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PH 212 07-24-2015
Physics 212 Exam-2
Solution NAME: _____________________________________
Write down your name also on the back of the package of sheets you turn in.
SIGNATURE and ID: ___________________________ Return this hard copy exam together with your other answer sheets.
Show all your procedures when giving the answers to the first 03 questions. For question 4 just provide the answer.
1. A solid, insulating sphere of radius a = 5 cm has a uniform charge density and a total
charge Q. Concentric with this sphere there is a conducting spherical shell whose inner and
outer radii are b = 20 cm and c =25 cm respectively.
The electric field at a point 10 cm from the center is 3.6 x 103 N/C pointing radially inward,
while the electric field at a point 50 cm from the center is 2 x 102 N/C pointing radially outward.
P b a
Q1= - 3 C
c
Insulator
Conductor
1A (10 points) Calculate the charge Q in the insulating sphere.
1B (10 points) Calculate the charge qin on the inner surface of the shell.
Calculate the charge qout on the outer surface of the shell.
1C (10 points) Calculate the electric potential difference Vb- Va.
2. A parallel plate capacitor with plate area A = 2.5 m2 and plate separation d = 3 mm is
connected to a 45 V battery.
2A Calculate the magnitude of the charge on each plate of the capacitor.
Calculate the electric field between the plates.
Calculate the energy stored in the capacitor.
45 V
A = 2.5 m2 d = 3 mm
+
With the capacitor still connected to the battery, a slab of plastic of dielectric K = 3.6 is placed between the plates of the capacitor.
2B Calculate the magnitude of the charge on each plate of the capacitor.
Calculate the electric field between the plates.
Calculate the energy stored in the capacitor.
45 V
A = 2.5 m2 d = 3 mm
K= 3.6
3. The figure shows the essentials of a mass spectrometer, which can be used to measure the mass of an ion.
An ion of mass m (to be measured) and charge q=1.6022 10-19 Coulomb is produced in the source S. The initially stationary ion is accelerated by the electric field due to a potential
difference (V=1000 Volts). The ion leaves S, passes through an aperture A, enters a chamber in which there exists a uniform magnetic field (B= 80 mT ) as indicated in the
figure, and strikes the detector at a point that lies at x =1.6254 m.
3A. (7.5 points) Calculate the electric potential difference between the points A and S (i.e. calculate VA – VS).
(VA – VS) = - 1000 Volts
(7.5 points Calculate the electric potential energy difference of the ion between A and S (i.e. calculate UA – US).
U = q ( V) = 1.6022 10-19 C (- 1000 Volts) = - 1.6022 10-16 Joules
3B. (15 points) Calculate the mass m of the individual ions; give your answer in atomic mass units ( 1 u = 1.66 10-27 kg).
Q1= - 3 C
A
S
1000 Volts
4. The figure below shows a parallel-plate capacitor. In what follows, assume for simplicity
that the electric field produced by the plates is equivalent to the electric field produced by
infinite plates charged uniformly all over their surface.
In the figures below, Q stands for the initial charge accumulated in the top plate, E is the
electric field in between the plates, and d is the distance between the plates.
4A (5 points) From left to right: While keeping the battery connected to the plates, an
external agent (not shown in the figure) increases the distance between the plates.
+ Ei Qi
2V - +
Ef 2V - di
Qf
df
At t=0 At t > 0
Which expression describes correctly what happens as the distance between the plates
is increased?
a) The electric field increases.
b) The energy stored in the capacitor decreases.
c) The magnitude of the charge in each plate increases.
d) The electric field remains constant.
e) All the expressions above are incorrect.
4B (5 points) In this case the capacitor is initially charged and then the battery is disconnected.
From left to right: an external agent (not shown in the figure) increases the distance between the plates.
Ei Qi
Ef di
Qf
df
At t=0 At t > 0
Which expression describes correctly what happens as the distance between the plates
is increased?
a) The electric potential between the plates decreases..
b) The electric potential energy stored in the capacitor decreases.
c) The magnitude of the charge in each plate decreases.
d) The electric field remains constant.
e) All the expressions above are incorrect.
3A. (b
(o
3B. ( g
(7.5 pointsbetween the
(VA – VS) =
(7.5 points of the ion be
U = q ( V
(15 points) Cgive your an
s) Calculatee points A an
‐ 1000 Volts
Calculate thetween A an
V) = 1.6022
Calculate theswer in ato
e the elecnd S (i.e. cal
s
he electric pnd S (i.e. calc
2 10‐19 C (‐ 1
e mass m of mic mass un
ctric potentculate VA –
potential enculate UA – U
1000 Volts)
the individunits ( 1 u = 1.
tial differeVS).
ergy differeUS).
= ‐ 1.6022
ual ions; .66 10‐27 kg
ence
ence
10‐16 Joule
g).
es
Some helpful formulas: nano mm = 10‐3 m
Centripetal force = m v2/ r
mass of electron: 9.109 x 10‐31 kg
1 eV = 1.602 x 10‐19 J
Coulomb's Law: ur
qqF
2
12
04
1
, where
2
29
0 C
Nm1099.8
4
1
Electric field, along the z‐axis, due to a charge Q distributed uniformly along a thin ring of radius R.:
kR
QE ˆ
4
12/322
0
z
z
Electric field established by an infinite and uniformly charged sheet: E = /2o
Gauss' Law
= S
sdE
= q /o, where q is the net charge inside the Gaussian surface.
Potential energy difference
UB - UA ≡ Wext (A ext
F B)
Work done by the external force to take the particle from A to B at constant speed
Definition of the Electric Potential 0
)()( 0
q
rWrV
qext
Electric potential difference 0
)(0
q
BAWVV
qext
AB
0q
AB UU
Potential difference sdEVVB
AAB
Electric potential due to a point charge q:
r
qV
04
1
Relationship between E and V: Ex = - dV/dx
About capacitance
Q = C V
For a parallel‐plate capacitor C = Ao /d
U = CV2 / 2 = Q2 / 2 C
Capacitors connected in parallel Cequiv = C1 + C2 + C3
Capacitors connected in series 1/Cequiv = (1/C1) + (1/C2) +(1/C3)
sincossin dCosd
MAGNETISM
BF
vq
F = force, q = charge, v = velocity, B = magnetic field
1 Tesla = 104 gauss
rvB
x4 3
0
r
q
Magnetic field produced by a charge q that moves with velocity v
= L i = Magnetic flux, L = inductance, i = current Inductive reactance XL = L
B
0 I
4
1
R Magnetic field at the center of a semi‐circle of radius "R"
B 0 I4 R
Magnetic field at the center of an arc of angle f (in radians) and
radius "R".
B 0 I
2 r Magnetic field produced by a infinitely long wire at a distance "r" from it.
F
L 0
IaIb
2 d Force per unit length between two parallel long wires,
carrying currents Ia and Ib respectively, separated by a distance "d"
Faraday's Law t
,
where = Magnetic flux and = electromotive force
Definition of the magnetic dipole moment of a loop of area A, carrying a current I: = I A n
where A = area, I current, n = unit vector perpendicular to the loop
