Lab: AP Review Sheets Chapter 24: Gauss’s Law AP Physics By Katherine Berry Background: This chapter focuses on Gauss’s Law and Electric Flux. Summary: Gauss’s Law helps understand and verify properties of conductors in electrostatic equilibrium. Electric flux is related to the number of field lines penetrating a given surface. EQUATIONS: φ E = EA⊥ = EA cos (θ ) φE =
∫ E dA φE =
qin
ε
∫ E dA =
qin
ε
Gauss’s Law: The net electric flux through any closed Gaussian surface is equal to the net charge inside the surface divided by ε . Electric Flux: It is proportional to the number of electric field lines. ELECTROSTATIC EQUILLIBRIUM: When there is no net charge on a conductor. It will have these properties: 1. The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow. 2. If the conductor is isolated and carries a charge, the charge resides on its surface. 3. The electric field at a point just outside a charged conductor is perpendicular σ
to the surface of the conductor and has a magnitude where σ is the ε
surface charge density at that point. 4. On an irregularly shaped conductor, the surface charge density is greater at locations where the radius curvature of the surface is smallest. PROBLEMS: 1. A -­‐2Q D C +Q -­‐Q B Determine the electric flux in each sphere Sphere A q
φ E = in
ε
Count the −2Q + Q
charges and add =
ε
−Q
=
ε
Sphere B q
φ E = in
ε
Count the charges −Q + Q
and add =
ε
=0
Sphere C qin
ε
Count the charges −2Q + Q − Q
=
and add ε
−2Q
=
ε
Sphere D q
φ E = in
No charges in the sphere ε
0
=
ε
=0
2. Lead-­‐208 atom A lead-­‐208’s nuclei has a volume 208 times that of a proton with a radius of 1.2x 10 −15 m. Determine the electric field. q
∫ E dA = εin 82
E ( 4π r 2 ) =
ε
Substitute k in 82
82
E=
= k 2 ε 4π r 2
r
4 3
Vn = π r 3
Volume of !4
$
Vp = 208 # π r 3 & proton needs to "3
%
be multiplied y !4
$
4 2
208 to get π r = 208 # π r 3 &
"3
%
3
nuclei 3
!4
$
4 2
π r = 208 # (1.2x10 −15 ) &
"3
%
3
r = 7.1x10 −15 φE =
E=k
82
r2
E = (9e9)
Plug in 82 (1.6x10 −15 )
7.1x10 −15
E = 2.3x10 21
3. A wire is surrounded by a hollow metal cylinder. The wire has a charge per unit length λ , while the cylinder has a net charge of 2 λ . a. Charge per unit length on inner surface of cylinder. Electric field inside any metallic structure is 0 so there has to be -­‐ λ on the inner surface b. Charge per unit length on outer surface of the cylinder. λnet = λ + 2 λ = 3λ c. Electric field outside the cylinder a distance r. q
∫ E dA = εin 3λ L
E ( 2π rL ) =
ε 3λ
E=
ε 2π r