Applications of Gauss’ Law to Charged Insulators

1. Applications of Gauss’ Law to Charged Insulators AP Physics C Montwood High School R. Casao

2. Gauss’ law is useful when there is a high degree of symmetry in the charge distribution. • The surface should always be chosen so that it has the same symmetry as that of the charge distribution. • Electric Field due to a Point Charge: For a point charge, choose a spherical gaussian surface of radius R that is centered on the point charge.

3. Electric Field Due to a Point Charge • The electric field of a positive point charge is directed radially outward by symmetry and is normal to the surface at every point. • E is parallel to the area vector at each point. The angle between the area vector and the electric field vector is zero. • E is constant everywhere on the surface of the gaussian sphere and can be pulled out in front of the integral sign. • Applying Gauss’ Law: • Integrating over the surface area of the sphere gives us the area, 4·p·r2.

4. Electric Field Due to a Point Charge • The equation for the electric field of a point charge is: • The flux equation becomes: • Remember that Q refers to the entire charge located inside the sphere (Qin or qin).

5. Spherically Symmetric Charge Distribution • Consider an insulating sphere of radius R having a uniform charge density r and a total positive charge Q. • Charge will remain in its position when on an insulating sphere, therefore, the charge on the insulating sphere will be uniformly distributed throughout the volume of the sphere (spherically symmetric). • For determining the electric field within the insulating sphere of radius R, (for situations in which r < R):

6. Apply a spherical gaussian surface of radius r, centered on the center of the insulating sphere. The amount of charge enclosed within the gaussian sphere will be less than the total charge Q. The symmetric charge distribution allows for a proportional relationship to be set up between the total charge Q, total volume V, enclosed charge qin, and the volume of the gaussian sphere. Spherically Symmetric Charge Distribution

7. Spherically Symmetric Charge Distribution • Volume of a sphere: • Relationship: • Solve for qin: • The electric field is constant in magnitude everywhere on the spherical Gaussian surface and is parallel to the area vector at each point. The angle between the area vector and the electric field vector is zero.

8. Spherically Symmetric Charge Distribution • The area vector is parallel to the electric field vector at the surface of the spherical Gaussian surface. • Applying Gauss’ law:

9. Spherically Symmetric Charge Distribution • As r increases to R, the magnitude of the electric field will increase as more charge is included within the volume of the spherical Gaussian surface. • The maximum electric field will be obtained when r = R. • For a spherical Gaussian surface of radius r  R, we can consider the charged insulating sphere as a point charge.

10. Spherically Symmetric Charge Distribution • Conclusion: the electric field E inside a uniformly charged solid insulating sphere varies linearly with the radius until reaching the surface. The electric field E outside the sphere varies inversely with 1/r2. The diagram on the next slide illustrates this relationship.

11. Spherically Symmetric Charge Distribution

12. Consider a thin spherical shell of radius R with a total charge Q distributed uniformly over its surface. For points outside the shell, r  R: the spherical Gaussian surface will enclose the thin spherical shell, so qin = Q. Thin Spherical Shell

13. Thin Spherical Shell • The area vector is parallel to the electric field vector at the surface of the spherical Gaussian surface. The angle between the area vector and the electric field vector is zero.

14. Thin Spherical Shell • For points inside the thin spherical shell, r < R: • The charge lies on the surface of the thin spherical shell. • No charge is found within the shell, so qin = 0 and the electric field and flux within the thin spherical shell is also 0.

15. Cylindrically Symmetric Charge Distribution • Consider a uniformly charged cylindrical rod of length L.

16. Cylindrically Symmetric Charge Distribution

17. Cylindrically Symmetric Charge Distribution • Instead of the shortened Gaussian cylinder indicated in the diagrams, I use a Gaussian cylinder that is the same length as the charged cylindrical rod. • The electric field lines are perpendicular to the charged cylindrical rod and are directed outward in all directions. • The electric field is perpendicular to the Gaussian cylinder and is parallel to the area vector at those points where it passes through the Gaussian cylinder. The angle between the area vector and the electric field vector is zero.

18. Cylindrically Symmetric Charge Distribution • The equation for the area of a cylinder is: A = 2··r·L • Applying Gauss’ law:

19. Nonconducting Plane Sheet of Charge • Consider a plane sheet of charge with a uniform charge density s.

20. Nonconducting Plane Sheet of Charge • The diagrams show a small cylinder through the plane sheet of charge; I use a rectangle that encloses the entire sheet of charge. The faces of the rectangle will have the same area as the plane sheet of charge. • The electric field E is perpendicular to the plane sheet of charge and the area vector for the rectangular face on either side of the plane sheet of charge is in the same direction as the electric field vector. • The angle between the area vector and the electric field vector is 0.

21. Nonconducting Plane Sheet of Charge • The electric field lines near the top of the plane of charge will run parallel to the sides of the rectangle and will not result in any flux through the rectangular surface. • There are two surfaces of the rectangle through which the electric field lines pass. • Applying Gauss’ law:

22. Nonconducting Plane Sheet of Charge • Textbooks will have:

23. Nonconducting Plane Sheet of Charge • Here is the conversion:

24. Two Parallel Nonconducting Sheets • The situation is different if you bring two nonconducting sheets of charge close to each other. • In this case, the charges cannot move, so there is no shielding, but now we can use the principle of superposition. • The electric field on the left due to the positively charged sheet is canceled by the electric field on the left of the negatively charged sheet, so the field there is zero. • The electric field on the right due to the negatively charged sheet is canceled by the electric field on the right of the positively charged sheet. • The electric field in between the two plates is equal to 2·E.