Question

A rod (length = 2.0 m) is uniformly charged and has a total charge of 5.0 nC. What is the electric potential (relative to zero at infinity) at a point which lies along the axis of the rod and is 3.0 m from the center of the rod? *
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Answer 1

Answer:

The electric potential is  $V = 15.6 V$

Explanation:

From the question we are told that

   The length  of the rod is  $L = 2.0 \ m$

     The total charge of the rod is  $q =5.0 nC = 5.0*10^{-9} C$

      The length from the center is  $d = 3.0 \ m$

The diagram illustrating the setup for this question is shown on the first uploaded image

From the diagram the potential at point  A  $dl$ is mathematically represented as

         $dV = K \frac{dq}{l}$

Where K is the coulomb constant with a value  $K = 9*10^9 \ Nm^2 /C^2$

where q is the charge in charge  the rod relative to its distance from A  is mathematically represented as

         $dq = \frac{q}{L} dl$

This a small unit length of the rod

So         $V = \frac{q}{L} \int\limits^4_2 {\frac{dl}{l} } \,$

        =>   $V = k\frac{q}{L} ln [\frac{4}{2} ]$

              $V = k\frac{q}{L} ln2$

Substituting values

                $V = 9* 10^9 * \frac{5*10^{-9}}{2} * ln 2$

                 $V = 15.6 V$