the fields. Three equations can be set up based on the electric fields in the three regions to solve for the surface charge densities. Understanding the contributions from each plane is crucial for correctly calculating σr.
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Homework Statement


The electric field equals 19j for points above plane 'r', -31j between the planes, and -19j below plane 's', where j is the unit vector in the +y direction and the fields are in V/m. Calculate σr, the suface charge density for plane 'r'.

positive y- straight up
positive x-to the right.


Homework Equations



E=Qenc/Epsilon0

The Attempt at a Solution



i tried adding the 19j + -31j then multiplying that by Epsilon 0. that didnt work. not sure how to add or apply the electric fields. any pointers on how to solve that?
 
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anyone? i don't know which field to add or subtract. any help?
 
You're using the wrong relationship between charge and E. While your relationship is strictly correct, what you need is

E=\frac{\sigma}{2\epsilon_0}

where E>0 means that it points away from the plane where \sigma is (and \sigma>0), and E<0 means that it points toward the plane where \sigma is (and \sigma<0). Then, you need to realize that the E in each of the three regions is the sum of the E's from the individual planes, and you have to be careful about the + and - signs. So, you can write 3 equations for the three regions in terms of the three E's and two \sigma's.
 
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