Find the Potential as a function of position

In summary, to find the potential as a function of position along the x-axis for a rod of length L with a uniformly distributed charge Q along its length lying along the y-axis with one end at the origin, you need to integrate along the rod from y = 0 to +L. The expression for dq can be worked out in terms of dy and the expression for r can be worked out in terms of x and y using Pythagoras.
  • #1
prace
102
0

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


[tex]dV=\vec{E}\cdotp d\vec{l}[/tex]

[tex]V=\frac{kq}{r}[/tex]


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?
 
Physics news on Phys.org
  • #2
prace said:

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


[tex]dV=\vec{E}\cdotp d\vec{l}[/tex]

[tex]V=\frac{kq}{r}[/tex]

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.

[tex]V(x) = \int_{0}^{L} \frac{k}{r}dq[/tex]

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
 
Last edited:
  • #3


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.
 

1. What is the concept of "potential" in physics?

In physics, potential refers to the amount of energy that a system has due to its position or configuration. It is a scalar quantity and is used to describe the energy of a system in relation to a reference point.

2. How is potential related to position?

Potential is directly related to position, as it describes the energy of a system at a specific location. This means that the potential at a particular point will vary depending on the position of the system.

3. How do you find the potential as a function of position?

To find the potential as a function of position, you can use the formula V(x) = -∫F(x)dx, where V(x) is the potential as a function of position, F(x) is the force acting on the system at that position, and ∫ indicates integration. This formula can be used to calculate the potential at any point in a system.

4. What are the units of potential?

The SI unit for potential is joules (J), which is the same unit as energy. However, potential can also be expressed in other units such as electron volts (eV) or ergs (erg) depending on the system being studied.

5. How is potential used in practical applications?

Potential is a fundamental concept in physics and is used in many practical applications. In electrical circuits, potential is used to describe the energy of electrons as they move through a circuit. In gravitational fields, potential is used to calculate the potential energy of objects at different heights. It is also used in fields such as chemistry, biology, and engineering to describe the energy of systems in different configurations.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
876
  • Introductory Physics Homework Help
Replies
16
Views
712
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
334
  • Introductory Physics Homework Help
Replies
9
Views
905
  • Introductory Physics Homework Help
Replies
13
Views
720
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top