Download 1402-Week Two-2017.pptx

Physics 1402.001 - Week Two
•  Electric Fields – Chapter 22
•  Continuous Charge Distributions
•  Gauss’ Law
Clicker Question No. One
Three charges +q, +Q, and –Q are placed at the corners of an
equilateral triangle as shown. The net force on charge +q due to
the other two charges is
A. up.
B. down.
C. along a diagonal.
D. to the left.
E. to the right.
Clicker Question No. Two
The force between two very small charged bodies is
found to be F. If the distance between them is tripled
without altering their charges, the force between them
becomes
A.  9F
B.  3F
C.  F/3
D.  F/9
E.  1/F 3
Clicker Question No. One
Three charges +q, +Q, and –Q are placed at the corners of an
equilateral triangle as shown. The net force on charge +q due to
the other two charges is
A. up.
B. down.
C. along a diagonal.
D. to the left.
E. to the right.
Clicker Question No. Two
The force between two very small charged bodies is
found to be F. If the distance between them is tripled
without altering their charges, the force between them
becomes
A.  9F
B.  3F
C.  F/3
D.  F/9
E.  1/F 3
Electric Field Lines for
Two Positive Charges
Enet
Enet
Ex
Ey
Electric Field Lines for
Two Positive Charges
Electric Fields from Continuous
Charge Distributions
E1 = kq/r1p2
E2 = kq/r2p2
R1p = sinθ1/yp
R2p = sinθ2/yp
E1 = kq sin2θ1/yp2
E2 = kq sin2θ2/yp2
Electric Fields from Continuous
Charge Distributions
Electric Fields from Continuous
Charge Distributions
Ex = kλ [1/r2 – 1/r1]
Electric Fields from Continuous
Charge Distributions
• 
• 
• 
• 
Ez = kλ [1/r2 – 1/r1]
r1 = z + L/2
r2 = z – L/2
Ez = kλ [1/(z – L/2) –
1/(z + L/2)]
•  Ez = kλL [1/(z2 – (L/2)2] and when z >>L/2
•  Ez ~ kQ/z2
Electric Fields from Continuous
Charge Distributions
Electric Fields from Continuous
Charge Distributions
Clicker Question No. Three
A conducting circular disk has a uniform positive surface charge
density. Which of the following diagrams best represents the electric
field lines from the disk? (The disk is drawn as a cross–section.)
A. 1
B. 2
C. 3
D. 4
E. None of the diagrams.
Clicker Question No. Three
A conducting circular disk has a uniform positive surface charge
density. Which of the following diagrams best represents the electric
field lines from the disk? (The disk is drawn as a cross–section.)
A. 1
B. 2
C. 3
D. 4
E. None of the diagrams.
Electric Fields from Continuous
Charge Distributions
Electric Fields for Infinite
Line of Charge
Ez = (kλ/R) (sinθ2 – sinθ1)
Ez = (kλ/R) (sinθ2 – sinθ1)
ER = (kλ/R) (cosθ2 – cosθ1)
ER = - (kλ/R) (cosθ2 – cosθ1)
θ1 = 0
θ2 = π
Ez = 0
ER = 2 (kλ/R)
Flux of Electric Field Through a Surface
Φ=E•A
Flux of Electric Field Through a Surface
Φ=E•A
Φ2 = Φ1
Flux of Electric Field Through a Surface
•  The Electric Flux Φ is defined as the number
of Electric Field E lines passing through an
Area A
•  That is Φ = E • A
•  If the area A is perpendicular to the E field,
then Φ = E x A
•  If the area A is at an angle θ to the E field, then
•  Φ = E • A = E A cosθ
Flux Through a Closed Surface
•  For a continuously closed surface that defines
an ‘inside’ and an ‘outside’ the total flux is
defined by
•  Φ = ∫S E • dA = ∫S E • n dA
•  Any charge that resides outside surface S will
therefore contribute Φ = 0 to calculation of
total flux
Flux of Electric Field Through a Surface
Φ = ∫E • dA
Only charges inside bounded surface
contribute a net Electric flux passing
through surface boundary
Electric Field Lines – Electric Dipole
Clicker Question No. Four
Thefigureshowsasurfaceenclosingthechargesqand
–q.Thenetfluxthroughthesurfacesurroundingthe
twochargesis
A. q/ε0
B. 2q/ε0
C. –q/ε0
D. zero
E. –2q/ε0
Clicker Question No. Four
Thefigureshowsasurfaceenclosingthechargesqand
–q.Thenetfluxthroughthesurfacesurroundingthe
twochargesis
A. q/ε0
B. 2q/ε0
C. –q/ε0
D. zero
E. –2q/ε0
Gauss’ Law – Point Charge
Φ = E • A = Qnet/εo
Φ = E(4πR2) = +Q/εo
E = Q/[(4πεo) R2]
E = kQ/R2
Gauss’ Law
The net Electric
Field passing
through any surface
fully enclosing
electric charges is
proportional to the
net charge residing
inside the surface.
Φ = E • A = Qnet/εo
Gauss’ Law
Electric Field Lines for
Two Positive Charges
Clicker Question No. Five
The figure shows a surface enclosing the
charges 2q and –q. The net flux through the
surface surrounding the two charges is
Clicker Question No. Five
Gauss’ Law
The figure shows a surface enclosing the
charges 2q and –q. The net flux through the
surface surrounding the two charges is
Φ = E • A = Qnet/εo
Φ  = (+2q – q)/εo = q/εo
Clicker Question No. Six
The figure shows a surface, S, with two
charges q and –2q. The net flux through the
surface is
Clicker Question No. Six
The figure shows a surface, S, with two
charges q and –2q. The net flux through the
surface is
Electric Fields from Line of Charge
Gauss’ Law – Infinite Line of Charge
Ez = (kλ/R) (sinθ2 – sinθ1)
ER = - (kλ/R) (cosθ2 – cosθ1)
θ1 = 0
θ2 = π
Ez = 0
ER = 2 (kλ/R)
ER = 2/4πεo (λ/R)
ER = 1/2πεo (λ/R)
Gauss’ Law – Infinite Line of Charge
Gauss’ Law – Distributed Charge
ER (2π R L) = λ L / εo
ER = λ L / (2π R L εo ) = λ / (2π εo R )
ER = 2 k λ / R
Gauss’ Law – Uniformly Charged Slab
Gauss’ Law – Uniformly Charged Slab
Symmetry of Gaussian Surfaces
Gauss’ Law – Thin Shell of Charge
•  Line symmetry – choose a Cylindrical Surface
•  Plane symmetry – choose a Cubic or
Cylindrical surface
•  Point symmetry – choose a Spherical surface
Gauss’ Law – Thin Shell of Charge
Gauss’ Law – Uniformly Charged Sphere
Gauss’ Law – Uniformly Charged Sphere
Clicker Question No. Seven
A solid conducting sphere of radius ra is placed concentrically inside
a conducting spherical shell of inner radius rb1 and outer radius rb2.
The inner sphere carries a charge Q while the outer sphere does not
carry any net charge. The electric field for ra < r < rb1 is
Clicker Question No. Seven
A solid conducting sphere of radius ra is placed concentrically inside
a conducting spherical shell of inner radius rb1 and outer radius rb2.
The inner sphere carries a charge Q while the outer sphere does not
carry any net charge. The electric field for ra < r < rb1 is
Clicker Question No. Eight
A solid conducting sphere of radius ra is placed concentrically inside
a conducting spherical shell of inner radius rb1 and outer radius rb2.
The inner sphere carries a charge Q while the outer sphere does not
carry any net charge. The electric field for rb1 < r < rb2 is
Clicker Question No. Eight
Clicker Question No. Nine
A solid conducting sphere of radius ra is placed concentrically inside
a conducting spherical shell of inner radius rb1 and outer radius rb2.
The inner sphere carries a charge Q while the outer sphere does not
carry any net charge. The electric field for rb1 < r < rb2 is
A solid conducting sphere of radius ra is placed concentrically
inside a conducting spherical shell of inner radius rb1 and outer
radius rb2. The inner sphere carries a charge Q while the outer
sphere does not carry any net charge. The electric field for r > rb2 is
Clicker Question No. Nine
A solid conducting sphere of radius ra is placed concentrically
inside a conducting spherical shell of inner radius rb1 and outer
radius rb2. The inner sphere carries a charge Q while the outer
sphere does not carry any net charge. The electric field for r > rb1 is