# kapoor exam 1 ```hernandez (uh286) – oldmidterm 01 – Turner – (59070)
1
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at (43 cm, 0.0). The Coulomb constant is
9 &times; 109 N m2 /C2 .
Calculate the magnitude of the force on the
charge at the origin.
001 10.0 points
Two spheres, fastened to “pucks”, are riding on a frictionless airtrack. Sphere “1” is
charged with 3 nC, and sphere “2” is charged
with 15 nC. Both objects have the same mass.
1 nC is equal to 1 &times; 10−9 C.
As they repel,
Explanation:
1. they have the same magnitude of acceleration. correct
2. sphere “2” accelerates 25 times as fast as
sphere “1”.
3. sphere “1” accelerates 5 times as fast as
sphere “2”.
4. they do not accelerate at all, but rather
separate at constant velocity.
5. sphere “2” accelerates 5 times as fast as
sphere “1”.
6. sphere “1” accelerates 25 times as fast as
sphere “2”.
Let : q = 7 &micro;C = 7 &times; 10−6 C ,
(x1 , y1 ) = (0, 25 cm) = (0, 0.25 m) ,
(x2 , y2 ) = (43 cm, 0) = (0.43 m, 0) and
(x0 , y0 ) = (0, 0) .
The electric fields are
ke q
+ y12
9 &times; 109 N m2 /C2 7 &times; 10−6 C
=
0 + (0.25 m)2
= 1.008 &times; 106 N/C , and
ke q
Ex = 2
x2 + y22
9 &times; 109 N m2 /C2 7 &times; 10−6 C
=
(0.43 m)2 + 0
= 3.40725 &times; 105 N/C ,
Thus q
Ey =
E=
x21
Ex2 + Ey2
Explanation:
and
The force of repulsion exerted on each mass
q
is determined by
F = q E = q Ex2 + Ey2
1 Q1 Q2
= 7 &times; 10−6 C
F =
= ma
q
4 π ǫ0 r 2
&middot; (1.008 &times; 106 N/C)2 + (3.40725 &times; 105 N/C)2
where r is the distance between the centers of
= 7.4482 N .
the two spheres.
~ 12 k = kF
~ 21 k
kF
Since both spheres have the same mass and
are subject to the same force, they have the
same acceleration.
002 10.0 points
Three equal charges of 7 &micro;C are in the x-y
plane. One is placed at the origin, another is
placed at (0.0, 25 cm), and the last is placed
003 10.0 points
Two identical small charged spheres hang in
equilibrium with equal masses as shown in
the figure. The length of the strings are equal
and the angle (shown in the figure) with the
vertical is identical.
The acceleration of gravity is 9.8 m/s2
and the value of Coulomb’s constant is
8.98755 &times; 109 N m2 /C2 .
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
vertical directions must separately add up to
zero:
X
Fx = T sin θ − Fe = 0
X
Fy = T cos θ − m g = 0 .
0. 2
5m
7◦
0.04 kg
0.04 kg
Find the magnitude of the charge on each
sphere.
Correct answer: 1.41013 &times; 10−7 C.
Explanation:
Let : L = 0.25 m ,
m = 0.04 kg ,
θ = 7◦ .
and
L
q
m
θ
a
m
q
From the right triangle in the figure above,
we see that
a
sin θ = .
L
Therefore
a = L sin θ
= (0.25 m) sin(7◦ )
= 0.0304673 m .
The separation of the spheres is r = 2 a =
0.0609347 m . The forces acting on one of the
spheres are shown in the figure below.
T
T cos θ
Fe
2
From the second equation in the system
mg
above, we see that T =
, so T can be
cos θ
eliminated from the first equation if we make
this substitution. This gives a value
Fe = m g tan θ
= (0.04 kg) 9.8 m/s2 tan(7◦ )
= 0.0481315 N ,
for the electric force.
From Coulomb’s law, the electric force between the charges has magnitude
|q|2
|Fe | = ke 2 ,
r
where |q| is the magnitude of the charge on
each sphere.
Note: The term |q|2 arises here because the
charge is the same on both spheres.
This equation can be solved for |q| to give
s
|Fe | r 2
|q| =
ke
s
(0.0481315 N) (0.0609347 m)2
=
(8.98755 &times; 109 N m2 /C2 )
= 1.41013 &times; 10−7 C .
004 10.0 points
A circular arc has a uniform linear charge
density of 6 nC/m.
The value of the Coulomb constant is
8.98755 &times; 109 N &middot; m2 /C2 .
y 14 ◦
6
θ
2. 8
θ
T sin θ
mg
Because the sphere is in equilibrium, the
resultant of the forces in the horizontal and
m
x
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
What is the magnitude of the electric field
at the center of the circle along which the arc
lies?
Explanation:
Let : λ = 6 nC/m = 6 &times; 10−9 C/m ,
∆θ = 146◦ , and
r = 2.8 m .
θ is defined as the angle in the counterclockwise direction from the positive x axis
as shown in the figure below.
73 ◦
◦
r
73
θ
~
E
First, position the arc symmetrically
around the y axis, centered at the origin. By
symmetry (in this rotated configuration) the
field in the x direction cancels due to charge
from opposites sides of the y-axis, so
Ex = 0 .
For a continuous linear charge distribution,
Z
~ = ke dq r̂
E
r2
In polar coordinates
dq = λ (r dθ) ,
where λ is the linear charge density. The
positive y axis is θ = 90◦ , so the y component
of the electric field is given by
dEy = dE sin θ .
Note: By symmetry, each half of the arc
about the y axis contributes equally to the
electric field at the origin. Hence, we may just
3
consider the right-half of the arc (beginning
on the positive y axis and extending towards
the positive x axis) and multiply the answer
by 2.
Note: The upper angular limit θ = 90◦ .
The lower angular limit θ = 90◦ − 73◦ = 17◦ ,
is the angle from the positive x axis to the
right-hand end of the arc.
!
Z ◦
λ 90
E = −2 ke
sin θ dθ ̂
r 17◦
= −2 ke
λ
[cos (17◦ ) − cos (90◦ )] ̂ .
r
Since
ke
λ
= (8.98755 &times; 109 N &middot; m2 /C2 )
r
(6 &times; 10−9 C/m)
&times;
(2.8 m)
= 19.259 N/C ,
E = −2 (19.259 N/C)
&times; [(0.956305) − (0)] ̂
= −36.835 N/C ̂
~ = 36.835 N/C .
kEk
~ in
Alternate Solution: Just solve for kEk
a straight forward manner, positioning the
beginning of the arc on the positive x axis (as
shown in the original figure in the question).
θ is still defined as the angle in the counterclockwise direction from the positive x axis.
!
◦
Z
ke λ 146
Ex = −
cos θ dθ ı̂
r 0◦
ke λ
[sin (146◦ ) − sin (0◦ )] ı̂
r
= −(19.259 N/C)
&times; [(0.559192) − 0.0] ı̂
= [−10.7695 N/C] ı̂ ,
!
◦
Z
ke λ 146
Ey = −
sin θ dθ ̂
r 0◦
=−
ke λ
[cos (0◦ ) − cos (146◦ )] ̂
r
= −(19.259 N/C)
&times; [1.0 − (−0.829038)] ̂
= [−35.2255 N/C] ̂ ,
=−
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
q
~ = E2 + E2
kEk
y
x
h
= (−10.7695 N/C)2
+ (−35.2255 N/C)2
= 36.835 N/C .
i1/2
005 10.0 points
From the electric field vector at a point, one
can determine which of the following?
I) the direction of the electrostatic force on
a test charge of known sign at that point;
II) the magnitude of the electrostatic force
exerted per unit charge on a test charge
at that point;
III) the electrostatic charge at that point.
1. I and II only correct
2. II and III only
3. III only
4. I, II and III
Let : m = 9.6 &times; 10−5 g = 9.6 &times; 10−8 kg ,
Ex = 4.6 N/C ,
Ey = Ez = 0 ,
vy = 2.3 &times; 106 m/s ,
vx = vz = 0 , and
t = 0.7 s .
According to Newton’s second law and the
definition of an electric field,
~ = m~a = q E
~.
F
Since the electric field has only an x component, the particle accelerates only in the x
direction
q Ex
ax =
.
m
To determine the x component of the final
velocity, vxf , use the kinematic relation
vxf = vxi + a (tf − ti ) = a tf .
Since ti = 0 and vxi = 0,
006 10.0 points
A particle of mass 9.6 &times; 10−5 g and charge
86 mC moves in a region of space where the
electric field is uniform and is 4.6 N/C in the
x direction and zero in the y and z direction.
If the initial velocity of the particle is given
by vy = 2.3 &times; 106 m/s, vx = vz = 0, what is
the speed of the particle at 0.7 s?
Correct answer: 3.68928 &times; 106 m/s.
Explanation:
q E x tf
m
(0.086 C) (4.6 N/C)(0.7 s)
=
(9.6 &times; 10−8 kg)
= 2.88458 &times; 106 m/s .
vxf =
5. I only
Explanation:
The definition of the electrostatic force is
~
~ = F . In another way, F
~ = q E.
~ This means
E
q
~ is in the same direction of, or opposite diF
~ depending on the sign of the
rection to E,
charge. And if we only consider the magnitude F = qE for a unit charge, the force on it
F
is just
= E. This is statement II.
q
4
vxf
No external force acts on the particle in the y
direction so vyi = vyf = 2.3 &times; 106 m/s. Hence
the final speed is given by
q
2 + v2
vf = vyf
xf
2
= 2.3 &times; 106 m/s
2
+ 2.88458 &times; 10 m/s
6
= 3.68928 &times; 106 m/s .
1/2
Note: This is analogous to a particle in a
gravitational field with the coordinates roπ
tated clockwise by (90◦ ).
2
007 (part 1 of 2) 10.0 points
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
A dipole (electrically neutral) is placed in an
external field.
(a)
−
+
−
+
(b)
+
−
−
(c)
+
5
Gauss’ law states
I
~ &middot; dA
~ = Q.
ΦS = E
ǫ0
Solutions: The electric dipole consists of
two equal and opposite charges separated by
a distance. In either situation (c) or (d), the
electric field is uniform everywhere between
the parallel infinite plates. Thus, the electric
force on one charge is equal but opposite to
that on another so that the net force on the
whole dipole is zero. By contrast, electric
fields are nonuniform for situations both (a)
and (b).
(d)
For which situation(s) shown above is the
net force on the dipole equal to zero?
1. Another combination
2. (c) and (d) correct
3. (a) and (c)
008 (part 2 of 2) 10.0 points
For which situation(s) shown above is the net
torque on the dipole equal to zero?
1. (a) only
2. (c) and (d)
3. other combination
4. (b) and (d)
4. (c) only
5. (c) only
5. (b) and (d)
6. (a) only
6. None of these
7. None of these
Explanation:
Basic Concepts: Field patterns of point
charge and parallel plates of infinite extent.
The force on a charge in the electric field is
given by
~ = qE
~
F
and the torque is defined as
~ = ~r &times; F
~
T
~ = k ∆q r̂
∆E
r2
X
~ =
~i.
E
∆E
Symmetry of the configuration will cause
some component of the electric field to be
zero.
7. (a) and (c) correct
8. (a) and (b)
Explanation:
A electric dipole can be regarded as a pair
of charges of opposite sign. Only in figures
(a) and (c), the electric fields are along the
direction of ~r, where ~r is the vector between
~ is
the pair of charges. Therefore the force F
also along ~r . This will lead to zero torque,
since
~ = ~r &times; F
~ ∝ ~r &times; ~r = 0 .
T
For figures (b) and (d), the torque on both
charges are nonzero and the resultant torques
are also nonzero.
009
10.0 points
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
Five charges are placed in a closed box. Each
charge (except the first) has a magnitude
which is twice that of the previous one placed
in the box. All charges have the same sign
and (after all the charges have been placed in
the box) the net electric flux through the box
is 9.5 &times; 107 N &middot; m2 /C.
What is the magnitude of the smallest
charge in the box?
6
r4
r3
r2
r1
q3
q2
q1
R1
R2 R3 R4 R5
Explanation:
Let : Φ = 9.5 &times; 107 N &middot; m2 /C
Let the first charge in the box be q. Then
the other four charges have magnitude 2 q, 4 q,
8 q, and 16 q. We can figure out a value for q
by considering Gauss’ law,
Φ=
qtotal
.
ǫ0
Our closed box is a Gaussian surface, and
we know the total flux through this closed
surface. Hence,
Φ=
(1 + 2 + 4 + 8 + 16) q
31 q
=
ǫ0
ǫ0
and solving for q,
Hint: Under static conditions, the charge
on a conductor resides on the surface of the
conductor.
What is the charge Qr3 on the inner surface
of the larger spherical conducting shell?
1. Qr3 = +q1 − q2
2. Qr3 = 0
3. Qr3 = −q1 − q2 − q3
4. Qr3 = +q1 + q2 + q3
5. Qr3 = −q1 − q2 + q3
6. Qr3 = +q1 + q2
q=
1
(9.5 &times; 107 N &middot; m2 /C)
31
&times; (8.85419 &times; 10−12 C2 /N &middot; m2 )
1 &times; 106 &micro;C
&times;
1C
= 27.1338 &micro;C .
010 10.0 points
A point charge q1 is concentric with two spherical conducting thick shells, as shown in the
figure below. The smaller spherical conducting shell has a net charge of q2 and the larger
spherical conducting shell has a net charge of
q3 .
7. Qr3 = +q1
8. Qr3 = −q1 − q2 correct
9. Qr3 = −q1 + q2
10. Qr3 = −q1
Explanation:
The net charge inside a Gaussian surface
located at r = R4 must be zero, since the field
in the conductor must be zero. Therefore,
the charge on the inner surface of the large
spherical conducting shell must be −(q1 + q2 ).
011
10.0 points
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
1
.
4 π ǫ0
A uniformly charged sphere (an insulating
sphere with radius R) is shown in the figure
below.
Given: k =
Q is the total
charge inside
the sphere.
R
p
R
2
If Q = 5.3 &times; 10−6 C, the magnitude of the
~ electric field at r = R is given by
kEk
2
~ = kQ .
1. kEk
16 R2
~ = k Q . correct
2. kEk
2 R2
~ = 2kQR.
3. kEk
~ = kQ .
4. kEk
8 R2
~ = kQR.
5. kEk
~ = 4kQR.
6. kEk
~ = kQ .
7. kEk
4 R2
~ = kQ.
8. kEk
R2
~ = kQ .
9. kEk
32 R2
~ = 8kQR.
10. kEk
Explanation:
Pick aspherical Gaussian
surface with ra
R
dius r, where r =
concentric with the
2
sphere of charge.
From Gauss’ law, we find
Φ = E &middot; 4 π r2 .
ρ=
4
3
Q
, so the net charge inside the Gausπ R3
7
R
is
2
4
3
πr
qen = ρ
3
Q
4
3
=
πr
4
3
π R3
3
r3
=Q 3.
R
Therefore the electric field is
Φc
E=
4 π r2
Q r3 1
=
ǫ0 R3 4 π r 2
Qr
=
,
4 π ǫ0 R3
at r. At r =
R
we have
2
Q
1
E=
4 π ǫ0 2 R2
kQ
.
=
2 R2
012 (part 1 of 2) 10.0 points
Consider the figure
+Q
A −Q
+
− y
+
−
+
−
+
−
+
−
x
+
−
+ C D −
+
−
+
−
+
−
+
−
#1
#2
B
Of the following elements, identify all that
correspond to an equipotential line or surface.
1. line AB only correct
2. neither AB nor CD
3. line CD only
4. both AB and CD
Explanation:
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
Consider the electric field
+Q
A −Q
+
− y
+
−
+
−
+
−
+
−
x
+ C D −
+
−
+
−
+
−
+
−
+
−
#1
#2
B
An equipotential line or surface (AB) is
normal to the electric field lines.
013 (part 2 of 2) 10.0 points
Consider the figure
C
B
+
+q
A
−
−q
8
Through what potential difference would an
electron need to be accelerated for it to
achieve a speed of 6 % of the speed of light
(2.99792 &times; 108 m/s), starting from rest?
Explanation:
Let : s = 6 % = 0.06 ,
c = 2.99792 &times; 108 m/s ,
me = 9.10939 &times; 10−31 kg ,
qe = 1.60218 &times; 10−19 C .
and
The speed of the electron is
v = 0.06 c
= 0.06 2.99792 &times; 108 m/s
= 1.79875 &times; 107 m/s ,
By conservation of energy
1
me v 2 = −(−qe ) ∆V
2
v2
∆V = me
2 qe
= 9.10939 &times; 10−31 kg
D
Of the following elements, identify all that
correspond to an equipotential line or surface.
1. both AB and CD
2
1.79875 &times; 107 m/s
&times;
2 (1.60218 &times; 10−19 C)
2. line CD only correct
3. line AB only
= 919.798 V .
4. neither AB nor CD
Explanation:
Consider the electric field:
C
A
B
−
+
D
An equipotential line or surface (CD) is
normal to the electric field lines.
015 (part 1 of 2) 10.0 points
The figure below shows two particles, each
with a charge of +Q, that are located at the
opposite corners of a square of side d.
Q
+
d
+
Q
P
What is the direction of the net electric field
at point P ?
1.
014
10.0 points
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
2.
3.
correct
017 10.0 points
The two charges Q are fixed at the vertices of
an equilateral triangle with sides of length a
as shown.
q
a
4.
5.
Q
Explanation:
At point P , the electric fields due to the two
positive charged particles have the same magnitude. One points downward, and the other
horizontally to the left. Thus the direction of
the net electric field points left and downward,
forming a 45◦ angle with the horizontal.
016 (part 2 of 2) 10.0 points
What is the potential energy of a particle of
charge q that is held at point P ?
√
2 qQ
1. U =
2 π ǫ0 d
2. U = 0
a
a
Q
The work required to move a charge q from
the other vertex to the center of the line joining the fixed charges is
√
2kQq
1. W =
a
kQq
2. W =
a
4kQq
3. W =
a
4. W = 0
6kQq
a
2kQq
correct
6. W =
a
Explanation:
5. W =
√
2 qQ
4 π ǫ0 d
1 qQ
4. U =
correct
2 π ǫ0 d
1 qQ
5. U =
4 π ǫ0 d
Explanation:
The potential energy of the particle charged
+q is the sum of the potential energies due to
the two +Q charged particles.
3. U =
9
U = U1 + U2
1 qQ
1 qQ
=
+
4 π ǫ0 d
4 π ǫ0 d
1 qQ
.
=
2 π ǫ0 d
Qq
Qq
Qq
+k
= +2 k
a
a
a
Qq
Qq
Qq
U2 = +k a + k a = +4 k
a
2
2
Qq
W = U2 − U1 = 2 k
.
a
U1 = +k
018 10.0 points
If the potential in a region is given by the
function
V = 2 x − y 2 − cos(z) ,
what is the y-component of the electric field
at the point P = (x′ , y ′ , z ′ )?
hernandez (uh286) – oldmidterm 01 – Turner – (59070)
1. Ey = −2 y ′
Let : λ = 2.4 &micro;C/m = 2.4 &times; 10−6 C/m ,
R = 2.4 m ,
2 R = 4.8 m , and
ke = 8.98755 &times; 109 N &middot; m2 /C2 .
2. Ey = − sin(z ′ )
3. Ey =
10
y ′3
3
4. Ey = 2
R
5. Ey = y ′2
p
2R
6. Ey = 2 x′
7. Ey = cos(z ′ )
8. Ey = 2 y ′ correct
9. Ey = sin(z ′ )
x′2
4
Explanation:
The electric field is the gradient of the potential, so the y-component of the E-field evaluated at P is
10. Ey =
∂V
Ey = −
∂y
∂ 2 x − y 2 − cos(x)
=−
∂y
= − [−2 y]
= 2y.
2R
Let p be the origin. Consider the potential
due to the line of charge to the right of p.
Z
Vright = dV
Z
dq
= ke
r
Z 3R
λdx
= ke
x
R
3R
= ke λ ln x
= ke λ ln 3 .
R
By symmetry, the contribution from the line
of charge to the left of p is the same. The
contribution from the semicircle is
Z π
λRdθ
Vsemi = ke
R
0
Z π
= ke λ
dθ
0
π
019 10.0 points
A wire that has a uniform linear charge density of 2.4 &micro;C/m is bent into the shape as
shown below, with radius 2.4 m.
2.4 m
4.8 m
p
4.8 m
The Coulomb constant is 8.98755 &times;
109 N &middot; m2 /C2 .
Find the electrical potential at point p.
Correct answer: 1.15159 &times; 105 V.
Explanation:
= ke λ θ
0
= ke λ π .
Hence the electric potential at p is
Vp = Vright + Vlef t + Vsemi
= 2 ke λ ln 3 + ke λ π
= ke λ (2 ln 3 + π)
= (8.98755 &times; 109 N &middot; m2 /C2 )
&times; (2.4 &times; 10−6 C/m)
&times; (2 ln 3 + π)
= 1.15159 &times; 105 V .
```