What is the correct electric potential of a rod?

[Jalo]

Homework Statement

Here's the problem:

Homework Equations

V = k_e ∫ dq/r
V is the electric potential, k_e Coulomb's constant, q the charge and d the distance.

λ = q / L , where λ is the charge density, q the charge and L the length of the rod.

The Attempt at a Solution

I have one solution for the problem. What I want to know is why is my answer incorrect.
Since electric potential is a scalar and not a potential what I calculated was the electric potential produced by the left part of the rod, from 0 to L/2, and multiplied the resulted for two due to the symmetry of the problem (I thought that the electric potential produced by the rod from L/2 to L was the same as from 0 to L/2)

q = λ*L ⇔ dq = λ*dx ⇔ dq = α*x*dx

V = 2*k_e*α ∫₀^L/2 x/sqrt(x²+b²) dx

I solved the integral and got:

V = 2*α*k_e*[ sqrt( (L/2)² + b² ) - b ]

It is incorrect tho.
Any help will be appreciated!
Thanks in advance.

 

[ehild]

Homework Statement

Here's the problem:

Homework Equations

V = k_e ∫ dq/r
V is the electric potential, k_e Coulomb's constant, q the charge and d the distance.

λ = q / L , where λ is the charge density, q the charge and L the length of the rod.

The Attempt at a Solution

I have one solution for the problem. What I want to know is why is my answer incorrect.
Since electric potential is a scalar and not a potential what I calculated was the electric potential produced by the left part of the rod, from 0 to L/2, and multiplied the resulted for two due to the symmetry of the problem (I thought that the electric potential produced by the rod from L/2 to L was the same as from 0 to L/2)

q = λ*L ⇔ dq = λ*dx ⇔ dq = α*x*dx

V = 2*k_e*α ∫₀^L/2 x/sqrt(x²+b²) dx

I solved the integral and got:

V = 2*α*k_e*[ sqrt( (L/2)² + b² ) - b ]

It is incorrect tho.
Any help will be appreciated!
Thanks in advance.

Read the problem text: λ=αx, the left end of the rod is at x=0. There is no symmetry.

ehild