Answer:
The electric potential is  [tex]V = 15.6 V[/tex]
Explanation:
From the question we are told that
  The length  of the rod is  [tex]L = 2.0 \ m[/tex]
   The total charge of the rod is  [tex]q =5.0 nC = 5.0*10^{-9} C[/tex]
   The length from the center is  [tex]d = 3.0 \ m[/tex]
The diagram illustrating the setup for this question is shown on the first uploaded image
From the diagram the potential at point  A  [tex]dl[/tex] is mathematically represented as
     [tex]dV = K \frac{dq}{l}[/tex]
Where K is the coulomb constant with a value  [tex]K = 9*10^9 \ Nm^2 /C^2[/tex]
where q is the charge in charge  the rod relative to its distance from A  is mathematically represented as
     [tex]dq = \frac{q}{L} dl[/tex]
This a small unit length of the rod
So     [tex]V = \frac{q}{L} \int\limits^4_2 {\frac{dl}{l} } \,[/tex]
    =>  [tex]V = k\frac{q}{L} ln [\frac{4}{2} ][/tex]
       [tex]V = k\frac{q}{L} ln2[/tex]
Substituting values
        [tex]V = 9* 10^9 * \frac{5*10^{-9}}{2} * ln 2[/tex]
         [tex]V = 15.6 V[/tex]
    Â
