- #1

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ϕ=[tex]\oint[/tex]EdA=EA

ϕ=[tex]\frac{q}{ϵ}[/tex]=[tex]\frac{σA}{ϵ}[/tex]

E1=[tex]\frac{σ1}{ϵ}[/tex]

E2=[tex]\frac{σ2}{ϵ}[/tex]

E=E1+E2

lets make our x axis parallel to plate 1 giving E1 a 0 value on x axis

E

_{x}= 0 + [tex]\frac{σ2}{ϵ}[/tex]*sinβ

E

_{y}= [tex]\frac{σ1}{ϵ}[/tex] + [tex]\frac{σ2}{ϵ}[/tex]*cosβ

|E|

^{2}=|E

_{x}|

^{2}+ |E

_{y}|

^{2}

=([tex]\frac{σ2}{ϵ}[/tex]*sinβ)

^{2}+([tex]\frac{σ1}{ϵ}[/tex] + [tex]\frac{σ2}{ϵ}[/tex]*cosβ)

^{2}

=[tex]\frac{1}{ϵ}[/tex]*(σ2

^{2}sin

^{2}β + σ1

^{2}+ σ2

^{2}cos

^{2}β + 2(σ1)(σ2)cosβ )

=[tex]\frac{1}{ϵ}[/tex]*(σ2

^{2}(sin

^{2}β + cos

^{2}β ) + σ1

^{2}+ 2(σ1)(σ2)cosβ)

|E|

^{2}=[tex]\frac{1}{ϵ}[/tex]*(σ1

^{2}+ σ2

^{2}+ 2(σ1)(σ2)cosβ)

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

|E|

^{2}=[tex]\frac{1}{ϵ}[/tex]*(σ1

^{2}+ σ2

^{2}- 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