Mastering Physics Solutions Chapter 19 Electric Charges, Forces, and Fields
Chapter 19 Electric Charges, Forces, and Fields Q.1CQ
When an object that was neutral becomes charged. does the total charge of the universechange? Explain
Solution:
No. charging of a neutral object does not change the total charge of the universe If a neutral object is charged with another charged object the transfer of charge takes place between them
and then the net charge remains constant Hence, the total charge in the universe remains same
Chapter 19 Electric Charges, Forces, and Fields Q.1P
CE PredictlExplain An electrically neutral object is given a positive charge. (a) In principle. does the object’s mass increase, decrease, or stay the same as a result of being charged? (b) Choose the best explanation from among the following:
I To give the object a positive charge we must remove some of its electrons: this will reduce its mass
ll Since electric charges have mass, giving the object a positive charge will increase its mass
llI Charge is conserved, and therefore the mass of the object will remain the sama
Solution:
(a) In order to give the positive charge to any body we should remove the electrons from the body. So some of the mass of the object will Idecreased
(b) As the removing of electrons leads to acquiring the positive charge this leads to decrease in mass of the object Hence best explanation is ( I).
Chapter 19 Electric Charges, Forces, and Fields Q.2CQ
The fact that the electron has a negative charge and the proton has a positive charge is due to a convention established by Benjamin Fran kIin Would there have been any significant consequences if Franklin had chosen the opposite convention? Is there any advantage to naming charges plus and minus as opposed to. say. A and B?
Solution:
No.
When the current existing convention for the charge in the physics for electron & proton is interchange then there would be no effect at all. because the net charge in the universe will be same.
If the + & – charge are replaced by A & B. then there would be a problem for the defining of O (zero) charge because O(zero) charge means, the object has equal number of positive & negative charge
Chapter 19 Electric Charges, Forces, and Fields Q.2P
CE PredictlExplain An electrically neutral object is given a negative charge. (a) In principle. does the objects mass increase, decrease, or stay the same as a result of being charged? (b) Choose the best explanation from among the following:
I To give the object a negative charge we must give it more electrons, and this will increase its mass
II A positive charge increases an object’s mass: a negative charge decreases its mass
IlI Charge is conserved, and therefore the mass of the object will remain the same
Solution:
(a)
An electrically neutral object acquires negative charge by addition of electrons These electrons will increase the mass of the object Hence, the total mass of the object will ¡n crease
(b)
Addition of more number of electrons results in acquiring negative charge by the object Therefore, the best explanation is
Chapter 19 Electric Charges, Forces, and Fields Q.3CQ
Explain why a comb that has been rubbed through your hair attracts small bits of paper. even though the paper is uncharged
Solution:
The plastic comb gets charged when it rubbed through our hair Here, the plastic comb gets negative charge and the hair gets positive charge. Like charges repel each other and the unlike charges attract each other.
When this negatively charged comb is placed near bits of paper then the electrons in each molecule of the bits of papers shill away from the combS As a result, the positive charge in the bit of paper comes close to the comb and charge separation takes place inside the bits of paper Hence, the bit of paper is polarized. Due to the electrostatic force between the negatively charged comb and the positive side of the paper they attracted by the comb
Chapter 19 Electric Charges, Forces, and Fields Q.3P
CE (a) Based on the materials listed in Table 19—1, is the charge of the rubber balloon shown on page 655 more likely to be positive or negative? Explain, (b) If the charge on the balloon is reversed, will the stream, of water deflect toward or away from the balloon? Explain
Solution:
(a)
From the table 19-1. one can infer that a rubber on rubbing acquires negative charge and readily accepts more number of electrons In the table specified, four minus signs were given
for the material rubber which means it is negatively charged. So. a balloon made up of rubber will also possesses the same charge Thus, the rubber balloon on rubbing acquires negative charge
(b)
Water is a polar molecule that does not possess net charge. as one end of the water molecule possesses partial positive charge and the other end possesses partial negative charge Assume that the charge on the balloon is reversed, that is. the charge on the balloon is positive If this positively charged balloon is brought near the stream of water, then the negative ends in the water molecule tend to align towards the positively charged balloon and positive ends in the water molecule move towards the opposite end As a result, the stream of water deflects towards the balloon In this case, the attractive
force between the negatively charged particles of water and positively charged balloon is greater when compared to the repulsive force between the positively charged particles of water and balloon.
Thus, whatever be the charge of the balloon, the water deflects towards the balloon because of its polarity nature.
Chapter 19 Electric Charges, Forces, and Fields Q.4CQ
Small bits of paper are attracted to an electrically charged comb, but as soon as they touch the comb they are strongly repelled Explain this behavior
Solution:
Bits of paper which are initially uncharged are attracted towards the electrically charged comb by polarization effect, but when they come in contact with the comb, the polarization effect will disappear & paper receives the same charge which shows the repulsion between them
Chapter 19 Electric Charges, Forces, and Fields Q.4P
CE This problem refers to the information given in Table 19—1 (a) If rabbit fur is rubbed against glass. what is the sign of the charge each acquires? Explain. (b) Repeat part (a) for the case of
glass and rubber. (C) Comparing the situations described in parts (a) and (b). in which case is the magnitude of the tribo— electric charge greater? Explain
Solution:
(a) Since the rabbit fur losses more electrons while the glass gains the electrons, so, when rabbit fur is rubbed against glass, the fur attains positive charge and the glass rod acquires negative charge
(b) Since the rubber has a tendency to gain the electrons so. when it rubbed against glass. the glass acquires + ve charge and rubber — ve charge
(c) Rabbit fur and glass are adjacent in the table where as glass and rubbers are widely separated So we can conclude that the magnitude of triboelectric charge is greater in the glass — rubber case than in rabbit fur — glass case
Chapter 19 Electric Charges, Forces, and Fields Q.5CQ
A charged rod is brought near a suspended object. which is repelled by the rod Can we conclude that the suspended object is charged? Explain
Solution:
Since the charged rod gets a repulsive force when it brought near a suspended object this observation clearly Indicates the object had the same Kind of charge on R Because the charges having lice signs are repel to each other
Chapter 19 Electric Charges, Forces, and Fields Q.5P
Find the net charge of a system consisting of 4.9 × 107 electrons.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.6CQ
A charged rod is brought near a suspended object. which is attracted to the rod Can we conclude that the suspended object is charged? ExpIain
Solution:
May or may not have the charge
Since the charged rod gets an attractive force when it brought near a suspended object. this observation clearly indicates the object had either the opposite kind of charge or electrically neutral
Reason for first case: the charges having unlike signs are attracted to each other Reason for second case: the charges attracts because of the polarization effect
Chapter 19 Electric Charges, Forces, and Fields Q.6P
Find the net charge of a system consisting of (a) 6.15 × 10 electrons and 7.44 × 106 protons or (b) 212 electrons and 165 protons.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.7CQ
Describe some of the similarities and differences between Coulombs law and Newton’s law of gravity.
Solution:
The Coulomb force is proportional to the product of charges and the force is inversely proportional to the square of the distance between the two charges. Here, k is the coulomb’s constant. It is coulomb’s law.
The gravitational force is proportional to the product of two different masses and inversely proportional to the square of the distance between the two masses. Here. G is the gravitational constant. It is Newton’s law of gravitation.
Similarities between ‘s law of gravity and Coulomb’s law:
1. Both the laws depend on the product of certain special properties of the objects.
Z Both the gravitational force and the electrostatic force are inversely proportional to the square of the distance between the objects
Dissimilarities:
1. In Newton’s law of gravity, mass is relevant, and in Coulomb’s law, the charge of the objects is irrelevant
2. Gravitational force is only attractive, whereas electrostatic force can be attractive or repulsive
Chapter 19 Electric Charges, Forces, and Fields Q.7P
How much negative electric charge is contained in 2 moles of carbon?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.8CQ
A point charge +Q is fixed at a height H above the ground. Directly below this charge is a small ball with a charge −q and a mass m. When the ball is at a height h above the ground, the net force (gravitational plus electrical) acting on it is zero. Is this a stable equilibrium for the object? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.8P
Find the total electric charge of 1.5 kg of (a) electrons and (b) protons.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.9CQ
Four identical point charges are placed at the corners of a square. A 111th point charge placed at the center of the square experiences zero net force. Is this a stable equilibrium for the fifth charge? Explain.
Solution:
An object is said to be in mechanical equilibrium, if the object is in a state of rest or balance under the action of two or more external forces. This essentially means that there is no unbalanced force or net force and net torque acting on the object And hence there is no acceleration of the object Consider equal charges q is placed at the tour corners of a square and consider the charge p be placed at the center of square Then, the charge p experiences 2 pairs of torces along the 2 diagonals of the square and the torces by each pairs of charges are equal and opposite in direction Thus, they cancel each other and experience no net torca This implies that the point charge in the center is not accelerating Hence, the point charge in the center is said to be in stable equilibrium Therefore, the this is a case of stable equilibrium for the fifth charge as the effect by all the charges on the fifth charge cancel
Chapter 19 Electric Charges, Forces, and Fields Q.9P
A container holds a gas consisting of 1.85 moles of oxygen molecules. One in a million of these molecules has lost a single electron. What is the net charge of the gas?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.10CQ
A proton moves in a region of constant electric field. Does it follow that the proton’s velocity is parallel to the electric field? Does it follow that the proton’s acceleration is parallel to the electric field? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.10P
The Charge on Adhesive Tape When adhesive tape is pulled from a dispenser, the detached tape acquires a positive charge and the remaining tape in the dispenser acquires a negative charge. If the tape pulled from the dispenser has 0.14 μC of charge per centimeter, what length of tape must be pulled to transfer 1.8 × 1013 electrons to the remaining tape?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.11CQ
Describe some of the differences between charging by induction and charging by contact
Solution:
(i) Charging by induction does not involve any physical contact between the charging object and the object being charged Where as the charging by contact involves physical contact to transfer charge form one object to the other
(ii) When an object is charged by induction, the object acquires a charge opposite to that of the charging object. Charging by contact gives the object to be charged the same charge as that of the charging object
Chapter 19 Electric Charges, Forces, and Fields Q.11P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.12CQ
A system consists of two charges of equal magnitude and opposite sign separated by a distance d Since the total electric charge of this system is zero, can we conclude that the electric field produced by the system is also zero? Does your answer depend on the separation if? Explain
Solution:
Consider two equal and opposite charges separated by a finite distance d The net charge of the system will be zero However it cannot be concluded that the net electric field produced by this system will be zero. For example. consider a point which is closer to the positive charge. Now, the electric field at this point will be the vector sum of electric fields due to both positive and negative charge Since the point is closer to positive charge than the negative charge. magnitude of electric field due to positive charge will be greater than magnitude of electric field due to negative charge. Therefore. after the vector sum, the effective electric field will be positive
However. if the separation becomes negligible (like the case when distance of point under consideration is much more than the separation distance d). the net electric field can be said to be zero as both the charge will have equal and opposite electric fields at the point under consideration
Chapter 19 Electric Charges, Forces, and Fields Q.12P
A system of 1525 particles, each of wliich is either an electron or a proton, has a net charge of −5.456 × 10−17 C. (a) How many electrons are in this system?(b) What is the mass of this system?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.13CQ
The force experienced by charge 1 at point A is different in direction and magnitude from the force experienced by charge 2 at point B. Can we conclude that the electric fields at points A and B are different? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.13P
CE A charge +q and a charge −q are placed at opposite corners of a square. Will a third point charge experience a greater force if it is placed at one of the empty corners of the square, or at the center of the square? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.14CQ
Can an electric tÏeId exist in a vacuum? Explain.
Solution:
Electromagnetic fields exist in solids. liquids, gases. and the vacuum In the vacuum, the electromagnetic held propagates this has a component of electric & magnetic held
Yes. the electric field exists in a vacuum.
Best example is sun. The sun is a giant ball of nuclear reactions which creates not just light but a large range of electromagnetic fields. Atmosphere absorbs some of this energy and the spinning of the earth’s gigantic core creates a huge magnetic field around the earthS This field points in the direction of one of the magnetic poles. At the poles of the earth, the magnetic field passes through the earth. Therefore, the field is smaller and causes the aurora borealis, the interaction of the sun’s electromagnetic energy with our atmosphere.
Chapter 19 Electric Charges, Forces, and Fields Q.14P
CE Repeat the previous question, this time with charges +q and +q at opposite corners of a square.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.15CQ
Explain why electric field lines never cross
Solution:
Electric field lines are imaginary lines to show the presence of the tield A tangent drawn at any point on the field line gives the direction of electric field at that point.
Consider two lines intersect at a point then in such case two tangents can be drawn at that point which indicates two directions of the electric field Now if a unit charge is placed at that point then force acting on the unit charge due to field, is in two direction accordingly which is not possible. Since the electric force could not act in two directions at a position. Therefore. electric field lines can never cross to each other
Chapter 19 Electric Charges, Forces, and Fields Q.15P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.16CQ
Charge ql is inside a closed Gaussian surface: charge q2 is just outside the surface Does the electric flux through the surface depend on ql?Does it depend on q2?Explain
Solution:
According to the Gauss Law, the electric flux through a surface depends on the charge that is enclosed by the surface So the electric flux through the Gaussian surface depends on q1
Since the charge q2 is outside the Gaussian surface so. it has no effect on the total electric flux through the surface
Chapter 19 Electric Charges, Forces, and Fields Q.16P
CE Predict!Explain Suppose the charged sphere in Active Example 19—2 is made from a conductor, rather than an insulator (a) Do you expect the magnitude of the force between the
point charge and the conducting sphere to be greater than, less than, or equal to the force between the point charge and an insulating sphere? (b) Choose the best explanation from among the following:
I The conducting sphere will allow the charges to move, resulting in a greater force.
II The charge of the sphere is the same whether it is conducting or insulating, and therefore the force is the same.
III The charge on a conducting sphere will move as far away as possible from the point charge. This results in a reduced force.
Solution:
(a)
The charge inside the conducting sphere comes on its surface. Hence, the resultant charge value on sphere remains sama Therefore, the magnitude of electrostatic force between the conducting sphere and point charge remains same as between an insulating sphere and point charge
There is no change in the value of charge on conducting and insulating sphere. Hence, the magnitude of electrostatic force also remains unchanged
Hence, option is the best explanation
Chapter 19 Electric Charges, Forces, and Fields Q.17CQ
In the previous question, does the electric field at a point on the Gaussian surface depend on q1?Does it depend on q2. Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.17P
At what separation is the electrostatic force between a +11.2-μC point charge and a +29.1-μCpoint charge equal in magnitude to 1.57 N?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.18CQ
Gauss’s law can tell us how much charge ¡s contained within a Gaussian surface Can it tell us where inside the surface it is located? Explain
Solution:
No.
The electric flux through a surface depends on the total charge enclosed by the surface but it is completely independent of location of the enclosed charges Hence Gauss law cannot tell us where inside the surface the charge is located
Chapter 19 Electric Charges, Forces, and Fields Q.18P
The attractive electrostatic force between the point charges +8.44 × 10−6 C and Q has a magnitude of 0.975 N when the separation between the charges is 1.31 m. Find the sign and magnitude of the charge Q.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.19CQ
Explain why Gauss’s law is not very useful in calculating the electric field of a charged disk.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.19P
If the speed of the electron in Example 19−1 were 73× 105 m/s, what would be the corresponding orbital radius?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.20P
IP Two point charges, the first with a charge of +3.13 × 10−6C and the second with a charge of −4.47 × 10−6 C, are separated by 25.5 cm. (a) Find the magnitude of the electrostatic force experienced by the positive charge, (b) Is the magnitude of the force experienced by the negative charge greater than, less than, or the same as that experienced by the positive charge? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.21P
When two identical ions are separated by a distance of 6.2 × 10−10 m, the electrostatic force each exerts on the other is 5.4 × 10−9 N. How many electrons are missing from each ion?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.22P
A sphere of radius 4.22 cm and uniform surface charge density +12.1 μC/m2 exerts an electrostatic force of magnitude 46.9 × 10−3 N on a point charge of +1.95 μC. Find the separation between the point charge and the center of the sphere.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.23P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.24P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.25P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.26P
IP Given that q = +12 μCand d = 19 cm, (a) find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in Figure 19−29. (b) How would your answers to part (a) change if the distance d were tripled?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.27P
Suppose the charge q2 in Figure 19−29 can be moved left or right along the line connecting the charges q1 and q3. Given that q = +12 μC, find the distance from q1 where q2 experiences a net electrostatic force of zero. (The charges q1 and q3 are separated by a fixed distance of 32 cm.)
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.28P
Find the orbital radius for which the kinetic energy of the electron in Example 19−1 is 1.51 eV. (Note: 1 eV = 1 electron volt = 1.6 × 10−19J.)
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.29P
Apoint charge q = −0.35 nC is fixed at the origin. Where must a proton be placed in order for the electric force acting on it to be exactly opposite to its weight? (Let the y axis be vertical and the x axis be horizontal.)
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.30P
A point charge q = −0.35 nC is fixed at the origin. Where must an electron be placed in order for the electric force acting on it to be exactly opposite to its weight? (Let the y axis be vertical and the x axis be horizontal.)
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.31P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.32P
IP(a) Find the direction and magnitude of the net electrostatic force exerted on the point charge q3 in Figure 19−32. Let q = +2.4 μC and d = 27 cm. (b) How would your answers to part (a) change if the distance d were doubled?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.33P
IP Two point charges lie on the x axis. A charge of +9.9 μC is at the origin, and a charge of −5.1 μC is at x = 10.0 cm. (a) At what position x would a third charge q3 be in equilibrium? (b) Does your answer to part (a) depend on whether q3 is positive or negative? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.34P
A system consists of two positive point charges, q1 and q2> q1. The total charge of the system is +62.0 μC, and each charge experiences an electrostatic force of magnitude 85.0 N when the separation between them is 0.270 m. Find q1 and q2
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.35P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.36P
Referring to Problem 35, suppose that the magnitude of the net electrostatic force exerted on the point charge q2 in Figure 19−33 is 0.65 N. (a) Find the distance d. (b) What is the direction of the net force exerted on q2?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.37P
IP (a) If the nucleus in Example 19−1 had a charge of +2e (as would be the case for a nucleus of helium), would the speed of the electron be greater than, less than, or the same as that found in the Example? Explain. (Assume the radius of the electron’s orbit is the same.) (b) Find the speed of the electron for a nucleus of charge +2e.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.38P
Four point charges are located at the corners of a square with sides of length a. Two of the charges are +q, and two are −q, Find the magnitude and direction of the net electric force exerted on a charge +Q, located at the center of the square, for each of the following two arrangements of charge: (a) The charges alternate in sign (+q, −q, +q, −q) as you go around the square; (b) the two positive charges are on the top corners, and the two negative charges are on the bottom corners.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.39P
IP Two identical point charges in free space are connected by a string 7.6 cm long. The tension in the string is 0.21 N. (a) Find the magnitude of the charge on each of the point charges, (b) Using the information given in the problem statement, is it possible to determine the sign of the charges? Explain, (c) Find the tension in the string if +1.0 μC of charge is transferred from one point charge to the other. Compare with your result from part (a).
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.40P
Two spheres with uniform surface charge density, one with a radius of 7.2 cm and the other with a radius of 4.7 cm, are separated by a center−to−center distance of 33 cm. The spheres have a combined charge of +55 μCand repel one another with a force of 0.75 N. What is the surface charge density on each sphere?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.41P
Point charges, q1 and q2, are placed on the x axis, with q1 at x = 0 and q2 at x = d. Athird point charge, +Q, is placed at x = 3d/4. if the net electrostatic force experienced by the charge +Q is zero, how are q1 and q2 related?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.42P
CE Two electric charges are separated by a finite distance. Somewhere between the charges, on the line connecting them, the net electric field they produce is zero, (a) Do the charges have the same or opposite signs? Explain, (b) If the point of zero field is closer to charge 1, is the magnitude of charge 1 greater than or less than the magnitude of charge 2? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.43P
What is the magnitude of the el ectric field produced by a charge of magnitude 7.50 μCat a distance of (a) 1.00 m and (b) 2.00 m?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.44P
A +5.0 μC charge experiences a 0.44−N force in the positive y direction. If this charge is replaced with a − 2.7μCcharge, what force will it experience?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.45P
Two point charges he on the x axis. A charge of +6.2 μC is at the origin, and a charge of −9.5 μCis at x = 10.0 cm. What is the net electric field at (a) x = −4.0 cm and at (b) x = +4.0 cm?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.46P
CE The electric field on the dashed Une in Figure 19−28 vanishes at infinity, but also at two different points a finite distance from the charges. Identify the regions in which you can find E = 0 at a finite distance from the charges: region 1, to the left of point A; region 2, between points A and B; region 3, between points B and C; region 4, to the right of point C.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.47P
An object with a charge of −3.6 μC and a mass of 0.012 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight, (a) Find the direction and magnitude of the electric field, (b) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.48P
IP Figure 19−33 shows a system consisting of three charges, q1 = +5.00 μC, q2 = +5.00 μC, and q3 = −5.00 μC, at the vertices of an equilateral triangle of side d = 2.95 cm. (a) Find the magnitude of the electric field at a point halfway between the charges q1 and q2. (b) Is the magnitude of the electric field halfway between the charges q2 and q3 greater than, less than, or the same as the electric field found in part (a)? Explain, (c) Find the magnitude of the electric field at the point specified in part (b).
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.49P
Two point charges of equal magnitude are 7.5 cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 45 N/C. Find the magnitude of the charges.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.50P
IP A point charge q1 = +4.7 μCis placed at each corner of an equilateral triangle with sides 0.21 m in length, (a) What is the magnitude of the electric field at the midpoint of any of the three sides of the triangle? (b) Is the magnitude of the electric field at the center of the triangle greater than, less than, or the same as the magnitude at the midpoint of a side? Explain.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.51P
IP Four point charges, each of magnitude q, are located at the corners of a square wi th sides of length a. Two of the charges are +q, and two aie −q. The charges are arranged in one of the following two ways: (1) The charges alternate in sign (+q, −q, +q, −q) as you go around the square; (2) the top two comers of the square have positive charges (+q, +q), and the bottom two corners have negative charges (−q, −q). (a) In which case will the electric field at the center of the square have the greatest magnitude? Explain, (b) Calculate the electric field at the center of the square for each of these two cases.
Solution:
.
Chapter 19 Electric Charges, Forces, and Fields Q.52P
The electric field at the point x = 5.00 cm and y = 0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x = 10.0 cm and y = 0 the electric field points in the positive x direction with a magnitude of 15.0 N/C. Assuming this electric field is produced by a single point charge, find (a) its location and (b) the sign and magnitude of its charge.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.53P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.54P
Make a qualitative sketch of the electric field lines produced by two equal positive charges, +q, separated by a distance d.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.55P
Make a qualitative sketch of the electric field lines produced by two charges, +q and −q, separated by a distance d.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.56P
Referring to Figure 19-34, suppose q2 is not known. Instead, it is given that q1 + q2 = −2.5 μC. Find q1 q2, and q3.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.57P
Make a qualitative sketch of the electric field lines produced by the four charges, +q, −q, +q, and −q, arranged clockwise on the four corners of a square with sides of length d.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.58P
Sketch the electric field lines for the system of charges shown in Figure 19-29.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.59P
Sketch the electric field lines for the system of charges described in Problem 35.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.60P
Suppose the magnitude of the electric field between the plates in Example 19-6 is changed, and a new object with a charge of −2.05 μC is attached to the string. If the tension in the string is 0.450 N, and the angle it makes with the vertical is 16°, what are (a) the mass of the object and (b) the magnitude of the electric field?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.61P
CE Predict/Explain Gaussian surface 1 has twice the area of Gaussian surface 2. Both surfaces enclose the same charge Q. (a) Ts the electric flux through surface 1 greater than, less than, or the same as the electric flux through surface 2? (b) Choose the best explanation from among the following:
I. Gaussian surface 2 is closer to the charge, since it has the smaller area. It follows that it has the greater electric flux.
II. The two surfaces enclose the same charge, and hence they have the same electric flux.
III. Electric flux is proportional to area. As a result, Gaussian surface 1 has the greater electric flux.
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.62P
CESuppose the conducting shell in Figure 19-25−which has a point charge +Q at its center−has a nonzero net charge. How much charge is on the inner and outer surface of the shell when the net charge of the shell is (a) −2Q, (b) −Q, and (c) +Q?
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.63P
Solution:
Chapter 19 Electric Charges, Forces, and Fields Q.64P
A uniform electric field of magnitude 25,000 N/C makes an angle of 37° with a plane surface of area 0.0153 m2, What is the electric flux through this surface?
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Chapter 19 Electric Charges, Forces, and Fields Q.65P
A surface encloses the charges q1 = 3.2 μC, q2 − 6.9 μC, and q3 = −4.1 μC. Find the electric flux through this surface.
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Chapter 19 Electric Charges, Forces, and Fields Q.66P
IPA uniform electric field of magnitude 6.00 × 103 N/C points upward. An empty, closed shoe box has a top and bottom that are 35.0 cm by 25.0 cm, vertical ends that are 25.0 cm by 20.0 cm, and vertical sides tha t are 20.0 cm by 35.0 cm. (a) Which side of the box has the greatest positive electric flux? Which side has the greatest negative electric flux? Which sides have zero electric flux? (b) Calculate the electric flux through each of the six sides of the box.
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Chapter 19 Electric Charges, Forces, and Fields Q.67P
BIO Nerve Cells Nerve cells are long, thin cylinders along which electrical disturbances (nerve impulses) travel. The cell membrane of a typical nerve cell consists of an inner and an outer wall separated by a distance of 0.10 μ m. The electric field within the cell membrane is 7.0 × 105 N/C. Approximating the cell membrane as a parallel-plate capacitor, determine the magnitude of the charge density on the inner and outer cell walls.
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Chapter 19 Electric Charges, Forces, and Fields Q.68P
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Chapter 19 Electric Charges, Forces, and Fields Q.69P
Consider a spherical Gaussian surface and three charges: q1 = 1.61 μC,q2 = −2.62 μC, and q3 = 3.91 μC. Find the electric flux through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges, (d) Suppose a fourth charge, Q, is added to the situation described in part (c). Find the sign and magnitude of Q required to give zero electric flux through the surface.
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Chapter 19 Electric Charges, Forces, and Fields Q.70P
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Chapter 19 Electric Charges, Forces, and Fields Q.71GP
CE Predict/Explain An electron and a proton are released from rest in space, far from any other objects. The particles move toward each other, due to their mutual electrical attraction. (a) When they meet, is the kinetic energy of the electron greater than, less than, or equal to the kinetic energy of the proton? (b) Choose the best explanation from among the following:
I. The proton has the greater mass. Since kinetic energy is proportional to mass, it follows that the proton will have the greater kinetic energy.
II. The two particles experience the same force, but the light electron moves farther than the massive proton. Therefore, the work done on the electron, and hence its kinetic energy, is greater.
III. The same force acts on the two particles. Therefore, they will have the same kinetic energy and energy will be conserved.
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Chapter 19 Electric Charges, Forces, and Fields Q.72GP
CE Predict/Explain In Conceptual Checkpoint 19−3, suppose the charge to be placed at cither point A or point B is +q rather than − q, (a) Is the magnitude of the net force experienced by the movable charge at point A greater than, less than, or equal to the magnitude of the net force at point B? (b) Choose the best explanation from among the following:
I. Point B is farther from the two fixed charges. As a result, the net force at point B is less than at point A. II. The net force at point A cancels, just as it does in Conceptual Checkpoint 19−3. Therefore, the nonzero net force at point B is greater in magnitude than the zero net force at point A.
III. The net force is greater in magnitude at point A because at that location the movable charge experiences a net repulsion from each of the fixed charges.
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Chapter 19 Electric Charges, Forces, and Fields Q.73GP
CE An electron (charge = −e) orbits a helium nucleus (charge = +2e). Is the magnitude of the force exerted on the he− Hum nucleus by the electron greater than, less than, or the same as the magnitude of the force exerted on the electron by the helium nucleus? Explain.
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Chapter 19 Electric Charges, Forces, and Fields Q.74GP
CE In the operating room, technicians and doctors must take care not to create an electric spark, since the presence of the oxygen gas used during an operation increases the risk of a deadly fire. Should the operating−room personnel wear shoes that are conducting or non-conducting? Explain.
Solution:
The operating room personnel should wear shoes that are conducting so that any charge they transform while walking, can flow into the ground.
Chapter 19 Electric Charges, Forces, and Fields Q.75GP
CE Under normal conditions, the electric field at the surface of the Earth points downward, into the ground. What is the sign of the electric charge on the ground?
Solution:
We know that electric field lines point in the direction of negative charges.So the charge on Earth must be negative
Chapter 19 Electric Charges, Forces, and Fields Q.76GP
CE Two identical spheres are made of conducting material. Initially, sphere 1 has a net charge of +35Q and sphere 2 has a net charge of −26Q. If the spheres are now brought into contact, what is the final charge on sphere 1? Explain.
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Chapter 19 Electric Charges, Forces, and Fields Q.77GP
CE A Gaussian surface for the charges shown in Figure 19−35 has an electric flux equal to +3q/ε0. Which charges arc contained within this Gaussian surface?
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Chapter 19 Electric Charges, Forces, and Fields Q.78GP
A proton is released from rest in a uniform electric field of magnitude 1.08 × 105 N/C. Find the speed of the proton after it has traveled (a) 1.00 cm and (b) 10.0 cm.
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Chapter 19 Electric Charges, Forces, and Fields Q.79GP
BIO Ventricular Fibrillation If a charge of 0.30 C passes through a person’s chest in 1.0 s, the heart can go into ventricular fibrillation−a nonrhythmic “fluttering” of the ventricles that results in little or no blood being pumped to the body. If this rate of charge transfer persists for 4.5 s, how many electrons pass through the chest?
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Chapter 19 Electric Charges, Forces, and Fields Q.80GP
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Chapter 19 Electric Charges, Forces, and Fields Q.81GP
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Chapter 19 Electric Charges, Forces, and Fields Q.82GP
Find (a) the direction and (b) the magnitude of the net electric field at the center of the equilateral triangle in Figure 19−31. Give your answers in terms of the angle θ, as defined in Figure 19−31, and E, the magnitude of the electric field produced by any oneof the charges at the center of the triangle.
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Chapter 19 Electric Charges, Forces, and Fields Q.83GP
At the moment, the number of electrons in your body is essentially the same as the number of protons, giving you a net charge of zero. Suppose, however, that this balance of charges is off by 1% in both you and your friend, who is 1 meter away. Estimate the magnitude of the electrostatic force each of you experiences, and compare it with your weight.
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Chapter 19 Electric Charges, Forces, and Fields Q.84GP
A small object of mass 0.0150 kg and charge 3.1 μChangs from the ceiling by a thread. A second small object, with a charge of 4.2 μC, is placed 1.2 m vertically below the first charge. Find (a) the electric field at the position of the upper charge due to the lower charge and (b) the tension in the thread.
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Chapter 19 Electric Charges, Forces, and Fields Q.85GP
IP Consider a system of three point charges on the x axis. Charge 1 is at x = 0, charge 2 is at x = 0.20 m, and charge 3 is at x = 0.40 m. In addition, the charges have the following values: q1 = −19 μC, q2 = q3 = +19 μC(a)The electric field vanishes at some point on the x axis between x = 0.20 m and x = 0.40 m. Ts the point of zero field (i) at x = 0.30 m, (ii) to the left of x = 0.30 m, or (iii) to the right of x = 0.30 m? Explain, (b) find the point where E = 0 between x = 0.20 m and x = 0.40 m.
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Chapter 19 Electric Charges, Forces, and Fields Q.86GP
IP Consider the system of three point charges described in the previous problem, (a) The electric field vanishes at two different points on the.x axis. One point is between x = 0.20 m and x = 0.40 m. Is the second point located to the left of charge 1 or to the right of charge 3? Explain, (b) Find the value of x at the second point where E = 0.
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Chapter 19 Electric Charges, Forces, and Fields Q.87GP
The electric field at a radial distance of 47.7 cm from the thin charged wire shown in Figure 19−36 has a magnitude of 35,400 N/C. (a) Using the result given in Problem 70, what is the magnitude of. the charge per length on this wire? (b) At what distance from the wire is the magnitude of the electric field equal to ?
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Chapter 19 Electric Charges, Forces, and Fields Q.88GP
A system consisting entirely of electrons and protons has a net charge of 1.84 × 10−15 C and a net mass of 4.56 × 10−23 kg. How many (a) electrons and (b) protons are in this system?
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Chapter 19 Electric Charges, Forces, and Fields Q.89GP
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Chapter 19 Electric Charges, Forces, and Fields Q.90GP
IP Consider the system of three charges described in the previous problem and shown in Figure 19−38. (a) Do you expect the net force acting on charge 1 to have a magnitude greater than, less than, or the same as the magnitude of the net force acting on charge 2? Explain, (b) Find the magnitude of the net force acting on charge 1. (c) Find the magnitude of the net force acting on charge 2.
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Chapter 19 Electric Charges, Forces, and Fields Q.91GP
IP BIO Cell Membranes The cell membrane in a nerve cell has a thictcness of 0.12μ m. (a) Approximating the cell membrane as a parallel−plate capacitor with a surface charge density of 5.9 × 10−6 C/m2, find the electric field within the membrane, (b) If the thickness of the membrane were doubled, would your answer to part (a) increase, decrease, or stay the same? Explain.
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Chapter 19 Electric Charges, Forces, and Fields Q.92GP
A square with sides of length L has a point charge at each of its four corners. Two corners that are diagonally opposite have charges equal to +2.25 μC; the other two diagonal comers have charges Q. Find the magnitude and sign of the charges Q such that each of the +2.25μC charges experiences zero net force.
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Chapter 19 Electric Charges, Forces, and Fields Q.93GP
IP Suppose a charge +Q is placed on the Earth, and another charge +Q is placed on the Moon. (a) Find the value of Q needed to “balance” the gravitational attraction between the Earth and the Moon. (b) How would your answer to part (a) change if the distance between the Earth and the Moon were doubled? Explain.
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Chapter 19 Electric Charges, Forces, and Fields Q.94GP
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Chapter 19 Electric Charges, Forces, and Fields Q.95GP
A small sphere with a charge of +2.44 μCis attached to a relaxed horizontal spring whose force constant is 89.2 N/m. The spring extends along the x axis, and the sphere rests on a frictionless surface with its center at the origin. A point charge Q = −8.55 μCis now moved slowly from infinity to a point x = d > 0 on the x axis. This causes the small sphere to move to the position x = 0.124 m. Find d.
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Chapter 19 Electric Charges, Forces, and Fields Q.96GP
Twelve identical point charges q are equally spaced around the circumference of a circle of radius R. The circle is centered al the origin. One of the twelve charges, which happens to be on the positive x axis, is now moved to the center of the circle. Find (a) the direction and (b) the magnitude of the net electric force exerted on this charge.
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Chapter 19 Electric Charges, Forces, and Fields Q.97GP
BIO Nerve Impulses When a nerve impulse propagates along a nerve cell, the electric field within the cell membrane changes from 7.0 × 105 N/C inone direction to 3.0 × 105 N/C in the other direction. Approximating the cell membrane as a parallel−plate capacitor, find the magnitude of the change in charge density on the walls of the cell membrane.
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Chapter 19 Electric Charges, Forces, and Fields Q.98GP
IP The Electric Field of the Earth The Earth produces an approximately uniform electric field at ground level. This electric field has a magnitude of 110 N/C and points radially inward, toward the center of the Earth. (a) Find the surface charge density (sign and magnitude) on the surface of the Earth. (b) Given that the radius of the Earth is 6.38 × 106 m, find the total electric charge on the Earth. (c) If the Moon had the same amount of electric charge distributed uniformly over its surface, would its electric field at the surface be greater than, less than, or equal to 110 N/C? Explain.
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Chapter 19 Electric Charges, Forces, and Fields Q.99GP
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Chapter 19 Electric Charges, Forces, and Fields Q.100GP
Four identical charges, +Q, occupy the corners of a square with sides of length a. A fifth charge, q, can be placed at any desired location. Find the location of the fifth charge, and the value of q, such that the net electric force acting on each of the original four charges, +Q, is zero.
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Chapter 19 Electric Charges, Forces, and Fields Q.101GP
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Chapter 19 Electric Charges, Forces, and Fields Q.102GP
Two identical conducting spheres are separated by a fixed center−to−center distance of 45 cm and have different charges. Initially, the spheres attract each other with a force of 0.095 N. The spheres are now connected by a thin conducting wire. After the wire is removed, the spheres are positively charged and repel one another with a force of 0.032 N. Find (a) the final and (b) the initial charges on the spheres.
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Chapter 19 Electric Charges, Forces, and Fields Q.103PP
How many electrons must be transferredaway from a bee to produce a charge of +93.0 pC?
A. 1.72 × 10−9
B. 5.81 × 108
C. 1.02 × 1020
D. 1.49 × 1029
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Chapter 19 Electric Charges, Forces, and Fields Q.104PP
Suppose two bees, each with a charge of 93.0 pC, are separated by a distance of 1.20 cm. Treatingthe bees as point charges, what is the magnitude of the electrostatic force experienced by the bees? (In comparison, the weight of a 0.140g bee is 1.37 × 10−3 N.)
A. 6.01 × 10−17 N
B. 6.48 × 10−9 N
C. 5.40 × 10−7N
D. 5.81 × 10−3N
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Chapter 19 Electric Charges, Forces, and Fields Q.105PP
The force required to detach a grain of pollen from an avocado stigma is approximately 4.0 × 10−5 N. What is the maximum distance at which the electrostatic force between a bee and a grain of pollen is sufficient to detach the pollen? Treat the bee and pollen as point charges, and assume the pollen has a charge opposite in sign and equal in magnitude to the bee.
A. 4.7 × 10−7 m
B. 1.9 mm
C. 4.4 cm
D. 220 m
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Chapter 19 Electric Charges, Forces, and Fields Q.106PP
The Earth produces an electric field of magnitude 110 N/C. What force does this electric field exert on a bee carrying a charge of 93.0 pC? (Again, for comparison, the weight of a bee is approximately 1.37 × 10−3 N.)
A. 1.76 × 10−17N
B. S.45 × 10−13 N
C. 1.02 × 10−8 N
D. 1.13 × 10−6 N
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Chapter 19 Electric Charges, Forces, and Fields Q.107IP
IP Referring to Example 19−5 Suppose q1 = +2.90μC is no longer at the origin, but is now on the y axis between y = 0 and y = 0.500 m. Thecharge q2 = +2.90 μCis at x = 0 and y = 0.500 m, and point 3 is at x = y = 0.500 m. (a) Is the magnitude of the net electric field at point 3, which we call Enct, greater than, less than, or equal to its previous value? Explain. (b) Is the angle θ that Enct makes with the x axis greater than, less than, or equal to its previous value? Explain. Find the new values of (c) Enct and (d) θ if q1 is at y = 0.250 m.
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Chapter 19 Electric Charges, Forces, and Fields Q.108IP
IP Referring to Example 19−5 In this system, the charge q1 is at the origin, the charge q2 is at x = 0 and y = 0.500 m, and point 3 is at x = y = 0.500 m. Suppose that q1 = + 2.90 μC, but that q2 isincreased to a value greater than +2.90 μC. As a result, do (a) Enct and (b) θ increase, decrease, or stay the same? Explain. If Enct = 1.66 × 105 N/C, find (c) q2 and (d) θ.
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Chapter 19 Electric Charges, Forces, and Fields Q.109IP
IP Referring to Example 19−6 The magnitude of the charge is changed until the angle the thread makes with the vertical is θ = 15.0°. The electric field is 1.46 × 104 N/C and the mass of the object is 0.0250 kg. (a) Is the new magnitude of q greater than or less than its previous value? Explain. (b) Find the new value of q.
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Chapter 19 Electric Charges, Forces, and Fields Q.110IP
Referring to Example 19−6 Suppose the magnitude of the electric field is adjusted to give a tension of 0.253 N in the thread. This will also change the angle the thread makes with the vertical. (a) Find the new value of E. (b) Find the new angle between the thread and the vertical.
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