Potential Difference Problem

[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 $da = dl_{r}dl_{\varphi} = (dr)(r sinθ d\varphi) = r sin \frac{\pi}{4} dr d\varphi = \frac{1}{\sqrt{2}} r dr d\varphi$.

 

[paranormal]

Dear student,

Thank you for reaching out for help with your homework problem. Your attempt at the solution looks correct, but there may have been a misunderstanding with the notation used by your instructor. The notation "da" in this context refers to the infinitesimal area element on the surface of the cone. In spherical coordinates, this can be written as da = r^{2}sinθdθdrd\varphi. However, since we are only interested in the surface charge distribution on the conical surface, we can simplify this expression by substituting in the value for θ and simplifying the sinθ term. This results in da = \frac{1}{\sqrt{2}}r^{2}drd\varphi, which is what you had initially written. Therefore, your solution is correct and there may have been a misunderstanding with the notation used by your instructor. I would suggest clarifying this with them to ensure that you both are on the same page. Keep up the good work!