Charge distribution and linear quadrupole

[nosracsan]

Homework Statement

The charge distribution on the figure below is referred to as a linear quadrupole.
Show that the potential at a point on the x-axis where x>a is given by:


PLEASE go to the attached MS Word document to view the rest of the problem / diagrams. It is easier to post them that way!

 

[Doc Al]

Don't just post the problem statement. Post your work!

 

[kerimek]

Thank you for sharing this problem with me. I am always excited to delve into problems involving charge distributions and potential calculations. After reviewing the attached document and diagrams, I have a few thoughts and suggestions on how to approach this problem.

Firstly, it is important to understand the concept of a linear quadrupole and how it differs from other charge distributions. A linear quadrupole consists of two equal and opposite dipoles placed end to end, resulting in a total charge of zero. This creates a unique electric field pattern where the field is strongest along the axis of the quadrupole and decreases rapidly in the transverse direction. This is in contrast to a point charge or a dipole where the field strength decreases gradually with distance.

Now, let's move on to the problem at hand. The goal is to find the potential at a point on the x-axis where x>a. To solve this, we can use the principle of superposition. This states that the total potential at a point due to multiple point charges is equal to the sum of the potentials due to each individual charge. In this case, we have two dipoles, each consisting of two point charges, so we can break down the problem into four separate potentials and add them together.

The first step is to calculate the potential due to a single dipole at the given point. This can be done using the formula for the potential due to a dipole, V = kq/r, where k is the Coulomb constant, q is the charge of the dipole, and r is the distance between the dipole and the point. Since the dipoles in this problem have equal and opposite charges, the total potential due to a single dipole will be zero at the given point.

Next, we need to find the potential due to the second dipole. This will also be zero at the given point since the dipoles are symmetrically placed on either side of the point and their potentials cancel out.

Now, let's move on to the potential due to the individual point charges. Using the same formula, we can calculate the potential due to each point charge and add them together. However, since the charges are equal and opposite, their potentials will also cancel out at the given point.

Finally, we can combine all of these potentials to get the total potential at the given point. Since both dipoles and individual point charges cancel out, the only remaining contribution will