Gauss-Theorem on a solid dielectric sphere

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SUMMARY

The discussion focuses on applying Gauss's Theorem to determine the electric field inside and outside a solid dielectric sphere under a point load and a load distribution. For the region where r<1m, the electric field vector is derived from the vector pointing from the center to point P, specifically r→=(0.3,0.4,0) with a magnitude of r=0.5. The calculation of the first component of the electric field vector is confirmed to be [−36+9]⋅(0.3)/(0.5), although there is a noted discrepancy in the decimal placement of the final answer.

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  • Understanding of Gauss's Theorem in electrostatics
  • Familiarity with vector calculus
  • Knowledge of electric field concepts
  • Basic principles of dielectric materials
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  • Study the application of Gauss's Theorem in different geometries
  • Learn about electric field calculations in dielectric materials
  • Explore vector calculus techniques for field vector determination
  • Investigate common errors in electric field component calculations
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Students of physics, electrical engineers, and professionals working with electrostatics and dielectric materials will benefit from this discussion.

Guillem_dlc
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Homework Statement
We have an Q=2nC-load uniformly distributed on a solid dielectric sphere of R=1m-radius with center at the coordinate origin of an Oxyz axis system. At the O centre of this sphere there is a negative point charge q=−1nC. Let us consider the points of space P=(0.3,0.4,0)m and S=(1,2,2)m. Calculate:

(a) The electric field vector at points P and Q.

(b) The difference in potential between these points, i.e. V(S)−V(P).
Relevant Equations
Gauss-Law
The load system formed by the point load and the load distribution generates two regions in space corresponding to r<1m and r>1m, i.e. inside and outside the sphere. Given the symmetry of the distribution, by means of the Gaussian theorem we can find the modulus of the field at a distance r from the center.
  • Region r<1m:
    Captura de pantalla de 2020-01-01 01-32-28.png
To find the electric field vector at point P we need the vector that goes from point O to point P, i.e. r→=(0.3,0.4,0), whose module is worth r=0.5. Thus, the field vector is given by:
Captura de pantalla de 2020-01-01 01-32-04.png


Until the last equation I have it well but the result gives me different. The first component is: [−36+9]⋅(0.3)/(0.5) isn't it? Or I don't see it ...
 
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Yes, it looks like the final answer has the decimal points in the wrong place.
 
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