Download Sample Quiz 1 - U of M Physics

PHYS 1202W
– M ULTIPLE C HOICE Q UESTIONS –
Q UIZ # 1
Answer the following multiple choice questions on the bubble sheet. Choose the best answer, 5 pts each.
MC1 An uncharged metal sphere will
(A) be repelled by a charged metal surface.
(B) feel no force when brought close to a charged metal surface.
(C) be attracted by a charged metal surface.
(D) not enough information
MC2 Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger
than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and
finally removed. As a result, the electrostatic force between A and B, which was originally F, becomes:
(A) F/2
(B) F/4
(C) 3F/8
(D) F/16
(E) 0
MC3 For a gaussian surface through which the net flux is zero, the following four statements could be true. Which
statement must be true?
(A) No charges are inside the surface.
(B) The net charge inside the surface is zero.
(C) The electric field is the same everywhere on the surface.
(D) No electric field lines leave the surface.
(E) There is precisely one dipole enclosed by the surface.
MC4 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 have any net charge. The electric field for
rb1 < r < rb2 is
(A)
(B)
(C)
(D)
(E)
kQ
r̂
r2
kQ
r̂
r2
2kQ
r̂
r2
2kQ
r̂
r2
0
MC5 Two large parallel plates carry positive charge of equal magnitude that is distributed uniformly over their inner
surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
(A) 1, 2, 3, 4, 5
(B) 5, 4, 3, 2, 1
(C) 1, 4, and 5 tie, then 2 and 3 tie
(D) 2 and 3 tie, then 1 and 4 tie, then 5
(E) 2 and 3 tie, then 1, 4, and 5 tie
Physics 1202W.200
February 6, 2015
12:20 PM – 1:10 PM
PHYS 1202W
– M ULTIPLE C HOICE Q UESTIONS –
Q UIZ # 1
Answer the following multiple choice questions on the bubble sheet. Choose the best answer, 5 pts each.
MC1 An uncharged metal sphere will
(A) be repelled by a charged metal surface.
(B) feel no force when brought close to a charged metal surface.
(C) be attracted by a charged metal surface.
(D) not enough information
MC2 Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger
than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and
finally removed. As a result, the electrostatic force between A and B, which was originally F, becomes:
(A) F/2
(B) F/4
(C) 3F/8
(D) F/16
(E) 0
MC3 For a gaussian surface through which the net flux is zero, the following four statements could be true. Which
statement must be true?
(A) No charges are inside the surface.
(B) The net charge inside the surface is zero.
(C) The electric field is the same everywhere on the surface.
(D) No electric field lines leave the surface.
(E) There is precisely one dipole enclosed by the surface.
MC4 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 have any net charge. The electric field for
rb1 < r < rb2 is
(A)
(B)
(C)
(D)
(E)
kQ
r̂
r2
kQ
r̂
r2
2kQ
r̂
r2
2kQ
r̂
r2
0
MC5 Two large parallel plates carry positive charge of equal magnitude that is distributed uniformly over their inner
surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
(A) 1, 2, 3, 4, 5
(B) 5, 4, 3, 2, 1
(C) 1, 4, and 5 tie, then 2 and 3 tie
(D) 2 and 3 tie, then 1 and 4 tie, then 5
(E) 2 and 3 tie, then 1, 4, and 5 tie
Physics 1202W.200
February 6, 2015
12:20 PM – 1:10 PM
The Electric Field
26-25
(b) The field strength at the point z = R/ 2 is
Q
( R/ 2)
2
Q
=
4πε 0 ⎡( R/ 2) 2 + R 2 ⎤ 3/2 3 3 4πε 0 R 2
⎣
⎦
( E z ) max =
26.44. Model: Assume that the semicircular rod is thin and that the charge lies along the semicircle of radius R.
Visualize:
The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of charge
∆q and arc length ∆s. Segment i creates a small electric field Ei at the origin. The line from the origin to segment i
makes an angle θ with the x-axis.
Solve: Because every segment i at an angle θ above the axis is matched by segment j at angle θ below the axis, the
y-components of the electric fields will cancel when the field is summed over all segments. This leads to a net field
pointing to the right with
E x = ∑ ( Ei ) x =∑ Ei cosθi
i
E y = 0 N/C
i
Note that angle θi depends on the location of segment i. Now all segments are at the same distance ri = R from the
origin, so
∆q
Ei =
4πε 0 ri2
=
∆q
4πε 0 R 2
The linear charge density on the rod is λ = Q/L, where L is the rod’s length. This allows us to relate charge ∆q to the
arc length ∆s through
∆q = λ ∆s = (Q/L)∆s
Thus, the net field at the origin is
Ex = ∑
i
(Q/L)∆s
4πε 0 R 2
cosθi =
Q
4πε 0 LR 2
∑ cosθi ∆s
i
The sum is over all the segments on the rim of a semicircle, so it will be easier to use polar coordinates and integrate
over θ rather than do a two-dimensional integral in x and y. We note that the arc length ∆s is related to the small angle
∆θ by ∆s = R∆θ , so
Ex =
Q
4πε 0 LR
∑ cosθi ∆θ
i
With ∆θ → dθ, the sum becomes an integral over all angles forming the rod. θ varies from ∆θ = −π /2 to θ = +π /2.
So we finally arrive at
Ex =
Q
π /2
Q
π /2
2Q
cosθ dθ =
sinθ −π /2 =
4πε 0 LR Ñ−π /2
4πε 0 LR
4πε 0 LR
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