Tuesday, February 2, 1999
Announcements: Sample exams are available on the class web page . Review sessions for Tuesday's exam will be announced in class on Thursday. You will be allowed to use one sheet of 8 1/2" by 11" paper during the exam. Be sure to include any equations or constants you may need to use. Use of calculators on the exam is permitted and be sure to bring a pencil.

Lecture notes:
 

• What if the surface is curved and the electric field is not constant?
• 
• Then DF = E_i x DA_i   ;  F = SE_i x DA_i  = lim ( DA_i approaches 0) E_i x DA_i  = I E_i x DA_i
• I is  the integral symbol

• The net electric field flux, F , through a closed surface is proportional to the net charge enclosed by the surface.
• F = q/e_o
• Example: The net electric flux through each surface is the same.

• In the following situation, the net flux is zero.

• The charges on a neutral isolated conductor is separated by a nearby positively charged rod. What is the net flux through each of the five gaussian (closed) surfaces?
• The charges enclosed by s1, s2, and s3 are equal in magnitude.

• Flux through s1 = q / e_o
• Flux through s2  = - q / e_o
• Flux through s3 = q / e_o
• Flux through s4 = 0
• Flux through s5 = q / e_o
• Gauss' law is only interested in net change!

• Spherical Symmetry

• A charge q is enclosed at the center of a sphere;  I E_i dA_i =  q / e_o
• E is the same at all points on the spherical surface of radius r
• The electric field ,E, is parallel to the area, A.
• I E_i dA_i = E_i I  dA_i =  4p E r² =q / e_o
• Therefore, E = q/  4p r² e_o

• The charges are distributed on a spherical shell of radius R. What is the field E as a function of the distance from the center?
• Let us consider a gaussian surface at r < R, then this surface encloses no net charge. Therefore, E = 0 for r < R.
• Consider a gaussian surface r >R, then I E_i dA_i  =q / e_o
•   4p E r² =q / e_o
• E = q/  4p r² e_o

• The total charge of q is distributed uniformly in the spherical volume. What is the electric field as a function of distance from center?
• Consider a gaussian spherical surface with radius r, and r> R
• Then I E_i dA  =q / e_o
•  4p E r² =q / e_o
• E = q/  4p r² e_o
• Consider a gaussian spherical surface with radius r,  r < R.
• What is the amount of charge enclosed by this sphere? Assume it is q^l.
• Then, q¹/q = (4/3 p r³)/ (4/3 p R³)
• q¹ =  (r³/ R³) q
• By Gauss' Law, I E_i dA  =q¹ / e_o