Electric Fields PDF

Electric Fields  The electrical interaction between charged particles can be reformulated by using the concept of electric field.  Two charges exert a force on each other whose effect of one charge on the other is transmitted through space.  These charges also act even if t...

Electric Fields  The electrical interaction between charged particles can be reformulated by using the concept of electric field.  Two charges exert a force on each other whose effect of one charge on the other is transmitted through space.  These charges also act even if they are not in contact with each other.  Let us think of an electric charge as creating an electric field in the region of space surrounding it.  That field, in turn, exerts a force on any other charge in that region.  Thus, a charged particle produces an electric field around it.  Within the range of electric field, any charged body can experience a force acting on it.  The electric field around a charged body may be thought of as represented by lines of force.  Charges obey the rule that they repel each other if they are similarly charged and attract each other if they are unlike charges.  Take note that the lines of force do not interact with each other. The field lines have a symmetrical pattern. Since electric field has a direction, it is therefore a vector quantity. Fig, 1. Electric flux: Electric flux visualized. The ring shows the surface boundaries. The red arrows for the electric field lines The magnitude of the electric field can be calculated as: E = Where:F – force in newton (N) q – unit charge in coulomb (C) E – electric field intensity in Newton per Coulomb (N/C) Sample Problem: A charge of 0.50µC is placed in an electrostatic field whose intensity is 4.0 x 10 5 N/C. What is the electrostatic force acting on the charge? Given: E = 4.0 x 105 N/C Find: F q = 0.50 µC = 5.0 x 10-7 C Solution: E = F = Eq = (4.0 x 105 N/C) (5.0 x 10-7 C) F = 0.20 N Practice 1. Calculate the intensity of the electric field at a point if the force on the particle of 60C placed at that point is 0.02N. 2. If a charge of 5 C placed in an electric field experiences a force of 0.8 N, what is the electric field intensity? 3. A charged sphere with a mass m suspended between two oppositely charged parallel plates. The electric field between the plates is 2500 N/C downward. If the charge of the sphere is -3.92 x 10⁻⁹ C, a) what is the magnitude of the electrostatic force that acts on the sphere? b) Calculate the mass of the sphere. Electric Flux and Gauss’ Law  Electric Flux is the rate of flow of the electric field through a given area.  Electric flux is proportional to the number of electric field lines going through a virtual surface.  If the electric field is uniform, the electric flux passing through a surface of vector area A is ɸ = E. A = EA cos ϴ where: E is the magnitude of the electric field (having units of N/C) A is the area of the surface, and ϴ is the angle between the electric field lines and the normal (perpendicular) to A