- #1

BruteCoder

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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 can't 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.

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

dQ = ρsdS

E = Q(r-r')/4*pi*ε0*|r-r'|

E[/B] = 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'| = (-x

E1 = 1/(4*π*ε0) ∫ ∫ ρs(-xâx + (y-y')ây + (z-z')âz)/ (x

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)

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.

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 can't 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.

## Homework Statement

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

## Homework Equations

dQ = ρsdS

E = Q(r-r')/4*pi*ε0*|r-r'|

^{3}## The Attempt at a Solution

E[/B] = 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'| = (-x

^{2}+ (y-y')^{2}+ (z-z')^{2})^{1/2}E1 = 1/(4*π*ε0) ∫ ∫ ρs(-xâx + (y-y')ây + (z-z')âz)/ (x

^{2}+ (y-y')^{2}+ (z-z')^{2})^{3/2}(dydz), where both limits are from -infinity to infinityAt 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 infinityBut 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.

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