Charged Spherical Shell: Difference between revisions
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Claimed by | Claimed by Chianne Connelly | ||
==The Main Idea== | ==The Main Idea== | ||
Charged objects create electric fields. Each object creates a different electric field depending on its shape, charge, and the distance to the observation location. A charged spherical shell acts like a point charge, so it uses the same equation as the electric field from a point charge. | |||
===A Mathematical Model=== | ===A Mathematical Model=== | ||
For an observation location outside of the sphere, the equation E_sphere = (1/4πε_0)(q/r^2)rhat should be used, where q is the charge of the object and r is the magnitude of the distance from the observation location to the source. | |||
However, if your observation location is inside of the sphere, E=0. | |||
===A Computational Model=== | ===A Computational Model=== | ||
(I spent a good amount of time trying to put images in this section but I could not manage to do so -- I'm sorry!) | |||
If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1. | |||
If the observation location is anywhere inside of the spherical shell, then the electric field is zero. This is because all of the charges will cancel out. | |||
==Examples== | ==Examples== | ||
Revision as of 21:19, 5 December 2015
Claimed by Chianne Connelly
The Main Idea
Charged objects create electric fields. Each object creates a different electric field depending on its shape, charge, and the distance to the observation location. A charged spherical shell acts like a point charge, so it uses the same equation as the electric field from a point charge.
A Mathematical Model
For an observation location outside of the sphere, the equation E_sphere = (1/4πε_0)(q/r^2)rhat should be used, where q is the charge of the object and r is the magnitude of the distance from the observation location to the source. However, if your observation location is inside of the sphere, E=0.
A Computational Model
(I spent a good amount of time trying to put images in this section but I could not manage to do so -- I'm sorry!)
If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.
If the observation location is anywhere inside of the spherical shell, then the electric field is zero. This is because all of the charges will cancel out.
Examples
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