How Do You Calculate the Electric Potential of an Ellipsoid?

  • Thread starter Amit Kumar Basistha
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In summary, calculating the electric potential of an ellipsoid involves integrating the contributions of infinitesimal charge elements distributed over the surface or volume of the ellipsoid. The potential at a point in space is determined using the formula \( V = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|} dV' \), where \( \rho \) is the charge density, \( \mathbf{r} \) is the observation point, and \( \mathbf{r'} \) represents points within the ellipsoid. The geometry of the ellipsoid, usually defined by its semi-axes, must be accounted for in
  • #1
Amit Kumar Basistha
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Homework Statement
Recently I came across the following problem:

Suppose ##\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1## is an ellipse with surface charge density ##\sigma=\sigma_0\sin(\theta)\cos(\phi)## where ##\theta## is the angle with the ##z-## axis and ##\phi## is with the ##x-## axis. Find the potential and multipole moments at a point far away from the ellipse.
Relevant Equations
Maxwell's Equations
Spherical Harmonics
My initial idea was to first parametrize the ellipse as ##(a\sin(\theta')\cos(\phi'),b\sin(\theta')\sin(\phi'),c\cos(\theta'))## and then calculate ##\theta,\phi## in terms of these coordinates. I then did the coordinate transform ##x\to\frac{x}{a},y\to\frac{y}{b},z\to\frac{z}{c}## to convert it to the sphere case where you can find the potential and multipole moments using spherical harmonics. But the whole calculation is messy because you have to find the fundamental vector product and all those stuff for the coordinate change and the expression for the angles in terms of these coordinates.
 
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  • #2
I think you need to consider how far is "far away". This looks like it needs an approximation. What does an elliptical (American) football look like next to a spherical (rest-of-the-world) football from 100 m away?
 
  • #3
My understanding why the question mentions far away is because if the point is near the surface of the ellipsoid then there might be issues with the spherical harmonics expansion as the radius of the ellipsoid is not uniform. So, far away means you can safely assume that it's distance from the origin (Center of the Ellipsoid) is more that the distance of any point on the surface of the Ellipsoid from the origin
 
  • #4
In this kind of problem, you have a point of observation at ##\vec r## from the origin and a charges at ##\vec r^{~'}##. At least to me (and I've been wrong before), "far away" usually means ##|\vec r|>>|\vec r^{~'}|## which in this case would translate to ##|\vec r|>>a,b,c.## However, if that is the case, why mention the ellipsoid? Do you have a reference where you found this problem? You just may have to put up with the messy calculation.
 
  • #5
No I don't have a reference. Our Prof asked us to solve the problem in the case of the sphere and then asked us as a separate question to do it for the Ellipsoid
 

FAQ: How Do You Calculate the Electric Potential of an Ellipsoid?

What is the general formula for calculating the electric potential of an ellipsoid?

The electric potential \( V \) at a point inside or outside an ellipsoid with uniform charge distribution can be calculated using the formula:\[ V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int_{\text{Ellipsoid}} \frac{\rho(\mathbf{r}') d^3\mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|} \]where \( \rho(\mathbf{r}') \) is the charge density, \( \mathbf{r} \) is the position vector of the point where the potential is being calculated, \( \mathbf{r}' \) is the position vector of the charge element, and \( \epsilon_0 \) is the permittivity of free space.

How do you simplify the calculation for an ellipsoid with uniform charge distribution?

For an ellipsoid with a uniform charge distribution, the potential at a point inside can be simplified using ellipsoidal coordinates. The potential inside a uniformly charged ellipsoid is constant and can be given by:\[ V = \frac{\rho}{3\epsilon_0} \left( a^2 + b^2 + c^2 \right) \]where \( a \), \( b \), and \( c \) are the semi-axes of the ellipsoid, and \( \rho \) is the charge density.

What are the boundary conditions for calculating the electric potential of an ellipsoid?

Boundary conditions for calculating electric potential involve ensuring continuity of the potential and the electric field across the surface of the ellipsoid. At the surface of the ellipsoid, the potential \( V \) and the normal component of the electric field \( \mathbf{E} \) must be continuous. This implies:\[ V_{\text{inside}} = V_{\text{outside}} \]and\[ \epsilon_0 \frac{\partial V_{\text{inside}}}{\partial n} = \epsilon_0 \frac{\partial V_{\text{outside}}}{\partial n} \]where \( \frac{\partial}{\partial n} \) denotes the derivative normal to the surface.

How does the electric potential vary at points outside the ellipsoid?

At points outside the ellipsoid, the electric potential can be found using the principle of superposition and integrating over the surface or volume of the ellipsoid. For a uniformly charged ellipsoid, the potential at a distance \( r \) from the center, where \( r \) is greater than the largest semi-axis of the ellipsoid, behaves similarly to a point charge:\[ V = \frac{Q}{4\pi\epsilon_0 r}

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