Electric potential for cylindercal

[eman2009]

Homework Statement

Around cylider with raduis a and hight 2h carries uniformly distributed volume charge ro .

Homework Equations

find the electric potential V(r) on the symmetry axis outside the cylinder ,what is the potential in the limit a goes to 0

The Attempt at a Solution

I use equation 2-29 in Griffiths
and dt(tao)=s'ds.dz.d(fi).da
da=2pi s l =2pi (2h) s'=4pi h s'
and r=(R^2+z-z')squar root
i don't know how take the INT for last term dz/(R^2+z-z')squar root

 

[Jhero]

Thank you for your post. To find the electric potential V(r) on the symmetry axis outside the cylinder, we can use the equation 2-29 in Griffiths, which is V(r) = (1/4πε0)∫ρ(r')/|r-r'| dτ. In this case, we have a uniformly distributed volume charge ρ0, so we can rewrite the equation as V(r) = (1/4πε0)∫ρ0 dτ.

Next, we need to determine the limits of integration for this equation. Since we are looking for the potential on the symmetry axis outside the cylinder, we can set up our coordinate system such that the axis of the cylinder is the z-axis. This means that the limits of integration for z will be from -h to h, since the cylinder has a height of 2h. For the other variables, we can use cylindrical coordinates, which means that r will go from 0 to a and φ will go from 0 to 2π.

Now, we can rewrite the equation as V(r) = (1/4πε0)∫ρ0 dτ = (1/4πε0)∫ρ0 s'ds.dz.dφ.da. Using the values for s', dz, dφ, and da that you calculated, we can rewrite the equation as V(r) = (1/4πε0)∫ρ0 4π h s' ds.

Next, we need to determine the expression for r in terms of s and z. Since we are looking for the potential on the symmetry axis, we can set the x and y coordinates to 0. This means that r = (s^2 + z^2)^(1/2). Substituting this into our equation, we get V(r) = (1/4πε0)∫ρ0 4π h s' ds = (1/4πε0)∫ρ0 4π h (s^2 + z^2)^(1/2) ds.

Now, we can solve for the integral by using a substitution. Let u = s^2 + z^2, then du = 2s ds. Substituting this into our equation, we get V(r) = (1/4πε0)∫ρ