If steps are not taken to ground a gasoline pump, static electricity can be placed on gasoline when filling your car’s tank. Suppose a tiny drop of gasoline has a mass of 4.00×10–15kg4.00×10–15kg and is given a positive charge of 3.20×10–19C3.20×10–19C. (a) Find the weight of the drop. (b) Calculate the electric force on the drop if there is an upward electric field of strength 3.00×105N/C3.00×105N/C due to other static electricity in the vicinity. (c) Calculate the drop’s acceleration.

*Strategy*

To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example asks for weight. This is a topic of dynamics and is defined in Dynamics: Force and Newton’s Laws of Motion. Part (b) deals with electric force on a charge, a topic of Electric Charge and Electric Field. Part (c) asks for acceleration, knowing forces and mass. These are part of Newton’s laws, also found in Dynamics: Force and Newton’s Laws of Motion.

The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so on.

*Solution for (a)*

Weight is mass times the acceleration due to gravity, as first expressed in

w
=
mg
.
w
=
mg
.
size 12{w= ital "mg"} {}

(1)Entering the given mass and the average acceleration due to gravity yields

w
=
(
4.00
×
10
−
15
kg
)
(
9
.
80
m/s
2
)
=
3
.
92
×
10
−
14
N
.
w
=
(
4.00
×
10
−
15
kg
)
(
9
.
80
m/s
2
)
=
3
.
92
×
10
−
14
N
.
size 12{w= \( "44" "." "00" times "10" rSup { size 8{ - "15"} } `"kg" \) \( 9 "." "80""m/s" rSup { size 8{2} } \) =3 "." "92" times "10" rSup { size 8{ - "14"} } N} {}

(2)*Discussion for (a)*

This is a small weight, consistent with the small mass of the drop.

*Solution for (b)*

The force an electric field exerts on a charge is given by rearranging the following equation:

F
=
qE
.
F
=
qE
.
size 12{F= ital "qE"} {}

(3)Here we are given the charge (3.20×10–19C3.20×10–19C is twice the fundamental unit of charge) and the electric field strength, and so the electric force is found to be

F
=
(
3.20
×
10
−
19
C
)
(
3
.
00
×
10
5
N/C
)
=
9
.
60
×
10
−
14
N
.
F
=
(
3.20
×
10
−
19
C
)
(
3
.
00
×
10
5
N/C
)
=
9
.
60
×
10
−
14
N
.
size 12{F= \( 3 "." "20" times "10" rSup { size 8{ - "19"} } C \) \( 3 "." "00" times "10" rSup { size 8{5} } "N/C" \) =9 "." "60" times "10" rSup { size 8{ - "14"} } N} {}

(4)*Discussion for (b)*

While this is a small force, it is greater than the weight of the drop.

*Solution for (c)*

The acceleration can be found using Newton’s second law, provided we can identify all of the external forces acting on the drop. We assume only the drop’s weight and the electric force are significant. Since the drop has a positive charge and the electric field is given to be upward, the electric force is upward. We thus have a one-dimensional (vertical direction) problem, and we can state Newton’s second law as

a
=
F
net
m
.
a
=
F
net
m
.
size 12{a= { { ital "net"`F} over {m} } `} {}

(5)where
F
net=F−w
F
net=F−w size 12{F=F - w} {}. Entering this and the known values into the expression for Newton’s second law yields

a
=
F
−
w
m
=
9.60
×
10
−
14
N
−
3.92
×
10
−
14
N
4.00
×
10
−
15
kg
=
14
.
2
m/s
2
.
a
=
F
−
w
m
=
9.60
×
10
−
14
N
−
3.92
×
10
−
14
N
4.00
×
10
−
15
kg
=
14
.
2
m/s
2
.
alignl { stack {
size 12{a= { {F - w} over {m} } } {} #
size 12{ {}= { {9 "." "60" times "10" rSup { size 8{ - "14"} } N - 3 "." "92"` times "10" rSup { size 8{ - "14"} } N} over {4 "." "00" times "10" rSup { size 8{ - "15"} } ital "kg"} } } {} #
="14" "." 2m/s rSup { size 8{2} } {}
} } {}

(6)*Discussion for (c)*

This is an upward acceleration great enough to carry the drop to places where you might not wish to have gasoline.

This worked example illustrates how to apply problem-solving strategies to situations that include topics in different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles.

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