### Field on a spherical cavity - Corrected Exercises Gauss Theorem

*of*radius R, carrying a load volume

*r*that is uniformly distributed throughout the volume which occupies the exception of a radius of

*cavity a.*The center of this cavity is the distance

*of*the center of the sphere. The cavity is empty loads.

Using the Gauss theorem and the principle of superposition, calculate the field at all points of the cavity.

what is remarkable?

Moving from contact interactions to distance interactions, two aspects change. Since there is no point of contact between the two interacting objects, the specification of the object on which a given force is exerted becomes much less ambiguous: the object concerned must be well chosen. This is an advantage. But, as a complication, a given object becomes likely to interact with another one whatever their respective locations. The whole space is doubly concerned. In the analysis of such situations, one must expect to add forces "created", as it is often said, by many objects, and it is the principle of superposition - recalled later - that comes into play. It happens, on the macroscopic scale, that the set of actions at a distance exerted on a body can be summarized very simply via a sort of equivalent point of application: this is the case of gravity. In the case of interactions "in 1/r2 ", associated with conditions of spherical symmetry, the situation can also be simplified considerably.

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