Electric Flux PH 203 Professor Lee Carkner Lecture 4

HW 3, #1, HRW7, P 9  q 2 = q 1 at x 2 = 70 cm and q 1 = 2.1  C at x 1 = 20 cm, where is E = 0? +q 1 -q 2 q 2 > q 1 r 2 < r 1 so E 2 > E 1 E 2 and E 1 both point right

HW 3, #1, HRW7, P 9  q 2 = q 1 at x 2 = 70 cm and q 1 = 2.1  C at x 1 = 20 cm, where is E = 0?  E 2 = E 1 at position x  (1/4  0 )((q 1 /(x-x 1 ) 2 )=(1/4  0 )((q 2 /(x-x 2 ) 2 )  q 2 /q 1 = (x-x 2 ) 2 /(x-x 1 ) 2  (q 2 /q 1 )½ =(x-x 2 ) 2 /(x-x 1 ) 2  ±2.12 = (x-70)/(x-20)  x-70 = ±2.12(x-20)  x = 36cm and -24.6cm  Only cm will work

Flux   If many field lines pass through a given area the forces there are strong  flux = EA  This is only true if the field in perpendicular to the area  E and A are actually vectors with the direction of A defined as the line normal to the surface

Finding Flux  Flux = EA cos   The maximum flux occurs when the field is directly perpendicular to the surface:   = 0, Flux = EA    = 90, Flux = 0   For a closed area:  If the field lines leave, the sign is positive  If the field lines enter, the sign is negative

Defining the Flux   For a real surface, we divide it into a large number of very small surfaces of area dA   = ∫ E dA  Which is a closed path integral over the entire surface  The units of flux are (N m 2 / C)

Types of Fields   Generated by an unknown agency and occupying some specific space   Real fields are generated by charges   What is the flux associated with real charges?

Flux of a Single Charge   For a sphere of radius r with a charge q at its center, the field is perpendicular to the surface everywhere (cos  = 1)    = ∫ E dA = EA = (1/4  0 )(q/r 2 )(4  r 2 ) = q/  0  Thus, for a sphere around a point charge:  = q/  0

Gauss’s Law   We can use Gauss’s Law, which states:   = q/  0 (for any surface)  For a positive charge the flux is positive (leave the surface) for a negative charge the flux is negative (enter the surface)

Using Gauss’s Law  We can use Gauss’s Law to relate the field to the charge:  0 ∫ E dA = q   Note that if the net flux is outward, the net charge is positive and if the net flux is inward the net charge is negative

Gauss Notes  The shape of the surface does not matter   Charges outside the surface do not matter  Other stuff does not matter  The flux just depends on the net enclosed charge   = (q 1 + q 2 + q 3 ) /  0

Next Time  Read  Problems: Ch 23, P: 6, 9, 22, 31, 44

What direction will the dipoles rotate in? A)1 and 2 clockwise, 3 and 4 counterclockwise B)1 and 2 counterclockwise, 3 and 4 clockwise C)1 and 3 clockwise, 2 and 4 counterclockwise D)1 and 3 counterclockwise, 2 and 4 clockwise E)All clockwise

The field very far from a dipole : The field very far from a charged ring A)zero : zero B)like a point charge : like a point charge C)zero : like a point charge D)like a point charge : zero E)Neither reduce to a simple expression

If you turn a dipole from orientation 1 to orientation 2, the work you do is A)positive B)negative C)zero D)the sign depends on the magnitude of q E)the sign depends on the magnitude of E

If you turn a dipole from orientation 1 to orientation 4, how does the net work compare to the previous situation? A)it is greater B)it is less C)it is the same D)it depends on the magnitude of q E)it depends on the magnitude of E