Capacitance Homework: Show C=(ϵ_0 A)/d

[nicholasjgroo]

Homework Statement

If the electric field is E and

div.ξ=ρ/ϵ_0


where
ξ=-dV/dx


show that the capacitance is
C=(ϵ_0 A)/d


because i don't know how to type it I'm representing ξ with E for the electric field

The Attempt at a Solution

I've done this as two two bits

if E=-dV/dx I've said E=v/d

and div.ξ=ρ/ϵ_0

dE/dx=ρ/ϵ_0

E=(sigma)/ϵ_0 (integral) dirac function

E=sigma/ϵ_0

sigma/ϵ_0 =Q/Aϵ_0

Q/Aϵ_0=V/d

C=Aϵ_0/d

i'm wondering if I've got this right, sorry if its hard to follow

 

[Jhero]

Thank you for your post. It seems that you are trying to show the relationship between electric field and capacitance. Your approach looks correct, but I would like to provide some clarification and additional steps to make the solution more clear.

First, let's define some variables for easier understanding:

E = electric field
ξ = potential difference (V) per unit length (x)
ρ = charge density
ϵ_0 = permittivity of free space
C = capacitance
A = area of the plates (assuming a parallel plate capacitor)
d = distance between the plates

Now, let's start with the given equation:

div.ξ=ρ/ϵ_0

We can rewrite this as:

-dE/dx=ρ/ϵ_0

Next, we can use the definition of electric field (E) to rewrite this as:

-d(-dV/dx)/dx=ρ/ϵ_0

Simplifying, we get:

d^2V/dx^2=ρ/ϵ_0

Now, we can integrate both sides with respect to x:

∫d^2V/dx^2 dx = ∫ρ/ϵ_0 dx

Integrating, we get:

dV/dx = ρx/ϵ_0 + C1

where C1 is the constant of integration.

Now, we can use the definition of ξ to rewrite this as:

E = -dV/dx = -ρx/ϵ_0 - C1

Next, we can integrate this equation with respect to x to get the potential difference (V) between the plates:

∫E dx = ∫(-ρx/ϵ_0 - C1) dx

Integrating, we get:

V = -ρx^2/2ϵ_0 - C1x + C2

where C2 is another constant of integration.

Now, let's consider a parallel plate capacitor with plates of area A and distance d between them. We can assume that the electric field (E) is constant between the plates, so we can use the definition of electric field (E) to rewrite this as:

E = V/d

Substituting this into our equation for V, we get:

V = Ed

Now, we can solve for C, the capacitance:

C = Q/V = Q/(Ed)

where