The question is missing some parts. Here is the complete question.
A finite rod of length L has a total charge q, distributed uniformly along its length. The rod lies on the x-axis and is centered at the origin. Thus one endpoint is located at (L/2) and the other is located at (L/2). Define the electrical potential to be zero at an infinite distance away from the rod. Throughout this problem, you may use the constant k in place of the expression 1 / [tex]4\pi\epsilon_{0}[/tex]
Part A (image 1): What is VA, the electric potential at point A (see figure below), is located a distance d above the midpoint of the rod on the y-axis? Express your answer in terms of L, d, q and k.
Part B (image 2): What is VB, the electric potential at point B, located at distance d form one end of the rod (on the x-axis)? give your answer in terms of L, d, q and k.
Answer:
Part A: [tex]V = \frac{kq}{L}[ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} }) ][/tex]
Part B: [tex]V=\frac{kq}{L} ln(\frac{L}{d} )[/tex]
Explanation: Electric Potential (V) is the amount of work done per unit charge to move a charge from point A to B.
For a finite rod with charge uniformly distributed along its length
[tex]V=\frac{1}{4\pi\epsilon_{0}} \int\ {\frac{\lambda}{r} } \,dl[/tex]
where
λ is charge density and, in this case, is constant: [tex]\lambda=\frac{q}{L}[/tex]
dl is differential of the rod
r is the distance the point is from the rod
Part A: [tex]r = \sqrt{(a^{2}+y^{2})}[/tex]
[tex]V=k \int\ {\frac{\lambda}{\sqrt{(a^{2}+y^{2})}} } \,da[/tex]
[tex]V=k\lambda \int\ {\frac{1}{{\sqrt{(a^{2}+y^{2})}}} } } \,da[/tex]
[tex]V=k\lambda \int\ {\frac{1}{{\sqrt{(a^{2}+y^{2})}}} } } \,da[/tex]
[tex]V=\frac{kq}{L}\int\limits^\frac{L}{2} _\frac{-L}{2} {\frac{1}{\sqrt{(a^{2}+y^{2})} } } \, da[/tex]
[tex]V=\frac{kq}{L}ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} } )[/tex]
At a point located at y-axis, electric potential is [tex]V=\frac{kq}{L}ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} } )[/tex]
Part B: r = x
[tex]V=k \int\limits^L_d {\frac{\lambda}{x} } \, dx[/tex]
[tex]V=k \frac{q}{L} \int\limits^L_d {\frac{1}{x} } \, dx[/tex]
[tex]V=k \frac{q}{L} ln(L - d)[/tex]
[tex]V= \frac{kq}{L} ln(\frac{L}{d} )[/tex]
At a point d from one end, electric potential is [tex]V= \frac{kq}{L} ln(\frac{L}{d} )[/tex]
