Hey, Just wanted to thank everyone for their time to help solve people problems, I can generally find all my answers here but I made an account now because i cant seem to find a related problem to this. In general the textbook for these types of problems make the plane z=something. In this case i am able to use cylindrical coordinate system and solve the problem with ease. However i don't think I can apply that same principle that easily here. The answer for a) is 0. 1. The problem statement, all variables and given/known data Plane x = 0 has a uniform density ρs, while plane x = a has -ρs. Determine the electric field intensity in regions a) x<0 b) 0<x<a c) x>a 2. Relevant equations dQ = ρsdS E = Q(r-r')/4*pi*ε0*|r-r'|3 3. The attempt at a solution E = E1 + E2, where E1 is the electric field due to plane x = 0 and E2 is the electric field due to plane x = a: a) Because both are planes: dS = dydz For E1: Q = ∫ ∫ ρs dydz, where both limits are from -infinity to infinity r = -xâx + yây + zâz ,because x<0 r' = 0âx + y'ây + z'âz r - r' = -xâx + (y-y')ây + (z-z')âz |r-r'| = (-x2 + (y-y')2 + (z-z')2)1/2 E1 = 1/(4*π*ε0) ∫ ∫ ρs(-xâx + (y-y')ây + (z-z')âz)/ (x2 + (y-y')2 + (z-z')2)3/2(dydz), where both limits are from -infinity to infinity At this point I am not sure how to solve these limits, I do know that for infinite sheet charges the only direction that will remain is normal to plane so in this case âx. Am i able to say that (y-y') and (z-z') are equal to zero because regardless where the observation point may be be placed there is a corresponding point on the plane? If so the equation simplifies too: E1 = 1/(4*π*ε0) ∫ ∫ ρs(-âx / (x)2(dydz), where both limits are from -infinity to infinity But the answer to this would be infinity? and I don't understand how we could prove that adding this to E2 would result in 0. Any help would be appreciated.