Lecture 10EEE 3401 For a line charge Example 3-4: Infinitely long straight line with uniform charge density, find Solution. Because of the symmetry, we select a cylinder coordinate system. where Hence (3.36)

Lecture 10EEE 3402 In the previous equation the component of is zero. where we have used (3.40)

Lecture 10EEE : Gauss’ Law and Applications The total outward flux of the E-field over a closed surface in free space is equal to the total charge enclosed divided by  o. Gauss’ law can be used to find E-field quickly for simple and symmetric cases. (3-41)

Lecture 10EEE 3404 Example 3-5: Uniformly charged line Solution: No contribution from the top and bottom faces because has no components. From Gauss’ law (3.41) Therefore

Lecture 10EEE 3405 Example 3-6: Uniformly charged plate  s Solution. Make a flat box with a rectangular shape Because of symmetry, no horizontal E-field is expected. The top surface The bottom surface Apply Gauss’ law Conclusion (3.42)

Lecture 10EEE 3406 Example 3-7: E-field caused by a spherical cloud with constant charge density of  v for Solution. (1). Make a hypothetical Gaussian surface of sphere r<b. Total outward E-flux Total charge enclosed by the Gaussian surface Apply Gauss’ law

Lecture 10EEE 3407 (2). Make a Gaussian surface of sphere r>b. The E-flux has the same expression as before. Total charge enclosed: Apply Gauss’ law 0b Fig r ErEr

Lecture 10EEE 3408 This example is very similar to Earth gravity problem. It helps us to visualize the E-fields. However, the charge density can be given in different functions. In contrast, the mass density of the Earth cannot arbitrary functions.