# E-field of a Semi-spherical charge distribution with a hole

• sergioro
In summary, the conversation discusses computing the electric field at a point in a semi-spherical charge distribution with a spherical hole, using direct integration. The relevant equations are Coulomb's law in differential form and the position vectors for the field point and source point. The attempt at a solution involves performing an integral, but there is a divergence in the result. The solution is eventually found using a power series. Changing the origin to the center of the hole may make the integration easier, but it would also introduce asymmetry. The conversation concludes with a question about whether integrating over a charged half-sphere and subtracting an integration over the hole sphere with opposite charge density could be a potential solution.
sergioro

## Homework Statement

Using direct integration, compute the electric field at the point (0,0,Z) of a semi-espherical (z > 0) charge distribution of radius R < Z and density ##\rho=constant## having a spherical hole of radius ##r_h<R## centered at (0,0,##r_h##).

## Homework Equations

Coulomb's law in differential form:
## d\vec{E} = \frac{Kdq}{|\vec{r} - \vec{r}_{dq}|^2}\frac{\vec{r} - \vec{r}_{dq}}{|\vec{r} - \vec{r}_{dq}|}##

##\vec{r} = Z \hat{k} ## (the field point)

##\vec{r}_{dq} = r\sin(\theta)\cos(\phi)\hat{\imath} + r\sin(\theta)\sin(\phi)\hat{\jmath} + r\cos(\theta)\hat{k}## (the source point)

## r = 2r_h\cos(\theta) ## (the spherical hole in spherical coordinates )

## The Attempt at a Solution

By symmetry, the only relevant integral is:

## \vec{E}(z) = 2\pi K \rho \int_{\theta=0}^{\theta=\pi/2} d\theta \int_{r=2 r_h \cos{\theta}}^{r=R} dr \left[ \frac{ Z-r\cos(\theta) }{ (r^2 - 2 r Z \cos(\theta) + Z^2)^{3/2} } \right]r^2 \sin(\theta) ##

The question is how to perform the integral. Using brute force via the wolfram alpha site
finished with Standard computation time exceeded...

https://www.wolframalpha.com/input/...,+{r,+2*rh*Cos[theta],+R}],+{theta,+0,+Pi/2}]

Sergio
PS. By the way, the formulation seems to be right as one can get the right ##\vec{E}## field of the full semi-sphere

I wonder about the lower limit of ##dr##. Why not simply ##r_h## ?

It is not a fix lower limit. It is a spherical hole.

But surely all points on the surface of a spherical hole that is centered at ##(0,0,r_h) ## are at a distance ##r_h## away from the origin ?

BvU said:
But surely all points on the surface of a spherical hole that is centered at ##(0,0,r_h) ## are at a distance ##r_h## away from the origin ?

Yes, but the charge source vector is not measure respect the hole. The picture attached "might" help to make clear the narrative of the problem.

I found the solution via power expansion. Nevertheless, I am still curios on why direct integration does not seem to work here (there is a divergence in the result due to a logarithm term, that in spite the integrand behaves well in the region of integration).
This might be the reason why textbooks does not have this neither as a solved problem nor as proposed one. It is a very rich problem to ilustrate vectors, integration that does not seems to work (or are very difficult) and solution via power series.

Perhaps changing the origin to be at the center of the hole could make the integration less hard, but the asymmetry it generates does not seem to support that.Sergio

O, I see , pretty dumb of me, sorry. I take it the picture came with the exercise ?
By now I understand relevant equations 1,2,3 And I think I understand ##
r = 2r_h\cos(\theta)## as the lower bound of ##r##.
I know it says 'by direct integration', but does that exclude an integration over a charged half-sphere (which you apparently have already) minus an integration over the ##r_h## sphere with opposite charge density (which indeed is easier with the origin at ##(0,0,r_h)## ?

## 1. What is the E-field of a semi-spherical charge distribution with a hole?

The E-field of a semi-spherical charge distribution with a hole refers to the electric field strength at any point in space surrounding a half-sphere with a hole in the center, caused by the distribution of electric charge on the surface of the half-sphere.

## 2. How is the E-field calculated for a semi-spherical charge distribution with a hole?

The E-field can be calculated using the formula E = kQ/r^2, where k is the Coulomb constant, Q is the total charge of the half-sphere, and r is the distance from the center of the half-sphere to the point in space where the E-field is being measured. This formula assumes that the charge distribution is evenly spread across the surface of the half-sphere.

## 3. What factors affect the strength of the E-field in a semi-spherical charge distribution with a hole?

The strength of the E-field is affected by the total charge of the half-sphere, the distance from the center of the half-sphere, and the size of the hole in the center. The E-field will be stronger if there is a larger charge, closer distance, and smaller hole.

## 4. How does the presence of the hole in the center affect the E-field of a semi-spherical charge distribution?

The hole in the center of the semi-sphere creates a region of zero charge, which can cause a discontinuity in the E-field. This means that the E-field will be weaker in this region compared to the rest of the space surrounding the half-sphere.

## 5. What are the applications of studying the E-field of a semi-spherical charge distribution with a hole?

Understanding the E-field of a semi-spherical charge distribution with a hole is important in various fields such as electrostatics, electromagnetism, and electronics. It can also be applied in the design and analysis of devices such as capacitors, antennas, and sensors.

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