Electric Field at Point P on Spherical Shell Insulator

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Homework Help Overview

The problem involves calculating the electric field at a specific point P located within a spherical shell insulator defined by inner and outer radii, carrying a uniform charge. The context is electrostatics, focusing on the behavior of electric fields in relation to charged objects.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the volume of the spherical shell and the charge density (rho). There are attempts to clarify the formula for the electric field and the volume calculation method. Questions arise regarding the interpretation of the equations provided by the professor.

Discussion Status

Participants are exploring different methods to calculate the electric field and are questioning the accuracy of their calculations. Some have received guidance suggesting a review of fundamental concepts in electrostatics. There is acknowledgment of potential issues with the online problem's unit handling.

Contextual Notes

There is mention of a specific interactive online problem that provides feedback on answers, which may be influencing the participants' approach. The discussion reflects uncertainty about the correct application of formulas and the implications of unit discrepancies.

heartofaragorn
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Homework Statement


An insulator in the shape of a spherical shell is shown. The insulator is defined by an inner radius a = 4 cm and an outer radius b = 6 cm and carries a total charge of Q = +9 microC. You may assume that the charge is distributed uniformly throughout the volume of the insulator. What is Ey, the y-component of the electric field at point P which is located at (x, y) = (0, -5 cm)?


Homework Equations


Volume of a sphere: 4/3pi R^3
Electric Field: E=4pi*r^2 = Qen / sigma o
Rho = total charge / total shell volume
Electric field at point P = rho * volume with charge


The Attempt at a Solution


Consulted my physics prof on this, but I am still not getting the correct answer (is an interactive online problem that says "OK" or "NO"). He told me to calculate the volume of the sphere, which was .000637 due to the large hole in the middle (volume only applicable to an area between r = 4 cm and r = 6 cm). Calculated rho to be 0.014135 based upon a total charge of +9 micro Coulombs...is this where I'm going wrong? Based upon the answer I received for rho, I multiplied it by the volume of the area between 4 and 5 cm, which I thought was represented by the formula 4/3 pi r^3. That gave me an answer of 5.92085 e -08 for the electric field. There is no x-component and the point lies on the y-axis, so no angle is considered. I tried using both a positive and negative value and cannot get a happy little "OK". Where did I go wrong? Please help! :confused:
 
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heartofaragorn said:
Electric field at point P = rho * volume with charge

I didn't understand this eqn.

[Based upon the answer I received for rho, I multiplied it by the volume of the area between 4 and 5 cm, which I thought was represented by the formula 4/3 pi r^3.

I presume you did 4/3 pi (5^3 -4^3) to get the volume?

Otherwise, your method looks all right.
 
I hadn't tried using 5^3 - 4^3, but when I did, I still didn't get the correct answer. The equation of E = rho * volume with charge was given to me by my professor.
 
I think you need to know more about electrostatics before solving this problem. Read up on the field of a point charge, the field of a uniformly charged shell -- inside and outside. Then, if you still don't get the answer, you can again post your question here. We'll be here.

How long will that take to do? Maybe just a couple of days.
 
Ok, I'll look more at the book and see what happens!
 
Received some help from a classmate...apparently when they wrote the online question, they messed around with the units...A LOT. Thanks for all the help, though!
 
I think we are in same physics 212 class. Can you finish explaining how the units were messed up? Because I'm pretty certain I have the right answer but it's not correct according to the system.
 

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