Gauss’s Law and Electric Flux Preview Gauss’s Law relates number of E field lines entering and leaving a surface to the net charge inside the surface. Consider imaginary spheres c centered on: a a.) +q (blue) b.) -q (red) c.) midpoint (yellow)

Number of lines exiting: a.) (blue) 24 b.) -q (red) -24 b c.) midpoint (yellow) 0

More on Gauss’s Law later. Electric Flux •Flux: Let’s quantify previous discussion about field- line “counting”

Define: electric flux ΦΕ through the closed surface S

r r “S” is surface •≡Φ AdE of the box E ∫ S

E dA

S Electric Flux r r •≡Φ AdE E ∫ S •What does this new quantity mean? • The integral is over a CLOSED SURFACE r r • Since• AdE is a SCALAR product, the electric flux is a SCALAR quantity r • The integration vectorAd is normal to the surface and points OUT r r of the surface. • AdE is interpreted as the component of E which is NORMAL to the SURFACE • Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface. • The sign matters!! Pay attention to the direction of the normal component as it penetrates the surface… is it “out of” or “into” the surface? • “Out of” is “+” “into” is “-” Electric Flux

Special case: uniform E perpendicular to plane surface A. E r r =•≡Φ dAEAdE cos θ == EAdAE dA E ∫∫∫ S S S s

constant =1

Another: uniform E at angle to plane surface A. E constants θ s dA r r =•≡Φ dAEAdE cos = cos θθ= EAdAE cos θ E ∫∫ ∫ S S S CLICKER QUESTION

A flat disk with radius r = 0.1m, is oriented with its normal unit vector relative the a constant E field as shown above. What is the electric flux through the disk area? 2 r r o A π ()1.0) E D π ()E (30cos1.02) ) 2 r o r o B π ()E (30cos1.0) ) E π ()E ()30sin1.02) 2 r o C π ()E ()30sin1.0) How to think about flux • We will be interested in net flux in or out of a closed w surface like this box z “A” is surface of the box y • This is the sum of the flux x surface area vector: through each side of the box r – consider each side separately A •= yareaA ˆ = 2 yw ˆ •Let E-field point in y-direction r r r r 2 –thenandE A are parallel and =• || wEAE

• Look at this from on top –down the z-axis A UI3PF3:

6) A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube.

a) Фbottom < 0

b) Фbottom = 0

c) Фbottom > 0 Exercise UI3A2

2A •Imagine a cube of side a positioned in a region of constant electric field as shown a •Which of the following statements

about the net electric flux ΦE through the surface of this cube is true? a

2 2 (a) ΦE = 0 (b) ΦE ∝ 2a (c) ΦE ∝ 6a

• Consider 2 spheres (of radius R and 2R) 2B drawn around a single charge as shown. – Which of the following statements R about the net electric flux through the 2 surfaces (Φ2R and ΦR) is true? 2R

(a) ΦR < Φ2R (b) ΦR = Φ2R (c) ΦR > Φ2R Exercise UI3A2

2A •Imagine a cube of side a positioned in a region of constant electric field as shown a •Which of the following statements

about the net electric flux ΦE through the surface of this cube is true? a

2 2 (a) ΦE = 0 (b) ΦE ∝ 2a (c) ΦE ∝ 6a

r r • The electric flux through the surface is defined by: ∫ •≡Φ AdE r r • ∫ • AdE is ZERO on the four sides that are parallel to the electric field. r r • ∫ • AdE on the bottom face is negative. (dA is out; E is in) r r • ∫ • AdE on the top face is positive. (dA is out; E is out) • Therefore, the total flux through the cube is: r r Φ≡˜ EdS • =Φ +Φ +Φ =00 −Ea22 + Ea = A sides bottom top Exercise UI3A2

• Consider 2 spheres (of radius R and 2R) 2B drawn around a single charge as shown. – Which of the following statements R about the net electric flux through the 2 surfaces (Φ2R and ΦR) is true? 2R (a) ΦR < Φ2R (b) ΦR = Φ2R (c) ΦR > Φ2R

• Look at the lines going out through each circle -- each circle has the same number of lines. • The electric field is different at the two surfaces, because E is proportional to 1 / r 2, but the surface areas are also different. The surface area of a sphere is proportional to r 2. r r • Since flux = ∫ • AdE , the r 2 and 1/r 2 terms will cancel, and the two S circles have the same flux! • There is an easier way. Gauss’ Law states the net flux is proportional to the NET enclosed charge. The NET charge is the SAME in both cases. • But, what is Gauss’ Law ??? --You’ll find out next lecture! c a

b

r r q AdE =Φ=• enclosed ∫ E ε 0 Fundamental Law of Electrostatics

• Coulomb’s Law Force between two point charges

OR

• Gauss’ Law Relationship between Electric Flux and charges Gauss’ Law

• Gauss’ Law (a FUNDAMENTAL LAW): The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

r r q AdE =Φ=• enclosed ∫ E ε 0

• How do we use this equation?? • The above equation is ALWAYS TRUE but it may not be easy to use. • It is very useful in finding E when the physical situation exhibits massive SYMMETRY. UI4PF4:

dA dA

1 2

2) A positive charge is contained inside a spherical shell.

How does the electric flux dФE through the surface element dA change when the charge is moved from position 1 to position 2?

a) dФE increases

b) dФE decreases

c) dФE doesn’t change dA dA

1 2

3) A positive charge is contained inside a spherical shell.

How does the flux ФE through the entire surface change when the charge is moved from position 1 to position 2?

a) ФE increases

b) ФE decreases

c) ФE doesn’t change dA dA

1 2

3) A positive charge is contained inside a spherical shell.

How does the flux ФE through the entire surface change when the charge is moved from position 1 to position 2?

a) ФE increases

b) ФE stays the same

c) ФE = 0