Is the Setup for the Electric Scaler Potential Integral Correct?

[kosmocomet]

Homework Statement

Homework Equations

• V=¼*(1/(π∈) * ∫(ρ_s/(R')*ds' where R' is distance from point to surface
• R'=|R-Ri| distance from observation point to location of surface charge density.

The Attempt at a Solution

So my attempt was to define R' as R'=√((-r)²+(-Φ)²+(z)²). Then I said that ds' is = rdrdΦ. I put this into the integral. I am confused since the integral looks very involved ad wanted to know if my set up was correct.

Any help is much appreciated.

 

[tnich]

Homework Statement

View attachment 223189

Homework Equations

• V=¼*(1/(π∈) * ∫(ρ_s/(R')*ds' where R' is distance from point to surface
• R'=|R-Ri| distance from observation point to location of surface charge density.

The Attempt at a Solution

So my attempt was to define R' as R'=√((-r)²+(-Φ)²+(z)²). Then I said that ds' is = rdrdΦ. I put this into the integral. I am confused since the integral looks very involved ad wanted to know if my set up was correct.

Any help is much appreciated.

Why do you define R' that way?

 

[kosmocomet]

Why do you define R' that way?

Thanks for the response. I am using cylindrical coordinates and since the point P is (0,0,z) and the disk is(r,Φ,0) doing the substraction is -r,-Φ,z whcih is the vector R'. The magnitude is then R'=√((-r)2+(-Φ)2+(z)2).

 

[tnich]

Thanks for the response. I am using cylindrical coordinates and since the point P is (0,0,z) and the disk is(r,Φ,0) doing the substraction is -r,-Φ,z whcih is the vector R'. The magnitude is then R'=√((-r)2+(-Φ)2+(z)2).

OK. I see how you did it, but it is not correct. For one thing it is dimensionally inconsistent. It does not make sense to add $rad^2$ to $m^2$. Let's approach it another way. Let r and z be constant and consider two angles, $Φ_1 = 0$ and $Φ_2 = π/2$. Is the distance R' different for the different values of $Φ$?

 

[kosmocomet]

Well...I guess not mow that I think about it. Since, the distance would be based on r and z, correct?

 

[kosmocomet]

If this is true, would the surface integral be drdΦ or dΦdz? I think it would be drdΦ since it is defined in radial and angular direction. The surface, I mean.

 

[tnich]

Well...I guess not mow that I think about it. Since, the distance would be based on r and z, correct?

Yes.

 

[tnich]

If this is true, would the surface integral be drdΦ or dΦdz? I think it would be drdΦ since it is defined in radial and angular direction. The surface, I mean.

That's close, but what is the differential area in polar coordinates?