Thursday, January 28, 1999
Announcements:  The first exam will be on Tuesday, February 9 from 6:30 until 7:45. It will cover chapters 22, 23, and 24. It will be twenty multiple choice questions.  The locations are as follows:
102 Forum:  Sections 1, 2, 5, 9, 16
105 Forum:  Sections 3, 10, 12, 13, 15
108 Forum:  Sections 4, 7, 11, 14, 17
111 Forum:  Sections 6, 8, 18, 19, 20

Conflict Exams will be held on Tuesday, February 9 from 5:00 to 6:15 in 258 Willard and from 8:00 to 9:15 in 111 Boucke.

There will be a quiz on chapter 22 in the next recitation. The quiz will be multiple choice format.

Next week (February 1-5), there will be a Lab quiz on the concepts covered in labs one and two

Lecture notes:

• Charge per unit area = s, therefore the total amount go charge in a ring of radius r and width dr is

• The contribution to the electric field due to this ring is dE = (z(dq))/ (4pe_o(z² + r²)^3/2)
• dE = ((z)s (2pr dr))/ (4pe_o(z² + r²)^3/2)
• The contribution due to all rings with r ranges from 0 to R and equal the contribution due to the entire disk
• Hence, E =  I de ( Let I stand for the integral symbol)
•  E =  I  ((z)s (2pr dr))/ (4pe_o(z² + r²)^3/2)
•  E=  (zs)/(2e_o ) I(From 0 to R)(r dr))/ ((z² + r²)^3/2)
• E=  (zs)/(4e_o ) I(From 0 to R)(2r dr))/ ((z² + r²)^3/2)
• Assume that y = ((z² + r²)^3/2), then dy = 2rdr
• As r goes from 0 to R, y goes from z² to z² + R²
• So... E=  (zs)/(4e_o ) I( from z² to z² + R²) dy/ y^3/2
• E= (zs)/(4e_o )  [ y^-1/2/ (-1/2) ] (evaluated from  z² to z² + R²)
• E= (zs)/(4e_o ) [ {( z² + r²)^-1/2 / (-1/2)} - {( z²)^-1/2/ (-1/2)} ]  = (zs)/(4e_o ) [ (2/z) - (2/( z² + r²)^1/2)]
            =  (s)/(2e_o ) [ 1- (z/ ( z² + r²)^1/2)]
• If r goes to positive infinity, ( the disk is infinitely large), then E = (s)/(2e_o )
• What if R is finite, but point P is very close to the sheet, ie. z/r <<1
• E= (s)/(2e_o ) [ 1- (1/ ( z² + r²)^1/2)] = (s)/(2e_o )

• The force on a point charge in an electric field is F = qE
• What about a dipole?

• The dipole rotates in the clockwise direction.
• The net force equals zero. Why?
• The electric field and forces on the dipole are equal and opposite so the cancel.
• The forces exert a torque on the dipole.
• t = F (d/2) sin q + F (d/2) sin q
• q is the angle between the axis of the dipole and the electric field
• t = F d sin q and P ( the dipole moment) = qd
• Therefore, t = E P sin q  or t = E x P, ( the cross product)
• Use the right hand rule to find the direction
• The dipole tends to rotate clockwise, therefore t = -PE sin q

• It costs energy ( or it takes work) to rotate a dipole in an electric field.
• D U = - W = - I (from q_i to q_f) tdq
• W is the work done by the electric field on P
• U - U_i = - W = - I (from q_i to q_f)  - PE sinqdq
• The convention is that we choose U_i = 0 when q_i = 90^o
• Therefore, U = - W = I (from q to 90^o) PE sinqdq = -PE cos q
• The lowest potential energy is when the dipole moment, P , and the electric field, E, are parrallel.
• The highest potential energy occurs when P and E are 180^o from each other.
• In this situation, the dipole will flip. It will lose its potential energy to kinetic energy.

• Electric Flux (F )
• Consider an electric field that is uniform in both magnitude and direction. If E is perpendicular to the plane of the area A, then F = EA
• 
• The magnitude of the electric field is proportional to the number of field lines per unit area
• Graphically, the flux F, is represented by the number of lines penetrating the plane of area A.
• Units, E = f/q, Therefore E is expressed in Newtons/ Coulomb or N/C. F is in Nm²/C
• What if the electric field is not perpendicular to the area of interest?
• 
• What is the electric flux ( due to E) crossingA ? The electric field is perpendicular to the plane of A^l, the field lines that penetrate A^l also penetrate A.
• Therefore, F =  E A^l = E (L w cos q) = EA cos q
• We have F = EA cos q . This can be written as F =  E  dot A, where the direction of A is perpendicular to the surface of A. The product is a scalar and flux is a scalar quantity.