Chapter 17 Electrostatics Pretest

1
Chapter 17 Electrostatics Pretest
2
1. If a charged glass rod touches the knob of an
electroscope, the leaves of the electroscope
become A) positively charged by conduction,
B) negatively charged by conduction, C)
negatively charged by induction, D) positively
charged by induction.
3
1. If a charged glass rod touches the knob of an
electroscope, the leaves of the electroscope
become A) positively charged by conduction,
B) negatively charged by conduction, C)
negatively charged by induction, D) positively
charged by induction.
4
4. Four steps in producing a residual charge by
induction on an electroscope are 1. Provide a
conducting path between the electroscope and
earth. 2. Remove the conducting path between
the electroscope and earth. 3. Hold a
charged object near the electroscope
knob. 4. Take the charged object away from the
area of the electroscope knob.The proper order
for these steps is A) 1, 3, 4, 2, B)
3, 1, 2, 4, C) 3, 1, 4, 2, D) 1, 2,
3, 4.
5
4. Four steps in producing a residual charge by
induction on an electroscope are 1. Provide a
conducting path between the electroscope and
earth. 2. Remove the conducting path between
the electroscope and earth. 3. Hold a
charged object near the electroscope
knob. 4. Take the charged object away from the
area of the electroscope knob.The proper order
for these steps is A) 1, 3, 4, 2, B)
3, 1, 2, 4, C) 3, 1, 4, 2, D) 1, 2,
3, 4.
6
1. Describe the method of charging an
electroscope A) positively by conduction.
B) negatively by conduction.
7
Positive touch a positively charged object to
the electroscope.Negative - touch a negatively
charged object to the electroscope.
8
2. Calculate the force between charges of 150.0
mC and -75.0 mC that are 0.600 m apart in air.
The value of k for air is 8.93 x109 Nm2/C2.
9
F k q1 q2 / d2F (9 x 109)(150 x 10-6)(75 x
10-6)/0.62F 281 NThe charges are opposite so
the force is attractive.
10
3. Draw the lines of force around two equal but
oppositely charged point charges. Each charge is
4.0 x 10-6 C and they are 2.0 m apart. Calculate
the strength of field at a point halfway between
the two charges.
11
(No Transcript)
12
E kq/d2 E- kq/d2E (9 x 109)(4 x
10-6)/12 E- (9 x 109)(-4 x 10-6)/12 E
36000 N/C E- 36000 N/CThe net field is
36000 36000 72000 N/C directed away from the
charge and toward the - charge.
13
4. Describe the strength of electric field inside
a statically charged conductor.
14
Because the charge is distributed over the
surface of the conductor, the sum of all the
forces on a positive charge would be zero. The
electric field strength is zero.
15
(No Transcript)