Electric field of infinite plate

[Dell]

2 infinite plates are placed one next to another in a V formation with an angle β between them, they have charge densities of σ1 and σ2. what is the magitude of the field betweein them.

ϕ=\ointEdA=EA
ϕ=\frac{q}{ϵ}=\frac{σA}{ϵ}

E1=\frac{σ1}{ϵ}
E2=\frac{σ2}{ϵ}

E=E1+E2

lets make our x-axis parallel to plate 1 giving E1 a 0 value on x axis
E_x= 0 + \frac{σ2}{ϵ}*sinβ
E_y= \frac{σ1}{ϵ} + \frac{σ2}{ϵ}*cosβ

|E|²=|E_x|² + |E_y|²
=(\frac{σ2}{ϵ}*sinβ)² +(\frac{σ1}{ϵ} + \frac{σ2}{ϵ}*cosβ)²
=\frac{1}{ϵ}*(σ2²sin²β + σ1² + σ2²cos²β + 2(σ1)(σ2)cosβ )
=\frac{1}{ϵ}*(σ2²(sin²β + cos²β ) + σ1² + 2(σ1)(σ2)cosβ)

|E|²=\frac{1}{ϵ}*(σ1² + σ2² + 2(σ1)(σ2)cosβ)

which is almost right except that in my answers there is a second possible solution that

|E|²=\frac{1}{ϵ}*(σ1² + σ2² - 2(σ1)(σ2)cosβ)

where does this minus come from? originally i thought it might be because of a possible negative charge density but surely if that was so the sign would be part of σ and come automatically when i plug in values

 

[Dell]

i see the latex codes came all messed up, meant to read:

Ex= 0 + (σ2/ɛ)*sinβ
Ey=(σ1/ɛ) + (σ2/ɛ)*cosβ

|E|^2=|Ex|^2 + |Ey|^2

and at the end after calculations come to

|E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)cosβ)

where there is meant to be a second option of

|E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)cosβ)

and i can't see where from. originally i thought it might be because of a possible negative charge density but surely if that was so the sign would be part of σ and come automatically when i plug in values

 

[LowlyPion]

Depending on the angle cosβ has a range of +1 to -1 doesn't it?

 

[Dell]

correct, so does that mean that if 90<β<270 then i will use the minus option? could i then also say |E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)|cosβ|)

 

[LowlyPion]

correct, so does that mean that if 90<β<270 then i will use the minus option? could i then also say |E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)|cosβ|)

I think all it means is that there are 2 solutions depending on the angle of β.

Your representation above ignores the second solution where cosβ is (-).

 

[Dell]

what are the 2 options? is it

|E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)cosβ) when (cosβ>1)

and

|E|^2=(1/ɛ)*(σ1^2 + σ2^2 - 2(σ1)(σ2)cosβ) when (cosβ<1)

?

if so than can i not just say that it is
|E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)|cosβ|)

and then cos will always be positive, in effect that is what i am doing anyway, taking minus the value of the negative cos

 

[Dell]

looking back at the original question and the one i posted, i see that in the original the 2 plates formed an X like shape whereas i wrote in my post a V shape, originally i never thought anything of it but now looking back i think that the (- cosB) is for the area between the top and bottom points of the X on either side whereas the ( + cosB) is for the area between the top 2 points or bottom 2 points, (when B is an acute angle,) or alternatively the opposite(when B is obtuse).
is this a correct presumption?

 

[LowlyPion]

looking back at the original question and the one i posted, i see that in the original the 2 plates formed an X like shape whereas i wrote in my post a V shape, originally i never thought anything of it but now looking back i think that the (- cosB) is for the area between the top and bottom points of the X on either side whereas the ( + cosB) is for the area between the top 2 points or bottom 2 points, (when B is an acute angle,) or alternatively the opposite(when B is obtuse).
is this a correct presumption?

An X configuration does simultaneously create 2 regions (4 actually, but 2 sets of 2).

 

[Dell]

yes but is one the -cos option and the other the +cod option?

 

[LowlyPion]

yes but is one the -cos option and the other the +cod option?

One governs acute angles and the other obtuse doesn't it?