Conic charge distribution setting up integral

[electrifice]

Homework Statement

A uniform charge (with density D) is distributed over a cone of radius R and height z. The axis of the cone is coincident with the x-axis. Find the total electric field due to the cone along the x axis.

Homework Equations

E = kqq/r^2

The Attempt at a Solution

This is really just a hypothetical problem to help study for my midterm so I need to make sure that I have the right idea.
For the first part...

.../2(pi).../ z.../(-R/z)x + R
...|...|...|
kD |da...|dz...| (1/r^2)r(dr)
...|...|...|
.../ 0.../ 0.../ 0

(Sorry for the messy-looking integral)
All of that will be multiplied by the cosine of the angle that the cone's laterals make with the x-axis. Does that look reasonably correct? Now for the second part... I am not able to set up a proper integral to be able to describe a hollow cone. Would dQ just be D*2(pi)r*(dr)? I think the second part should be easier than the first, but I just can't figure out the integral.

 

[anathema]

Your approach for the first part looks reasonable. However, instead of integrating over the entire cone, you only need to integrate over the lateral surface of the cone since the electric field is zero at the base and tip of the cone. Also, you will need to take into account the fact that the charge density is not constant along the lateral surface of the cone. You can divide the lateral surface into small elements and integrate over each element, taking into account the varying charge density at each element.

For the second part, you are correct in assuming that the charge element will be D*2(pi)r*(dr). However, you will also need to consider the varying distance between the charge element and the point on the x-axis where you are calculating the electric field. This will affect the electric field magnitude and direction. You can also divide the lateral surface of the hollow cone into small elements and integrate over each element, taking into account the varying distance between the charge element and the point on the x-axis.

 

[Moetasim]

I would approach this problem by first understanding the physical setup and the variables involved. The cone is a three-dimensional object with a uniform charge density (D) distributed over its surface. The radius (R) and height (z) of the cone are given, and the axis of the cone is coincident with the x-axis. We are asked to find the total electric field along the x-axis due to this cone.

To solve this problem, we can use the principle of superposition, which states that the total electric field at a point due to multiple charges is the vector sum of the individual electric fields at that point. In this case, we can divide the cone into infinitesimally small sections, each with a charge dq = D*da, where da is the infinitesimal surface area of that section.

Now, let's focus on a single section of the cone. We can draw a small right triangle with base dr and height dz, as shown in the diagram below:

/|
/ |
/ |
/ |
/____|
dz/ |dr
/_____|

The hypotenuse of this triangle is r, which is the distance from the section to the point on the x-axis where we want to find the electric field. Using the Pythagorean theorem, we can express r in terms of dr and dz as r = √(dr^2 + dz^2).

Now, we can use Coulomb's law to find the electric field at this point due to this small section of the cone:

dE = k * dq / r^2 = k * D * da / r^2

Substituting for r, we get dE = k * D * da / (dr^2 + dz^2)

To find the total electric field, we need to integrate this expression over the entire surface of the cone. The limits of integration will be from 0 to R for r, and from 0 to z for z. The integral will be:

E = ∫∫ dE = ∫∫ k * D * da / (dr^2 + dz^2)

This integral can be solved using polar coordinates, where da = r * dr * dθ. The integral then becomes:

E = ∫∫ k * D * r * dr * dθ / (dr^2 + dz^2)

Integr