You'll have to provide a better problem description than that if you want to receive any meaningful help. As it stands I can sort of understand that you want to integrate over a charge distribution and determine the resulting electric field at some point.I-aM-Lost said:Homework Statement
Electric field at a distance 'a' from the end of a solid uniform charge volume
dq = Pdv
dv = (pi)(y)(dy)
A = (pi)(y)
dE = f(dq)
Homework Equations
What is the starting integral
The Attempt at a Solution
limits from 0 to b
(k)(p)(pi)(y)(dy)
E = --------------------
(b+a-y)^2
gneill said:You'll have to provide a better problem description than that if you want to receive any meaningful help. As it stands I can sort of understand that you want to integrate over a charge distribution and determine the resulting electric field at some point.
But I have no idea what the shape of the distribution is, so I can't check your setup in any way. Also, you didn't actually ask a question, merely presented some work.
gneill said:So this is a 3D problem? You'll have a triple integral to cover the volume if you want to do it all in one go. maybe choose cylindrical coordinates. Otherwise you might attack it as a stack of disks with a single integral, provided that you know the field from a uniformly charged disk...
gneill said:So where are you at with regards the field from a single disk? Have you derived an expression for the electric field on the axis of a charged disk?
## f(x) = \int_0^b\left( (k)(p)(pi-y)(dy) \right ) ## all over (b+a-y)^2gneill said:It would be better to write out your equations on a single line so that space preservation is not a concern. That or place your text inside code tags " [ code ] ... [ /code ] " (without the spaces inside the square brackets). Best of all learn to use Latex syntax and write your equations using it. For example:
## f(x) = \int_0^\pi \left( 3x^2 + 2x \right) dx ##
As for your problem, you need to describe what it is you're doing in detail. What the variables are, what the integral you're describing is meant to sum. The electric field on the axis of a uniformly charged disk is not as simple as that of a point charge! Take a look at the Hyperphysics website entry.
An electric field uniform refers to a region in space where the electric field strength is constant in both direction and magnitude. This means that the electric field lines are evenly spaced and parallel to each other.
An electric field uniform can be created by placing two parallel plates with opposite charges close together. The electric field lines between the plates will be evenly spaced and parallel, creating a uniform electric field in between the plates.
An electric field uniform is significant because it allows for easier calculations and predictions of the behavior of electric charges within the uniform field. It also simplifies the understanding of complex electric fields.
The strength of an electric field uniform is measured in volts per meter (V/m). This is the amount of electric potential energy per unit of charge at a given point in the field.
While it is difficult to create a perfectly uniform electric field in real life, it is possible to approximate a uniform field in certain situations, such as between two parallel plates or in the vicinity of a large, uniformly charged object.