Finding Potential (Spherical coordinates )

In summary, the conversation discusses the calculation of electric potential at a given point due to an electric dipole located at the origin in free space. The dipole has a moment of p = 3ax −2ay +az nC·m and the potential is calculated using the formula V = kp*r/r3, where r is the vector from the origin to the point of observation and r its magnitude. The conversation also mentions the use of unit vectors and the importance of translating to a Cartesian coordinate system before doing a dot product. It is also noted that for a dipole centered at the origin along the z-axis, the potential may have an azimuth dependence.
  • #1
hadez
2
0
1. . An electric dipole located at the origin in free space has a moment p = 3ax −2ay +az nC·m. Find V at r = 2.5, θ =30◦, φ =40◦.
I find it difficult to solve when its in spherical co-ordinates.2.Relevent Eq
V =P.(r-r')/( 4∏ε|r−r'|2)(|r-r'|)I am confused how to find a unit vector on spherical co-ordinate. ie |r-r'|This que is from william Hayt and page no 100.
 
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  • #3
Thanks for the reply,but my que
1)how to find unit vector in a spherical coordinate system?
2) dipole p is in rectangular coordinate system and distancw vector is in spherical coordinate system.How do i multiply both with a dot product.?
Haunts me.
I spent a whole day and and went mad with vectors buzzin in my brain. I am preparing for an test this fundas are essential.please look into this
 
  • #4
You can express the observation point for V in cartesian coordinates, then use just the cartesian coordinartes. The observation point is totally defined for you in spherical terms.

V = kp*r/r3

where r is the vector from the origin to the point of observation, and r its magnitude.

So you need to figure out how to express this potential in terms of your cartesian coordinate system, then do the dot-product with the dipole moment expression.

Don't try to do a dot-product in a spherical system. You have to translate to cartesian first.
 
Last edited:
  • #5
hadez said:
1. . An electric dipole located at the origin in free space has a moment p = 3ax −2ay +az nC·m.


Should say p = 3a i - 2a j + a k,
ijk unit vectors
 
  • #6
rude man said:
Should say p = 3a i - 2a j + a k,
ijk unit vectors

yes and do the dot product and expand using ##V = \frac{\vec{p} . \vec r}{4 \pi \epsilon_0 r^3}##

And for a dipole centrered at origin along z-axis for example, what can you say about azimuth dependence? Does the potential depend on ##\phi##?
 

Related to Finding Potential (Spherical coordinates )

What are spherical coordinates and why are they important in finding potential?

Spherical coordinates are a system of locating points in three-dimensional space using a distance (r), an angle from the positive z-axis (θ), and an angle from the positive x-axis (φ). They are important in finding potential because they provide a way to represent a point in space relative to a specified origin, which is necessary for calculating potential energy.

How do you convert Cartesian coordinates to spherical coordinates?

To convert Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, φ), you can use the following formulas:

r = √(x² + y² + z²)

θ = arccos(z/r)

φ = arctan(y/x)

What is the difference between potential energy and potential function?

Potential energy is a measure of the energy possessed by an object due to its position relative to a specified origin. Potential function, on the other hand, is a mathematical function that describes the potential energy at any point in space. In other words, potential energy is a physical concept while potential function is a mathematical representation of that concept.

How can spherical coordinates be used to find the potential of a point charge?

To find the potential of a point charge using spherical coordinates, you can use the formula:

V = kq/r

Where k is the Coulomb constant, q is the charge of the point charge, and r is the distance between the point charge and the specified origin. You can use spherical coordinates to determine the value of r and then plug it into the formula to find the potential at that point.

Are there any limitations to using spherical coordinates in finding potential?

While spherical coordinates are useful in many cases, they do have some limitations. One limitation is that they are not suitable for describing points on the z-axis, as this would result in an undefined value for θ. Additionally, spherical coordinates may not be the most efficient system for certain calculations, such as when dealing with objects that are not spherically symmetric. In these cases, other coordinate systems may be more appropriate.

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