Find the Potential as a function of position

[prace]

Homework Statement

A rod of length L carries a charge Q uniformly distributed along its length. The rod lies along the y-axis with one end at the origin. Find the potential as a function of position along the x-axis

Homework Equations

$$dV=\vec{E}\cdotp d\vec{l}$$

$$V=\frac{kq}{r}$$

The Attempt at a Solution

I think I am to use the first equation posted, but I am not sure how to relate it to the x-axis, if the rod lies along the y-axis. Also, what is the displacement here? I am assuming it is just dy?

 

[Andrew Mason]

Homework Statement

A rod of length L carries a charge Q uniformly distributed along its length. The rod lies along the y-axis with one end at the origin. Find the potential as a function of position along the x-axis

Homework Equations

$$dV=\vec{E}\cdotp d\vec{l}$$

$$V=\frac{kq}{r}$$

The Attempt at a Solution

I think I am to use the first equation posted, but I am not sure how to relate it to the x-axis, if the rod lies along the y-axis. Also, what is the displacement here? I am assuming it is just dy?

You have to integrate along the rod from y = 0 to +L.

$$V(x) = \int_{0}^{L} \frac{k}{r}dq$$

Work out the expression for dq in terms of dy. Work out the expression for r in terms of x and y (think: Pythagoras).

AM

 

[m2003]

To find the potential as a function of position along the x-axis, we can use the equation V=\frac{kq}{r}, where k is the Coulomb's constant, q is the charge on the rod, and r is the distance from the point on the x-axis to the point on the rod. Since the rod is uniformly distributed along its length, we can consider small elements along the rod and integrate the potential contributions from each element to find the total potential at a given point on the x-axis.

We can express the potential contribution from each element as dV=\vec{E}\cdotp d\vec{l}, where d\vec{l} is the displacement along the rod and \vec{E} is the electric field at that point. Since the rod is charged uniformly, the electric field at any point on the rod is given by \vec{E}=\frac{kq}{r^2}\hat{r}, where r is the distance from the point on the rod to the point on the x-axis and \hat{r} is the unit vector in the direction from the point on the rod to the point on the x-axis.

Integrating this potential contribution over the entire length of the rod, we get the total potential at a point on the x-axis as V=\int_{0}^{L}\frac{kq}{r^2}\hat{r}\cdot d\vec{l}=\int_{0}^{L}\frac{kq}{r^2}\cos\theta dl=\int_{0}^{L}\frac{kq}{r^2}\frac{x}{r}dl, where x is the distance from the origin to the point on the x-axis and \theta is the angle between the displacement vector d\vec{l} and the x-axis.

Since the rod lies along the y-axis, we can express the distance r as r=\sqrt{x^2+y^2}, where y is the distance from the origin to the point on the rod. Substituting this into the integral, we get V=\int_{0}^{L}\frac{kq}{(x^2+y^2)^{3/2}}xdl. This integral can be evaluated numerically to find the potential at any point on the x-axis.