Physics 30
Unit 2 Review
UNIT 2 REVIEW
1.
A neutral balloon is rubbed by a neutral cloth and as a result, the balloon gains a positive charge.
It is then placed near a neutral wall and it “sticks” to the wall. Using the principles of physics,
explain:
a) why the balloon gains a positive charge
b) why the positively-charged balloon “sticks” to the neutral wall
2.
There are three identical metal spheres with the following charges:
q1 = 57 mC , q 2 = 83 mC , and q3 = 41 mC.
q 2 is touched to q1 , and then q 2 is touched to q3 . Determine the final charges of the spheres.
3.
A 74.0 C metal sphere is touched to an identical metal
sphere with a charge Q. After they contact, they are moved
to a distance of 6.80 cm from each other and they experience
a repulsion force of 5.10 kN.
q1 
H
q2 
+ q3
Determine the two possible charges for Q.
4.
The point charges q1, q2, and q3 shown in the diagram
have charges of equal magnitude. Roughly sketch the
vector sum diagram that determines the net electric field at:
a)
5.
G
b) H
G
For the point charges shown,
determine the magnitude and
direction of the net electric field
at P.
q1 = 14.0 mC
8.40 cm
10.3 cm
P
q2 = +18.0 mC
6.
18
A 7.0 g pith ball is attached to a string to form a “pith ball
electroscope”. The pith ball is then given a charge q and when
it is placed between two parallel plates, the “pendulum” is
deflected from the vertical by 18, as shown.
q
If the plates are connected to a 12 V power supply and
they are 10 cm apart, then determine the charge q.
Be certain to identify whether the charge is positive or
negative. Explain your choice.
1
12 V
Physics 30
Unit 2 Review
7.
An alpha particle is originally moving at a horizontal velocity of 7.00  105 m/s. It then passes
through a potential difference of 4500 V (creating a parallel electric field), which causes it to
slow down. Determine the final speed of the alpha particle. Sketch a diagram of this situation.
8.
Two parallel plates are 24.0 cm apart.
A charge with 11 electrons in excess
is initially located 6.00 cm from the lower
plate. If it requires a minimum work
of 8.45  1016 J to move it to the
upper plate, then determine the
potential difference V across the plates.
24.0 cm
6.00 cm
(Ignore the effects of Fg)
9.
V
q
A proton travels through a perpendicular electric field, as shown below:
25.0 cm
7.00 cm
p+
7.00 cm
If the magnitude of the electric field between the plates is 8800 N/C, then determine the
initial speed of the proton.
10. An oil droplet has 6 electrons in deficit. It is then injected between two parallel plates that are
11 cm apart and the potential difference between the plates is adjusted until the oil droplet is
suspended. If the required potential difference is 900 V, then determine the mass of the oil droplet.
Sketch the device.
11. A neutral metal box is placed in an external
electric field, as shown. Explain how the net electric field
inside the metal box could still be zero.
E
E=0
metal box
2
Physics 30
Unit 2 Review
12. Consider the oscilloscope below:
Electron gun
Deflection region
D
cathode anode
5.00  106 m/s
heater
14.0 cm
8.00 cm
fluorescent
screen
Va
Determine:
a) the accelerating potential difference Va across the electron gun
b) the electric field (mag and dir) between the deflecting plates, if the vertical deflection
is 3.00 cm as the electrons travel across the plates
*c) the overall deflection D by the time it hits the fluorescent screen
SOLUTIONS
6.
Charging by friction: The cloth is higher on the electrostatic series, so it strips the excited eoff the balloon. The balloon ends up with a deficit of electrons (i.e. a positive charge).
b) Induced separation of charge: The positive balloon attracts the e- in the wall, inducing a
E3
negative charge in the closest part of the wall. Opposite charges attract.
q1 = 13 mC ; q2 = 27 mC ; q3 = 27 mC
E2
Magnitude of charge after contact: q = 51.2 C
net EG
If q = +51.2 C, then Q = 28.4 C ; If q = 51.2 C, then Q = 176 C
E1
See diagrams on right.

net E p = 2.35  1010 N/C at 49.5 N of E (40.5 E of N)
net EH

4
Fe  mg tan 18 = 0.0223 N ; E = 120 N/C left ; q = +1.9  10 C
E3
7.
E k f  E ki  E = 1.8925  1016 J ;
1.
2.
3.
4.
5.
a)
Using E k  0.5mv 2 , v = 2.39  105 m/s
E1
E2
q = 1.76  1018 C ; V = 480 V between initial and final positions ; V  4 V = 640 V
3
11
2
7
9. Vertical: a = 8.43  10 m/s ; t = 4.075 10 s
Horiz: v = 2.63  107 m/s

10. q = +9.60  1019 C ; E = 8.182  103 V/m ; Using Fe = Fg, m = 8.0  1016 kg
8.
11. See diagram. Ein is opposite to Eout, so Enet = 0.
12. a) Using E k   q V , Va = 71.2 V
b) Horiz: t = 1.6  108 s ; Vert: ay = 2.344  1014 m/s2 up
Electric field = 1.33 kV/m down
*c) Angle out of plates = 36.9 ; D = 10.5 cm + 3.0 cm = 13.5 cm
3





Ein
E=0
metal box
+
+
+
+
+
Eout