1. Charge and Electric Flux

2. Electric Flux A closed surface around an enclosed charge has an electric flux that is outward on inward through the surface.

3. Flux If the net charge in the closed surface is zero then the net flux is zero. There are just as many field lines out as there are field lines going in. Flux is directly proportional to the charge inside the surface.

4. Electric Flux We must define electric flux as: the product of the average perpendicular component of E and the area of that face; then add up the results form all the faces of the box Net flux due to a single point in a box is independent of the size of the box, it depends only on the charge inside

5. Gauss’ Law Whether there is a net outward or inward flux through the closed surface depends on the sign of the enclosed charge. Changes outside the surface do not give a net electric flux through the surface. The net electric flux is directly proportional to the net amount of charge enclosed within the surface but is otherwise independent of the size of the closed surface.

6. Gauss’ Law

7. Problem A positive point charge q = 3.0 microCoulombs is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge.

8. Solution E = kq/r 2 E = 9 x 10 9 (3 x 10 -6 )/(0.20 ) 2 E = 6.75 x 10 5 N/C Flux = ∫ E dA E is the same at every point and therefore constant Flux = EA = 6.75 x 10 5 N/C (4π) (0.20 m) 2 Flux = 3.4 x 10 5 Nm 2 /C

9. Rules for Using Gauss’ Law Select the surface (Gaussian surface) The surface may be an imaginary geometric surface. Usually, you can evaluate the integral in Gauss’ Law only if the Gaussian surface and the charge distribution have some symmetry property. If the charge distribution has cylindrical or spherical properties, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere.

10. Rules Continued Often, you can think of a closed Gaussian surface as being made up of several separate surfaces, such as the sides and ends of a cylinder. The integral over the entire closed surface is equal to the sum of the integrals over all the separate surfaces. If E is outward and has the same magnitude for each point on the surface then the integral equals EA and if inward then –EA.

11. Rules Continued If E is tangent to the surface at every point E perpendicular is zero then the integral is zero. If E = 0 at every point on the surface then the integral is zero.

12. Rules Continued Finally, in the integral E ┴ is always the perpendicular component of the total electric field at each point on the closed surface. In general, this field may be caused partly by charges within and partly by charges outside. Even when there is no charge within the surface, the field at points on the surface is not necessarily zero: in that case however the integral over the Gaussian surface is always zero. (Flux through the surface is zero.)