Transcript Test Review Jeopardy

Review Jeopardy
Electrostatics
1. Who was the first to determine the
electron’s charge & what is that
charge?
Qelectron = -1.6 x 10-19C
2. A piece of plastic has a net charge of
+2.00 μC. How many more protons than
electrons does this piece of plastic have?
1proton
13
2 10 C 
 1.25 10 protons
19
1.6 10 C
6
3. Two point charges, separated by 1.5 cm, have
charge values of +2.0 and –4.0 μC, respectively.
What is the magnitude of the electric force between
them?
kQ1Q2 (9 10 )(2 10 )(4 10 )
F

 320N
2
2
d
(0.015)
9
6
6
4. A force of 6.0 N acts on a charge of 3.0
μC when it is placed in a uniform electric
field. What is the magnitude of this electric
field?
F = QE
6.0 = (3.0 x 10-6C) E
E = 2.0 x 106 N/C
5. A metal sphere of radius 10.0 cm carries a
charge of +2.0 μC. What is the magnitude of
the electric field 5.0 cm from the sphere’s
surface?
kQ1 (9 10 9 Nm 2 /C 2 )(2.0 106 C)
6
E 2 

7.2
10
N /C
2
d
(0.05m)
6. Sketch the electric field lines
around a negative point charge
7. Determine the unknown
charge.
3C
?
3 3C  9C  9C
8. An uncharged conductor is supported by an
insulating stand. I pass a positively charged rod near
the left end of the conductor, but do not touch it. What
charge will the right end of the conductor have?
+
+
+
+
+
+ + + +
9. If the distance between two point charges is
increased by a factor of 5, the mutual force
between them will be changed by what factor?
kQ1Q2
F
2
(5d)
1
F _ decrease _ by _
25
10. An electron in an electric field will
experience a force acting:
a.
b.
c.
d.
e.
Parallel (same direction)
Anti-parallel (opposite direction)
Perpendicular
Along an equipotential line
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FINAL JEOPARY
• Write down how many points your team
is willing to bet.
– You can bet as low as zero points and as
high as your whole score.
Calculate the electric field (magnitude & direction) at
the origin generated by the following three charges. A
q2 = -2.00 C at (-5.00 m, 0), q1= -1.50 C at (3.5 m,
0.0), and q3 = +1.50 C at (4.00 m, 0).
q3= + 1.5x10-6C
d= 4.0 m
kQ (9 10 9 Nm 2 /C 2 )(2 106 C)
E2  2 
 720N /C 
2
5
q1= - 1.5x10-6C d
d= 3.5 m
 -6C
q2= - 2.0x10
d=5.00 m

kQ (9 10 9 Nm 2 /C 2 )(1.5 106 C)
E1  2 
 1,102N /C 
2
d
(3.5m)
kQ (9 10 9 Nm 2 /C 2 )(1.5 106 C)
E3  2 
 843N /C 
2
d
(4m)
E total  720N /C 1,102N /C  843N /C  461N /C 
