Answer:
The potential at the point 8 meters is approximately 48.98 Joules/C
Explanation:
Notice that we need to find the potential at a point aligned to the rod's axis (see attached figure), that is located 6 meters from one end of the rod. Notice as well, that the length (L) of the rod is 5 meters.
Since we have a uniformly distributed charge, the charge density per unit of length is defined as:
[tex]\lambda=\frac{Q}{L} \\\lambda=\frac{45}{5}\,\frac{10^{-9}\,C}{m} \\\lambda=9\,\frac{10^{-9}\,C}{m}[/tex]
In order to find the contribution of each little segment dx of charge
[tex]dq=\lambda\,dx[/tex]
to the potential at the requested point, we need to perform an integral:
[tex]V=\int\limits^{11}_{6} {} \, dV \\V=\int\limits^{11}_{6} {} \, k\,\frac{dq}{x} \\V=\lambda \,*k\,\int\limits^{11}_{6} {} \, \frac{dx}{x} \\V=8.98*9 \, \,ln(\frac{11}{6}) \,\frac{J}{C} \\V=80.82\,* \,0.606 \,\frac{J}{C}\\V=48.98 \frac{J}{C}[/tex]
