# 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=$$\oint$$EdA=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
Ex= 0 + $$\frac{σ2}{ϵ}$$*sinβ
Ey= $$\frac{σ1}{ϵ}$$ + $$\frac{σ2}{ϵ}$$*cosβ

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

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#### 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 cant 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

Homework Helper
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

Homework Helper
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

Homework Helper
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

Homework Helper
yes but is one the -cos option and the other the +cod option?
One governs acute angles and the other obtuse doesn't it?

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