How to Calculate the Potential Difference on a Charged Conical Surface?

[NoPhysicsGenius]

Homework Statement

A conical surface (an empty ice-cream cone) carries a uniform surface charge σ. The height of the cone is a, as is the radius of the top. Find the potential difference between the points P (the vertex) and Q (the center of the top).

Homework Equations

V(P) = \frac{1}{4\piε_{0}}∫\frac{σ}{\sqrt{z^{2}+2r^{2}-2zr}}da
da = r^{2}sinθdθdrd\varphi

The Attempt at a Solution

I am having difficulties in determining da for spherical coordinates.

Because the conical surface is a right circular cone, θ = \frac{\pi}{4}. Therefore, sinθ = sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}. Also, dθ = 1.

Apparently then, da = \frac{1}{\sqrt{2}}r^{2}drd\varphi.

However, according to my instructor, this is incorrect and the correct answer should be da = \frac{1}{\sqrt{2}}rdrd\varphi.

What have I done wrong? Thank you.

 

[NoPhysicsGenius]

Additional Problem Info

Note that in the relevant equation for V(P), the square root term represents the separation vector from the source point (the conical surface of uniform charge σ) and the point where the potential is to be measured. The letter z is to be taken as a constant; V(Q) = V(z = a) and V(P) = V(z = 0). We are then tasked with first finding V(z) and then finding V(Q) - V(P).

 

[vela]

Homework Statement

A conical surface (an empty ice-cream cone) carries a uniform surface charge σ. The height of the cone is a, as is the radius of the top. Find the potential difference between the points P (the vertex) and Q (the center of the top).

Homework Equations

V(P) = \frac{1}{4\piε_{0}}∫\frac{σ}{\sqrt{z^{2}+2r^{2}-2zr}}da
da = r^{2}sinθdθdrd\varphi

Consider the units. That isn't an area element; it's a volume element.

The Attempt at a Solution

I am having difficulties in determining da for spherical coordinates.

Because the conical surface is a right circular cone, θ = \frac{\pi}{4}. Therefore, sinθ = sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}. Also, dθ = 1.

Apparently then, da = \frac{1}{\sqrt{2}}r^{2}drd\varphi.

However, according to my instructor, this is incorrect and the correct answer should be da = \frac{1}{\sqrt{2}}rdrd\varphi.

What have I done wrong? Thank you.

 

[NoPhysicsGenius]

Surface element da

Thanks! I get it now ...

The surface element (da) is da = dl_{r}dl_{\varphi} = (dr)(r sinθ d\varphi) = r sin \frac{\pi}{4} dr d\varphi = \frac{1}{\sqrt{2}} r dr d\varphi.