# Field due to a solid hemisphere with uniform charge density

1. Jun 24, 2013

### PhizKid

1. The problem statement, all variables and given/known data
Given a solid hemisphere with radius R and uniform charge density $/rho$, find the electric field at the center.

2. Relevant equations
$E = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^{2}}$
$E = \frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho (x',y',z') \hat{r} dx' dy' dz'}{r^{2}}$

3. The attempt at a solution

The strategy is to slice the hemisphere into rings around the symmetry axis, then find the electric field due to each ring, and then integrate over the rings to obtain the field due to the entire hemisphere.

I am confused by the diagram, though:

It says that a rectangle with side lengths dr and r dθ is the cross-sectional area, and that this cross-sectional area multiplied by 2πrsinθ is the volume, since the radius of the ring is rsinθ. However, isn't this just the circumference multiplied by the cross-sectional area? There is no thickness of the ring, so is it a volume? Are there more rings within the ring, towards the axis? (Like smaller rings within larger rings.) How come the thickness isn't mentioned here?

2. Jun 24, 2013

### barryj

canceled

Last edited: Jun 24, 2013
3. Jun 24, 2013

### voko

It is.

Not sure what you mean here. The rings involved are 3D objects, and they definitely do have "thickness". The cross-section of a ring is a rectangle, as you said.

4. Jun 25, 2013

### PhizKid

Actually, I am now confused about the radius used to compute the circumference. In the 2 circles within the hemisphere for the drawing, the 2 circles have different radii. Is it because both are approximately the same radius?

5. Jun 25, 2013

### voko

Let's assume that the cross-section is trapezoidal. What is its area then?