112
Solutions Physics to Homework Questions 1
Homework 1 (solutions)
(2004 Fall)
Chapt15, Problem-1:
A 4.5 x 109 C charge is located 3.2 m from a 2.8 x 109 C charge. Find the electrostatic force exerted by one charge on the other.
Solution: Since the charges have opposite signs, the force is attractive . The magnitude so the force is given by Coulomb's Law, so making the substitutions we get
F=
k e q1q2 r
2
-9 -9 N m 2 4.5 10 C 2 .8 10 C -8 = 8.99 109 = 1.1 10 N 2 C2 (3.2 m )
(
)(
)
Chapt15, Problem-3:
An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electrical force acting on the alpha particle when it is 2.0 x 1014 m from the gold nucleus?
Solution: Since the charges have opposite signs, the force is repulsive The magnitude so the force is given by Coulomb's Law, so making the substitutions we get
F=
k e( 2 e ) (79 e ) r2
-19 2 (158 ) 1.60 10 C 9 N m = 8.99 10 2 C2 2 .0 10 -14 m
(
)
2
(
)
= 91 N
( repulsion )
Chapt15, Problem-5:
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons). (a) What is the force between the two alpha particles when they are 5.00 x 1015 m apart, and (b) what will be the magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u.
Solution: Since the charges have opposite signs, the force is repulsive The magnitude so the force is given by Coulomb's Law, so making the substitutions we get
k e ( 2e )2 N m2 (a) F = = 8.99 10 9 r2 C2
-19 C 2 4 1.60 10 = 36.8 N 2 5.00 10-15 m
(b) The mass of an alpha particle is m = 4.0026 u , where u = 1.66 10-27 kg is the unified mass unit. Applying Newton's 2nd law, the acceleration of either alpha particle is then
a=
F 36.8 N 27 2 = = 5.54 10 m s -27 m 4.0026 1.66 10 kg
(
)
Of course from Newton's 3rd law, both alpha particles experience the same force, and hence undergo the same acceleration.
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Physics 112
Chapt15, Problem-7:
Homework 1 (solutions)
(2004 Fall)
Suppose that 1.00 g of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at Earth's North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on Earth? Solution:
1.00 g of hydrogen contains Avogadro's number of atoms, each containing one proton and one electron. Thus, each charge has magnitude q = N A e . The distance separating these charges is r = 2 RE , where RE is the Earth's radius. Thus applying Coulomb's Law,
F=
k e NA e
(
)
2
(2 R E )
2
23 -19 N m 2 6.02 10 1.60 10 C = 8.99 109 2 C2 4 6.38 106 m
[(
)(
)]
2 5 = 5.12 10 N
Chapt15, Problem-8:
(
)
An electron is released a short distance above Earth's surface. A second electron directly below it exerts an electrostatic force on the first electron just great enough to cancel the gravitational force on it. How far below the first electron is the second?
Solution: The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus by using Coulomb;'s Law and Newton's 2nd Law (applied to gravity), we have
ke e2 r 2 = m e g
so rearranging this expression, and then making the substitutions, we get
ke e2 r= = me g
(8.99 10 N m C )(1.60 10 C ) (9.11 10 kg )( 9.80 m s )
-19 2 -31
9
2
2
2
= 5.08 m
Chapt15, Problem-11:
Three charges are arranged as shown in Figure P15.11. Find the magnitude and direction of the electrostatic force on the charge at the origin.
Solution: In the sketch to the right, FR is the resultant of the forces F and F3 that are exerted on the charge at 6 the origin by the 6.00 nC and the 3.00 nC charges respectively. Applying Coulomb's Law to each, we get
-9 -9 N m2 6.00 10 C 5.00 10 C F6 = 8.99 109 C2 (0.300 m )2
(
(
)(
)
)
= 3.00 10-6 N
-9 -9 2 3.00 10 C 5.00 10 C 9 N m F3 = 8.99 10 = 1.35 10-5 N C2 (0.100 m )2
)(
From the Superposition Principle, the resultant is
FR =
or
(F6 )2 + ( F3 )2 = 1.38 10 -5 N
at = tan -1
F3 = 77.5 , F6
-5 FR = 1.38 10 N at 77.5 below - x axis
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Physics 112
Chapt15, Problem-13:
Homework 1 (solutions)
(2004 Fall)
Three point charges are located at the corners of an equilateral triangle as in Figure P15.13. Calculate the net electric force on the 7.00 C charge.
Solution: The forces on the 7.00 C charge are shown in the sketch to the right. Applying Coulomb's Law to calculate each force, we get
2 7.00 10-6 C 2.00 10-6 C 9 N m F1 = 8.99 10 C2 (0.500 m )2
( (
)( )(
) )
= 0.503 N 2 7.00 10 -6 C 4.00 10-6 C 9 N m F2 = 8.99 10 C2 (0.500 m )2
From the superposition principle, we known Fx = ( F1 + F2 ) cos 60.0 = 0.755 N , and
= 1.01 N
Fy = ( F1 - F2 ) sin 60.0 = -0.436 N
So the resultant force on the 7.00 mC charge is
Fy = 0.872 N at = tan -1 = -30.0 , Fx or FR = 0.872 N at 30.0 below the + x axis FR =
(Fx )2 + ( Fy )
2
Chapt15, Problem-16:
A charge of 6.00x109 C and a charge of 3.00x109 C are separated by a distance of 60.0 cm. Find the position at which a third charge of 12.0 x 109 C can be placed so that the net electrostatic force on it is zero.
Solution: The required position is shown in the sketch to the right. Note that this places q closer to the smaller charge, which will allow the two forces to cancel. Applying Coulmob's Law, and requiring that F = F gives 6 3
k e (6.00 nC ) q
(x + 0.600 m )
x= 0.600 m 2-1
2
=
k e ( 3.00 nC) q 2 , or 2 x 2 = ( x + 0.600 m ) x2
Solving for x gives the equilibrium position as
= 1.45 m beyond the - 3.00 nC charge
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Physics 112
Chapt15, Problem-19:
Homework (solutions)
(2004 1 Fall)
An airplane is flying through a thundercloud at a height of 2 000 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If there are charge concentrations of +40.0 C at height 3 000 m within the cloud and 40.0 C at height 1 000 m, what is the electric field E at the aircraft?
Solution: We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions, one due to each concentration. The contribution due to the positive charge at 3000 m altitude is
2 ( 40.0 C) 9 N m 5 E+ = ke 2 = 8.99 10 2 2 = 3.60 10 N C r C ( 1000 m )
q
(downward)
The contribution due to the negative charge at 1000 m altitude is
E - = ke
N m2 ( 40.0 C) = 8.99 109 = 3.60 105 N C r2 C2 ( 1000 m ) 2 q
(downward)
From the Superposition Principle, the resultant field is then
5 E = E+ + E - = 7.20 10 N C ( downward )
Chapt15, Problem-27:
In Figure P15.27, determine the point (other than infinity) at which the total electric field is zero.
Solution: If the resultant field is zero, the contributions from the two charges must be in opposite directions and also have equal magnitudes. Choose the line connecting the charges as the x-axis, with the origin at the 2.5 C charge. Then, the two contributions will have opposite directions only in the regions x < 0 and x > 1.0 m . For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the point of zero resultant field is on the x-axis at x < 0 . Requiring equal magnitudes gives Thus, (1.0 m + d )
ke q1 r12
=
ke q2 r22
or
2 .5 C 6.0 C = . 2 d (1.0 m + d )2
2 .5 =d 6.0 d = 1.8 m ,
or
Solving for d yields
1.8 m to the left of the - 2 .5 C charge
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Physics 112
Chapt15, Problem-28:
Homework 1 (solutions)
(2004 Fall)
Figure P15.28 shows the electric field lines for two point charges separated by a small distance. (a) Determine the ratio q1 /q2 . (b) What are the signs of q 1 and q2 ?
Solution: The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines emerge from q2 as enter q1 . q2 = 3 q1 (a) (b) Then,
q1 q2 =-1 3
q2 > 0 because lines emerge from it,
and q1 < 0 because lines terminate on it
Chapt15, Problem-35:
If the electric field strength in air exceeds 3.0 x 106 N/C, the air becomes a conductor. Using this fact, determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius.
Solution: For a uniformly charged sphere, the field is strongest at the surface. Thus, Emax = or
k eqmax , R2 2 6 R 2 Emax ( 2.0 m ) 3.0 10 N C -3 qmax = = = 1.3 10 C ke 8.99 109 N m2 C2
(
)
Chapt15, Problem-41:
Solution:
A 40-cm-diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.2 x 105 N m2/C. Calculate the electric field strength in this region.
From the definition of Electric Flux Thus, E =
E = EAcos
and E = E , max when = 0
5 2 E , max E , max 4 5.2 10 N m C 6 = = = 4.1 10 N C 2 A d 2 4 (0.40 m )
(
)
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Physics 112
Chapt15, Problem-57:
Homework 1 (solutions)
(2004 Fall)
Two 2.0-g spheres are suspended by 10.0-cm-long light strings (Fig. P15.57). A uniform electric field is applied in the x direction. If the spheres have charges of 5.0 x 108 C and +5.0 x 108 C, determine the electric field intensity that enabl...