PH 212
07-24-2015
Physics 212 Exam-2
Solution
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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.
3
The electric field at a point 10 cm from the center is 3.6 x 10 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.
Q1= - 3 C
b
c
P
a
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.
A = 2.5 m2
d = 3 mm
45 V
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.
A = 2.5 m2
d = 3 mm
45 V
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
1000
Volts
S
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
2V -
Qi
Qf
+
di
Ef
2V -
At t=0
df
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
Qf
Qi
di
At t=0
Ef
df
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. ((7.5 pointss) Calculate
e the elecctric potenttial differeence between the
b
e points A an
nd S (i.e. calculate VA – VS). (VA – VS) = ‐ 1000 Voltss (7.5 (
points Calculate th
he electric potential p
en ergy differeence of the ion be
o
etween A an
nd S (i.e. calcculate UA – U
US). U = q ( V

V) = 1.6022
2  10‐19 C (‐ 1
1000 Volts) = ‐ 1.6022  10‐16 Joulees 3B. ((15 points) C
Calculate the
e mass m of the individuual ions; give your an
g
swer in atomic mass un
nits ( 1 u = 1..66  10‐27 kgg). Some helpful formulas: 
 nano mm = 10‐3 m

Centripetal force = m v / r

mass of electron: 9.109 x 10


1 eV = 1.602 x 10‐19 J 
F
Coulomb's Law: 
Electric field, along the z‐axis, due to a charge Q distributed uniformly along a thin ring of radius R.:
2
‐31 kg 2
q 2 q1 
1
9 Nm
u
, where 
8
.
99

10
4 0
C2
4 0 r 2
1

E
1
Qz

4 0 z 2  R 2

3/ 2
kˆ

Electric field established by an infinite and uniformly charged sheet:

Gauss' Law

 = E  ds = q /o,

E =  /2o
where q is the net charge inside the Gaussian surface.
S

Potential energy difference UB - UA ≡
 B)
Wext (A 
F
ext
Work done by the external force to take the particle from A to B at constant speed 
Definition of the Electric Potential

Electric potential difference

Potential difference V
B
V (r ) 
W ext ( 
r)
q
0
q0
VB  V A 
Wext ( A 
B)
q0
q0

UB UA
q0

1 q 4 r


 V   E  d s
B
A
A

Electric potential due to a point charge q: V
0

Relationship between E and V:

About capacitance Q = C V For a parallel‐plate capacitor C = Ao /d U = CV2 / 2 = Q2 / 2 C Ex = - dV/dx
Capacitors connected in parallel Cequiv = C1 + C2 + C3 Capacitors connected in series 1/Cequiv = (1/C1) + (1/C2) +(1/C3) 
 sin  d   cos
 Cos d  sin 
MAGNETISM




F  qv B

1 Tesla = 104 gauss 
  q  
B 0 3 vx r
4 r

 = L i 
Inductive reactance XL = L

B

B

B

F
II
 0 a b
L
2 d

0 I 1
4 R
 0 I
4 R
F = force, q = charge, v = velocity, B = magnetic field 
Magnetic field produced by a charge q that moves with velocity v  = Magnetic flux, L = inductance, i = current
Magnetic field at the center of a semi‐circle of radius "R" Magnetic field at the center of an arc of angle f (in radians) and radius "R". 0 I
2 r
Magnetic field produced by a infinitely long wire at a distance "r" from it.
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