Download Physics 2220 – Module 02 Homework

Physics 2220 – Module 02 Homework
01.
The cube in the figure contains negative charge. The electric field strengths in N/C are constant over
each face of the cube. Does the missing electric field vector on the front face point in or out? What
strength must this field exceed?
Recall:
• If the net flux flows out of a closed surface, the net enclosed charge is positive.
• If the net flux flows into a closed surface, the net enclosed charge is negative.
The electric field over each face of the cube is constant, therefore the electric flux can be calculated
using the following:
ΦE = ⃗
E⋅ ⃗
A = EAcos (θ)
Sum up all flux going out of the cube:
Φout = (20 N/C + 10 N/C + 10 N/C) A cos (0 o ) = 40A N m 2 / C
Sum up all flux going into the cube:
Φin = (20 N/C + 15 N/C) A cos (180o ) = −35A N m2 / C
Since the cube contains negative charge:
∣Φin∣ > ∣Φout∣
The flux on the other face of the cube must be going into the cube and be greater than:
5A N m 2 / C
02.
What is the electric flux through the surface shown in the figure?
For the constant electric field, the electric flux can be found using:
⃗⋅ ⃗
ΦE = E
A = EAcos (θ)
To find the angle between the electric field vector and the normal vector to the area, put the tails of the
two vectors together. From the diagram, this will make the angle 120 degrees.
Φ E = (180 N/C) (0.15 m) (0.15 m) cos (120o ) = −2.0 N m 2 / C
03.
How much charge is enclosed within the octahedron which has a net electric flux of -1000 N m 2/C?
Gauss's Law shows us the relation between flux and the enclosed charge
q
ΦE = ∮ ⃗
E ⋅d ⃗
A = ϵin
0
Since the electric flux is known, and it can be the flux through any closed surface, the enclosed charge
can be determined.
q
Φ E = ϵin
0
q in = ΦE ϵ0
q in = (−1000 N m2 / C) (8.85 × 10−12 C 2 / N m 2) = −8.85 × 10− 9 C = −8.85 nC
04.
The electric field strength just above one face of a copper penny is 2000 N/C. What is the surface
charge density on this face of the penny?
Since the penny is a conductor, the charge is located on the surface. Assuming a location very near the
surface of the conductor, and therefore it can be treated like an infinite surface, the electric field will be:
Solve for the surface charge density:
E = ϵσ
0
σ = E ϵ0 = (2000 N/C) (8.85 × 10 −12 C2 / N m2) = 1.77 × 10 −8 C/m 2
05.
The conducting box in the figure has been given an excess negative charge. The surface density of
excess electrons at the center of the top surface is 5.0 ⨯ 1010 electrons / m2. What are the electric field
strengths E1 to E3 at points 1 to 3?
Recall that for a conductor, all the charge will be on the surface.
Point 3: All the charge resides on the surface. Near the surface, the infinite surface
model can be used. Note a conversion from the number of electrons per
meter to Coulombs per meter must be done.
(5.0 × 10 10 electrons / m 2) (1.60 × 10−19 C / electron)
E = ϵσ =
= 904 N/C
0
8.85 × 10−12 C2 / N m2
Point 2: E2 is inside the conductor and will be zero. Consider a Gaussian surface
surrounding any portion inside the conductor. There will be no charge
enclosed, therefore E2 is zero.
Point 1: E3 is inside the conductor and will be zero. Consider a Gaussian surface
surrounding any location inside the hollow portion of the conductor . There
will be no charge enclosed, therefore E3 is zero.
06.
A spherically symmetric charge distribution produces the electric field E = (200/r)r N/C
200
N
⃗
E=
r̂
r
C
( )
where r is in meters.
(a)
What is the electric field strength at r = 10 cm?
Plug the radius into the given equation:
⃗
E=
(b)
( 0.10200m ) ̂r NC = 2000 ̂r NC
What is the electric flux through a 20-cm-diameter spherical surface that is concentric with the
charge distribution?
The electric flux through a closed surface will be the same regardless of the shape or size of the
surface as long as it encloses all the charge. Therefore we can calculate the electric flux at r =
10 cm and it will be the same as r = 20 cm
Start with Gauss's Law:
q
⃗ ⋅d ⃗
ΦE = ∮ E
A = ϵin
0
Since the charge distribution is spherically symmetric, and the surface is spherical, the direction
of the electric field and the area element will always be parallel. Thus the angle between them is
zero. The integral then becomes:
Φ E = ∮ EdA
The magnitude of the electric field is constant at all points along the surface area, so it can be
taken out of the integral, which will just leave an integral over the surface
ΦE = E ∮ dA = E (4 π r 2)
Φ E = (2000 N/C) (4 π ) (0.10 m)2 = 250 N m 2 / C
(c)
How much charge is inside the 20-cm-diameter spherical surface?
Once again use Gauss's Law since the electric flux through a closed surface is independent of
distance from the charge.
q
⃗ ⋅d ⃗
ΦE = ∮ E
A = ϵin
0
q
ΦE = ϵin
0
q in = (250 N m 2 / C) (8.85 × 10−12 C2 / N m2 ) = 2.2 × 10 −9 C = 2.2 nC
07.
An infinite slab of charge of thickness 2z0 lies in the xy-plane between z = -z0 and z = +z0. The volume
charge density is a constant.
(a)
Use Gauss's Law to find an expression for the electric field strength inside the slab.
Use a Gaussian pill box inside the slab to determine the electric field strength inside. In this
case:
−z ≤ z ≤ z
0
0
Use Gauss's Law and add up the electric flux from each portion of the surface.
⃗ ⋅d ⃗
ΦE = ∮ E
A=
∫
⃗
E ⋅d ⃗
A+
bottom
∫ ⃗E ⋅d
⃗
A+
top
∫
curved
q
⃗
E ⋅d ⃗
A = ϵin
0
The electric field will always be parallel to the sides of the curved surface, so with respect to the
area vector direction, the curved integral will be zero since the cosine of 90 degrees is zero.
The pillbox will only contain the charge it encloses. The enclosed charge will need to be put in
terms of the charge density. Note from the image above, the pillbox is centered at z = 0 and its
length is 2z. This means the magnitude of the electric field will be the same on both the top and
bottom surfaces of area A.
q in = ρ V = ρ (2zA)
The electric field is constant on each surface and always parallel to the area vector. It can be
removed from the integral which will just leave adding up all the surface area in the integral.
∫
EdA +
bottom
E
∫
∫ E dA +
top
dA + E
bottom
q
0 = ϵin
0
ρ
∫ dA = 2zA
ϵ0
top
2zA ρ
ϵ0
2zA ρ
2EA = ϵ
0
zρ
E= ϵ
0
EA + EA =
(b)
Find an expression for the electric field strength above the slab.
Now all the possible charge is contained in the pillbox up to z0, so the enclosed charge
becomes:
q in = ρ V = ρ (2z 0 A)
Same reasoning in part (a) applied here.
∫
EdA +
bottom
E
∫
∫ E dA +
top
dA + E
bottom
q
0 = ϵin
0
∫ dA =
top
2z0 A ρ
ϵ0
2z 0 A ρ
ϵ0
2z A ρ
2EA = 0ϵ
0
z0 ρ
E= ϵ
0
EA + EA =
(c)
Draw a graph of E from z = 0 to z = 3 z0.
Drawing should look something like the following: