Exercise Pro

∎ Return to the exercise 1. Expression of the electrostatic field.   All the planes containing the center O of the sphere and the point P are the load distribution of planes of symmetry. The electric field must belong simultaneously to all of these plans, it is therefore carried by their intersection which is the line OM.  we obtain.: As this distribution has a rotational invariance around the point O, the electric field does not depend on the angular variables. This result does not depend on the position of this point P. We choose a Gaussian surface centered in O and radius r. The Gauss theorem is: Then there are two regions of space: ● For r <R (one can put an equal sign here because the density distribution is what ensures the continuity of the normal component (here radial) of the electrostatic field). For the domestic load: Is obtained by applying the Gauss theorem: ...

►Voir solution Considering in the vacuum a sphere of radius R, the center O, with a volume charge distribution : and k are constants and 1. Determine the expression of the electrostatic field at any P in space. Note OP = r. 2. Show that within the sphere, the electrostatic field has a maximum for a ratio r / R given. Calculate the value of k in the case where the field is extremum for r / R = 1/2 ►Voir solution ∎ See the list of electromagnetism corrected exercises  

Considering a half sphere of center O, of radius R, uniformly charged surface with the surface density s. Determine, using only the symmetries The invariance, the Gauss theorem and the superposition principle the direction of the electrostatic field at any point M of the diametrical plane.   ∎ solution soon ∎ See the list of electromagnetism exercises

∎ Return to the exercise   We Can modeled within a cavity hollowed out of the sphere R as the superposition of a charged sphere of radius a volume density - ρ   of center O2 and a full sphere of volume density ρ of radius R and center O1. The principle of superposition is applied at a point M of the cavity: field created by the distribution ρ field created by the distribution - ρ Symmetry and invariance of each source can be concluded for each radial field : Using the Gauss, taking for each distribution a sphere of radius r and center Oi closed surface and passing through the point M.   We get : The field is uniform at any point inside the cavity. ∎ Return to the exercise ∎ Back to the list of electromagnetism corrected exercises

►Voir solution A sphere 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? ►Voir solution ∎ See the list of electromagnetism exercises

∎ Back to exercise     The Load distribution can be seen as the result of the superposition of an infinite plane of a surface density of charge s and a disc with a density - s. The field at a point M of the axis  according to the principle of superposition is the vector sum of the fields created by each of these distributions taken separately. Either:   Noting the unit vector normal to the plane and oriented to the point M, z the coast of the point M from the center of the circular opening ,we have , using conventional demonstrated results in progress (demonstrate need to know!): Finally : ∎ Back to exercise ∎ See the list of electromagnetism exercises

∎ See the solution Consider a uniformly charged surface plane which is practiced in an empty hole of radius R and fillers. Determine the electrostatic field at a point M of the axis of the hole. We note s the surface density loads.   ∎ See the solution   ∎ See the list of electromagnetism corrected exercises