For this question... A conical capacitor is formed by having electrodes on conical surfaces. The positive electrode is at θ=θ_+, r_a≤r≤r_b, and 0≤ϕ≤ϕ_0. The negative electrode is at θ=θ_-, r_a≤r≤r_b, and 0≤ϕ≤ϕ_0.

here is my working so far...

An electric field is established when a voltage is applied across the electrodes, with the field lines pointing from the positive electrode to the negative electrode (i.e. ‘outwards’). The strength of the electric field is inversely proportional to the distance between the electrode, and therefore, the geometry of the capacitor. In this example, with a conical capacitor, there is azimuthal symmetry (ϕ), but not radial symmetry (r) nor angular symmetry (θ) (image link). Therefore, the strength of the electric field varies with the angle, θ, specifically, the difference in the angle size, θ_d, between the two electrodes, and consequently, also the radial distance, r. The electric field will be stronger at the apex of the cone (where the electrodes are closer together) compared to the base of the cone (where the electrodes are further apart). Imagine a conical Gaussian surface located between the two electrodes (i.e. within the dielectric material). As seen in Lecture 2, Slide 4, the electric flux density, D ⃗, and the electric field, Ε ⃗, are related by the equation below. D ⃗=ϵ_0 Ε ⃗ The total electric flux exiting a closed surface is related to that of the charge enclosed within the surface by Gauss’ Law (also defined in Lecture 2, Slide 4). ∯▒〖D ⃗∙ds ⃗ 〗=∯▒〖ϵ_0 Ε ⃗∙ds ⃗ 〗=ϵ_0 ∯▒〖Ε ⃗∙ds ⃗ 〗=Q_enclosed ⇒∯▒〖ℇ ̅∙ds ⃗ 〗=Q_enclosed/ϵ_0 Given the aspects of symmetry discussed above, we infer that the electric field will be in the form Ε ⃗=E(r,θ) r ̂ In other words, the electric field is radial, but can vary with both r and θ. As seen in this online source, ds ⃗ can be expressed as ds ⃗=r^2 sin⁡(θ)dθ dϕ Therefore, the LHS of the above Gauss’ Law equation is ∯▒〖Ε ⃗∙ds ⃗ 〗=Ε ⃗∯▒〖ds ⃗ 〗 =E(r,θ) ∫0^(ϕ_0)▒〖∫(r_a)^(r_b)▒〖r^2 sin⁡(θ) 〗 dθ dϕ〗 =E(r,θ) r^2 ϕ_0 (cos⁡(r_a )-cos⁡(r_b ) ) In the above equation, I assume that E(r,θ) is the electric field for a specific r and θ within that infinitesimally small differential area, ds ⃗, for simplicity. The integral when then calculated using Mathematica.

For the RHS of the above equation, we can use the surface charge density, ρ_s, on the positive electrode to calculate Q_enclosed. Q_enclosed=ρ_s×(surface area of cone) =ρ_s×∫0^(ϕ_0)▒〖∫(r_a)^(r_b)▒〖r sin⁡(θ_+ ) 〗 dr dϕ〗 =ρ_s ((r_b^2-r_a^2)/2) ϕ_0 sin⁡(θ_+ ) This was calculated using Mathematica. Steps taken as similar to those seen in this online source.

Therefore, we now have ∯▒〖Ε ⃗∙ds ⃗ 〗=Q_enclosed/ϵ_0 E(r,θ) r^2 ϕ_0 (cos⁡(r_a )-cos⁡(r_b ) )=(ρ_s ((r_b^2-r_a^2)/2) ϕ_0 sin⁡(θ_+ ))/ϵ_0 E(r,θ)=(ρ_s ((r_b^2-r_a^2)/2) ϕ_0 sin⁡(θ_+ ))/(ϵ_0 r^2 ϕ_0 (cos⁡(r_a )-cos⁡(r_b ) ) )

now could you please show me how to find / derive the voltage?