3. We use E A , where A A 140m .
(a) 6.00 N C 1.40 m 0.
(b) 2.00 N C 1.40 m 3.92 N m 2 C.
(c) 3.00 N C 400 N C k 1.40 m 0 .
(d) The total flux of a uniform field through a closed surface is always zero.
4. The flux through t
First Midterm Solutions, Phyx 135-2, Lecture 02, Version 2, Fall 2013
1) (10 points) Suppose three charges are placed at the corners of an equilateral triangle. The
length of each side of the triangle is l = 8 m. The two bottom charges are both +2 C; the
Physics 1352, Midterm 1, 10 am sec
1) (5 points) A dipole with q = 0.1 mC is placed in a uniform electric field of 23 N/C. The
electric field is oriented along the xaxis; the dipole is oriented at = 22 to the xaxis. The
dipole is then turn
This weeks experiment examined the electrostatic forces between two
charged objects, as described by Coulombs Law. Coulombs Law states that
the force between two charged objects is proportional to the product of the
charges and inversely proportional to t
Synthesis of 2,5-dichloro-2,5-dimethylhexane from 2,5-dimethylhexane-2,5-diol
Objective To synthesize 2,5-dichloro-2,5-dimethylhexane by a concentrated acid-catalyzed SN1 process and then
characterize the product by NMR
Experiment 6: Conservation of
Energy and Momentum 3
The primary goal of this weeks experiment was to examine elastic collisions
in two dimensions. Using two air hockey pucks, pointed electrode, and a
piece of carbon paper the motion of the co
Experiment 7: Newtons Second
Law for Rotational Motion
This weeks experiment examined the rotational form of Newtons Second Law of
Motion. The procedure used two different components to verify the rotational form
of Newtons Second Law of Moti
Experiment 8: Oscillations and
This weeks experiment examined the rotation of a rigid body about a fixed
axis. For this laboratory, the periodic motion of a physical pendulum under
varying amplitudes was measured and analyzed usi
Experiment 1: A Modern Galileo
The primary goal of this experiment was to investigate the relationship between position, velocity, and
time under constant acceleration. Through graphical analysis the kinematic relationship between
Conservation of Energy and
This weeks experiment focused on the Principle of Conservation of Momentum. The
principle states that the total momentum of an isolated system remains constant.
This principle is especially important when
Experiment 4: Conservation of
Energy and Momentum 1
This weeks experiment was focused on the Principle of Conservation of Energy. The
principle states, a loss of kinetic energy must be accompanied by an equivalent
gain of potential energy and
Experiment 3: Newtons Second
For this experiment, the motion of two objects, a glider and hanging weight, on a
horizontal air track was analyzed to verify Newtons second law of mechanics. First,
the acceleration of the glider was measured
Experiment 2: Kinematics and
The primary goal of this experiment was to analyze the dynamics of twodimensional projectile motion under the effects of gravity. The position of a puck as
it slid down an inclined table and boun
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a
distance r from the wire, is given by
B 0 .
With r = 20 ft = 6.10 m, we have
2 b mg
T m A 100 A
g 3.3 10
T 3.3 T.
(b) This is about one-sixth the magn
14. For a free charge q inside the metal strip with velocity v we have F q E v B .
We set this force equal to zero and use the relation between (uniform) electric field and
potential difference. Thus,
3.90 109 V
E Vx Vy d xy
0.382 m s .
19. (a) The area of a sphere may be written 4R2= D2. Thus,
2.4 106 C
4.5 107 C/m 2 .
(b) Equation 23-11 gives
4.5 107 C/m 2
5.1 104 N/C.
0 8.85 1012 C2 / N m 2
39. The forces acting on the ball are shown in the diagram below. The grav
29. We assume the charge density of both the conducting cylinder and the shell are
uniform, and we neglect fringing effect. Symmetry can be used to show that the electric
field is radial, both between the cylinder and the shell and outside the shell. It i
3. If the electric potential is zero at infinity then at the surface of a uniformly charged
sphere it is V = q/40R, where q is the charge on the sphere and R is the sphere radius.
Thus q = 40RV and the number of electrons is
q 4 R V
1.0 106 m 400 V
15. (a) The electric potential V at the surface of the drop, the charge q on the drop, and
the radius R of the drop are related by V = q/40R. Thus
N m 2 / C2 30 1012 C
5.4 104 m.
(b) After the drops combine the total volume is t
28. Consider an infinitesimal segment of the rod, located between x and x + dx. It has
length dx and contains charge dq = dx, where = Q/L is the linear charge density of the
rod. Its distance from P1 is d + x and the potential it creates at P1 is
10. The equivalent capacitance is
10.0 F 5.00 F
10.0 F 5.00 F
11. The equivalent capacitance is
C1 C2 C3 10.0 F 5.00 F 4.00 F 3.16 F.
C1 C2 C3
10.0 F 5.00 F 4.00 F
14. (a) The potential difference across C1
46. Each capacitor has 12.0 V across it, so Eq. 25-1 yields the charge values once we
know C1 and C2. From Eq. 25-9,
and from Eq. 25-27,
This leads to
= 2.21 1011 F ,
= 6.64 1011 F .
q1 = C1V1 = 8.00 1010 C, q2 = C2V2 = 2.66 1010 C.
11. (a) The current resulting from this nonuniform current density is
J a dA
r 2 rdr R 2 J 0 (3.40 103 m) 2 (5.50 104 A/m 2 )
(b) In this case,
J b dA J 0 1 2 rdr R 2 J 0 (3.40 103 m) 2 (5.50 104 A/m 2 )
2. The current in the circuit is
i = (150 V 50 V)/(3.0 + 2.0 ) = 20 A.
So from VQ + 150 V (2.0 )i = VP, we get VQ = 100 V + (2.0 )(20 A) 150 V = 10 V.
3. (a) The potential difference is V = + ir = 12 V + (50 A)(0.040 ) = 14 V.
(b) P = i2r = (50 A)2(0.040
23. Let i1 be the current in R1 and take it to be positive if it is to the right. Let i2 be the
current in R2 and take it to be positive if it is upward.
(a) When the loop rule is applied to the lower loop, the result is
2 i1 R1 0 .
The equation yields
58. (a) = RC = (1.40 106 )(1.80 106 F) = 2.52 s.
(b) qo = C = (12.0 V)(1.80 F) = 21.6 C.
(c) The time t satisfies q = q0(1 et/RC), or
t RC ln 0 2.52s ln
21.6 C 16.0 C
62. The time it takes for the voltage difference across the
3. (a) The force on the electron is
FB qv B q vx v y Bx By j q vx By v y Bx k
= 1.6 1019 C 2.0 106 m s 0.15 T 3.0 106 m s 0.030 T
6.2 1014 N k.
Thus, the magnitude of FB is 6.2 1014 N, and FB points in the positive z direction.
(b) This amounts to
Image of Formation
Images can result when light rays encounter flat or curved surfaces between two
Images can be formed either by reflection or refraction due to these surfaces.
Mirrors and lenses can be d