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Solutions to Problems : Chapter 19
Problems appeared on the end of chapter 19 of the Textbook
- Picture the Problem: Two point charges exert an electrostatic force on each other.
Strategy: Solve Coulomb’s law (equation 19-5) for the separation distance r.
Solution: Solve equation 19-5 for r
( )( )( ) 9 2 2 6 6 1 2 8.99^ 10 N m^ / C^ 11.2^10 C^ 29.1^10 C^ 1.29 m 1.77 N
kq q r F
× ⋅ × −^ × −
Insight: Although 1.77 N is only about 6.4 ounces of force, these microcoulomb-sized charges can exert such a force over the substantial distance of 1.29 m or 4.23 ft.
- Picture the Problem: The honeybee acquires an electrostatic charge in active flight.
Strategy: Use the magnitude of an electron’s charge e to find the number of electrons that correspond to the 93.0 pC total charge. Then use Coulomb’s law to find the magnitude of the force between the two charged bees, and compare it with the weight of a single bee.
Solution: 1. (a) Find the number of electrons:
12 8 19
93.0 10 C
5.81 10 electrons 1.60 10 C/electron
Q
N
e
− −
×
= = = ×
×
2. (b) Use Coulomb’s law to find the force between two charged bees: ( )
( )
12 2 1 2 9 2 2 2 2
7
93.0 10 C
8.99 10 N m / C 0.0120 m
5.40 10 N
kq q F r
−
−
×
= = × ⋅
= ×
3. Determine the weight of a bee: (^) W = mg = (^) ( 0.140 × 10 −^3 kg (^) )( 9.81 m/s^2 )= 1.37 × 10 −^3 N 4. Calculate the ratio of the forces:
7 4 3
5.40 10 N 1
1.37 10 N 2540
F
mg
− − −
×
= = × =
×
Insight: The electrical force between the bees is a tiny fraction of their weight because the amount of electrical charge is quite small. It would require a charge of only 4.68 nC on each bee for the electrical force to equal the weight!
- Picture the Problem: An electric field exists around a 5.00 μ C charge at the origin.
Strategy: Use the definition of the electric field (equation 19-10) to determine its magnitude.
Solution: 1. (a) Apply equation 19- 10 directly:
( )( )
9 2 2 6 4 2 2
8.99 10 N m / C 5.00 10 C 4.50 10 N/C 1.00 m
kq E r
× ⋅ ×^ −
= = = ×
2. (b) Repeat for the new distance: ( )( )
9 2 2 6 4 2 2
8.99 10 N m / C 5.00 10 C 1.12 10 N/C 2.00 m
kq E r
× ⋅ ×^ −
= = = ×
Insight: When the distance from the charge is doubled, the field is cut to a fourth.
- Picture the Problem: Two charges are placed on the x -axis as shown at right and create an electric field in the space around them. Strategy: Use equation 19-8 to find the magnitude and direction of the electric fields created by each of the two charges at the specified locations, then find the vector sum of those fields to find the net electric field. At x = −4.0 cm the field from q 1 will point in the − x ˆ direction and the field from q 2 will point in the + x ˆ direction.
Solution: 1. (a) Sum the fields produced by the two charges at x = −4.0 cm:
( )
( )
1 2 1 2 2 2 2 2 1 2 1 2 6 6 9 2 2 7 2 2
6.2 10 C 9.5 10 C
8.99 10 N m / C ˆ 3.0 10 N/Cˆ 0.040 m 0.140 m
k q k q q q k r r r r − −
⎡ × × ⎤
= × ⋅ ⎢ − + ⎥ = − ×
E x x x
x x
G
- (b) Repeat for x = 4.0 cm:
( )
( )
1 2 1 2 2 2 2 2 1 2 1 2 6 6 9 2 2 7 2 2
6.2 10 C 9.5 10 C
8.99 10 N m / C ˆ 5.9 10 N/Cˆ 0.040 m 0.060 m
k q k q q q k r r r r − −
⎡ × × ⎤
= × ⋅ ⎢ + ⎥ = ×
E x x x
x x
G
Insight: Although q 2 has a larger magnitude than q 1 , at x = −4.0 cm the closer distance to q 1 means its contribution to the field is larger than the contribution from q 2 , and the net field points in the − x ˆ direction.
- Picture the Problem: Three charges are positioned as shown at right.
r
Strategy: Each of the three charges produces its own electric field that surrounds it. The total electric field at any point is the vector sum of the fields from each charge. Use equation 19-10 and the component method of vector addition to find the magnitude electric field at the points indicated in the problem statement. Let q 1 be at the origin and q 3 be on the positive x -axis.
Solution: 1. (a) At a point halfway between charges q 1 and q 2 the vectors E 1
G
and E 2
G
cancel one another. The remaining contribution comes from q 3. First find the distance r from q 3 to the midpoint of the opposite side:
2 2 2
2
2
3 0.0275 m 4 0.0238 m
r d d
r d
r
2. Apply equation 19-10 to find E 3 :
( )( )
9 2 2 6 3 7 (^3 2 )
8.99 10 N m / C 5.00 10 C 7.94 10 N/C 0.0238 m
kq E r
× ⋅ ×^ −
= = = ×
3. (b) At this location, the electric fields of q 2 and q 3 add, and the resulting field points toward q 3. The field due to q 1 will have the same magnitude as found in part (a) and will be perpendicular to the
x −4.0 cm 0 4.0 cm 10.0 cm
E q 1 q 2
G
E
G
- Picture the Problem: The field produced by an unknown charge at an unknown position is given at two points, as depicted in the figure at the right. Strategy: Because both measurements of the electric field point in the positive x -direction, the point charge must be on the x -axis. Also, since the magnitude of an electric field is larger the closer it is measured to its source, the position at the point charge must be greater than x = 10.0 cm. Furthermore, since the vectors point toward the right, they must point toward the charge and the charge must be negative. Write out equation 19-10 for the two given fields and combine them to get a quadratic equation in x. Solve the expression to find the location of the charge. Then use either of the two field equations to find the magnitude of the charge. Solution: 1. (a) Use equation 19-10 to express
the field magnitudes: (^ )^ (^ )
(^1 2 ) 1 2
and
k q k q E E x x x x
2. Combine the equations to eliminate q , take the square root of both sides, and solve for x :
2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 2 10.0 N/C 5.00 cm 15.0 N/C 10.0 cm 32 cm 10.0 N/C 15.0 N/C
E x x kq E x x E x x E x x E x E x x E E
3. (b) Find the magnitude of q from E 1 :
2 2 (^1 1 ) 9 2 2
10.0 N/C 0.32 0.0500 m 8.1 10 C 81 pC 8.99 10 N m / C
E x x q k
= −^ = − = × − =
× ⋅
4. Because the field vector points toward the charge it must be negative: q = −81 pC
Insight: If the charge were positive, the field magnitudes would be the same but the vectors would point toward − x. The rules of subtraction limit the answers to only two significant figures.
- Picture the Problem: A proton is accelerated from rest in a uniform electric field and travels along a straight line. Strategy: Use Newton's Second Law to find the acceleration of the proton from the electric force F = qE exerted by the electric field. Once the acceleration is known we can find the speed of the particle from the equation for velocity as a function of position, v^2 = v 0^2 + 2 a Δ x (equation 2-12).
Solution: 1. (a) Apply Newton's Second Law to find a :
∑^ F^ =^ qE^ =^ ma^ ⇒^ a^ = qE m
2. Solve equation 2-12 for v when v 0 (^) = 0 :
v = v 02 + 2 a Δ x = 0 + 2 ( qE m )Δ x
3. Find v for Δ x = 0.0100 m:
( )( )(^ ) 19 5 5 27
2 1.60 10 C 1.08 10 N/C 0.0100 m 4.55 10 m/s 1.673 10 kg
v
− −
× ×
= = ×
×
4. (b) Find v for Δ x = 0.100 m:
( )( )(^ ) 19 5 6 27
2 1.60 10 C 1.08 10 N/C 0.100 m 1.44 10 m/s 1.673 10 kg
v
− −
× ×
= = ×
×
Insight: The large charge-to-mass ratio of elementary particles produces large accelerations when they
x 5.00 cm 10.0 cm x
E 1 E (^2) q
are immersed in an electric field. In this case the acceleration of the proton is an astounding 1.03×10^13 m/s^2!
- Picture the Problem: A point charge situated at the origin produces an electric field that completely surrounds it. Strategy: Use the definition of the electric field (equation 19-10) to find the magnitude of the charge that creates the stipulated electric field. Since the electric field points toward the origin and therefore toward the charge, the sign of the charge must be negative.
Solution: Solve equation 19-10 for q , assuming the charge is negative:
2 2 6 9 2 2
36, 000 N/C 0.50 m 1.0 10 C 8.99 10 N m /C 1.0 C
Er q k
q μ
= = = × −
× ⋅
Insight: If the sign of the charge had been positive, the field would have pointed in the − x ˆ direction.
- Picture the Problem: A cylindrical Gaussian surface is built around a long, thin wire that has a
charge per unit length λ on it.
Strategy: In problem 55 we used Gauss’s law to find that
E
r
where r is the distance from the wire and λ is the
wire’s charge per unit length. Use this expression
and the given information to find λ, and then use a
ratio to find r when E is reduced by a factor of 2. Solution: 1. (a) Solve
E
r
= for λ :
( )( )( )
0 12 2 2 7
2 8.85 10 C / N m 0.477 m 25, 400 N/C 6.74 10 C/m
λ πε rE
π −^ −
= × ⋅ = ×
- (b) Use a ratio to find the new r :
new 0 new old old 1 old 0 old new 2 old new old
2 2 0.477 m 0.954 m
r E E E r E E E r r
Insight: Another way to solve part (b) is to use the λ from part (a) to solve the given expression for r.
However, using a ratio can often be a time-saving step that also avoids the accumulation of rounding errors that can occur.
