Question

A rod of length L has a charge Q uniformly distributed along its length. The rod lies along the y axis with one end at the origin. (a) Find an expression for the electric potential as a function of position along the x axis.

Answer 1

Answer:

 V = k Q/l  ln [(l +√(l² + x²)) / x]

Explanation:

The electrical potential for a continuous distribution of charges is

      V = k ∫ dq / r

Let's apply this expression to our case, define a linear charge density for the bar

           λ = dq / dy

          dq = λ dy

The distance from a point on the bar to the x-axis is

            r = √ (x² + y²)

Let's replace

          V = K ∫ λ dy /√ (x² + y²)

     

We integrate

         V = k λ ln (y + √ (x² + y²))

       

Let's evaluate between y = 0 and y = l

         V = k λ [ ln (l +√(x² + l²) - ln x]

We substitute the linear density

           V = k Q/l  ln [(l +√(l² + x²)) / x]