Electric Flux

PH 203

Professor Lee Carkner

Lecture 4

HW 3, #1, HRW7, P 9 q2= - 4.50 q1 at x2 = 70 cm and q1 = 2.1

10-8 C at x1 = 20 cm, where is E = 0?

+q1 -q2

q2 > q1

r2 < r1

so E2 > E1

E2 and E1both point right

HW 3, #1, HRW7, P 9 q2= - 4.50 q1 at x2 = 70 cm and q1 = 2.1 10-

8 C at x1 = 20 cm, where is E = 0? E2 = E1 at position x

(1/40)((q1/(x-x1)2)=(1/40)((q2/(x-x2)2)

q2/q1 = (x-x2)2/(x-x1)2

(q2/q1)½ =(x-x2)2/(x-x1)2

±2.12 = (x-70)/(x-20) x-70 = ±2.12(x-20) x = 36cm and -24.6cm

Only -24.6 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 m2 / 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 = (1/40)(q/r2)(4r2) = 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 = (q1 + q2 + q3) / 0

Next Time

Read 23.5-23.9 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) positiveB) negativeC) zeroD) 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