Cours et Exercices

∎ 1. The potential field created by a disc at a point on its axis of revolution . ~ ∎ ( The Solution )   ∎ 2. Potential and field created by a charged hemisphere surface ~ ∎ ( the solution ) ∎ 3. The potential created by a portion of a cone ~ ∎ ( The solution ) ∎ 4. Field Lines ►See the list of electromagnetic corrected exercises

Considering the electrostatic field in a plane defined by its components in polar coordinates: Determine the equation of the field lines. Back to the list of electromagnetism corrected exercises

∎ Back to exercise   Consider a surface element S on the truncated cone centered about the point P. The Elemental electrostatic potential created at the point O of the axis Ox by a surface element dS centered about the point P is expressed as: so : We get :   By integration: We get : ∎ Back to exercise ∎ Back to the list of electromagnetism corrected exercises

Consider a portion of a cone ,of half-angle with the vertex a and  limit rays R1 and R2 (R1 <R2). This system is loaded on the surface with non-uniform density: a is a constant homogeneous to a length and r the radius of the cone at a point on its axis of symmetry. Determine the electrostatic potential  at the vertex O of the cone. ► See the solution ► Back to the corrected exercises of electromagnetism 

Considering a half sphere of center O, of radius R, uniformly charged on surface with the surface density σ. 1. Determine the electrostatic potential at a point M of the axis Oz of symmetry of this hemisphere. 2. Deduce the expression of the electrostatic field at that point Mr. 3. Determine the potential and the electrostatic field on O. ( think for limited development). ► See the solution ► Back to the list of electromagnetism exercises

∎ Back to exercise   1. Research of the potential.   Consider a surface element of the half sphere centered at a point P. The electrostatic potential created at a point M of the axis Oz is expressed as: By integration over the azimuthal angle and expressing r:    and : Can then be expressed: The potential M is then: 2. Electrostatic field at M     We have : 3. Potential and field on O . The potential is written in the form: Calculating the boundary of V where z tends to 0: For the electric field, the procedure is the same way: ∎ Back to exercise ∎ Back to the list of electromagnetism exercises

Consider a disk its center O and its radius R. This system is loaded on the surface with the surface density:   a is a constant homogeneous to a length and r is the distance of a point P on the surface of the disc to its center . Determine the electrostatic potential at a point M of the axis  of revolution Oz of this disc. For the calculation is carried out the change of variable r = z u sh. Deduce the expression of the electrostatic field in point M. ► See the solution ► Back to the exercises of electromagnetism list  

∎ Back to exercise 1. The potential at a point M of the axis. Consider a surface area element of the disk centered at a point P. The electrostatic potential created at a point M of the axis Oz is expressed as: Since r and θ variables are separated : For the calculation is performed following variable change: For : On the other hand : We get : Taking the zero potential at infinity is obtained: 2. Field at a point M of the axis. The z axis is axis of symmetry of the charge distribution. The field at a point M of the axis is carried by this axis: We must therefore calculate the derivative of the function . We have: x = y where sh As   ∎ Back to exercise ∎ Back to the list of electromagnetism exercises

► 1. Field the center of a ring having an opening. ~  ► (The solution ) ► 2. Field created by a hemisphere surface   charge . ~   ► ( The Solution )   ► 3. field created by a portion of a cone . ~   ► (The Solution ) ► 4. Field created by a disc at a point on its axis. ~    ► (The Solution )   ► 5. electrostatic field created by an electrified segment. ~   ► (The solution )   ► 6. Electrostatic field created by a hemisphere surface charge. ~   ► ( The solution )   ► 7. Electric field on the axis of a system (-q, + q) ~   ► ( the solution ) ►See the list of electromagnetic corrected exercises

►Back to the Exercise Field created by a portion of a cone. The planes passing through the axis Ox are the charge distribution of planes of symmetry. The electrostatic field must belong simultaneously to all of these plans therefore their intersection: the field has its direction carried by the axis Ox Consider a surface element S on the truncated cone centered about the point P. The figure in the plane of the sheet is as follows: This surface element creates an elementary field O: Only the projection of this vector onto axis Ox contributes to the field at the point O: L is expressed as a function of r: We get : The field O is: ►Back to exercise ►See the list of exercises