Sapling Questions
A certain lightning bolt moves 50.0 C of charge. How many units 𝑁 of fundamental charge 𝑒 is this?
In order to calculate the units 𝑁 of fundamental charge in a lightning bolt, use the relationship 𝑁=𝑄/𝑒 where 𝑄 is the total charge of the bolt and 𝑒 is the fundamental charge. 𝑒 = 1.60 × 10^−19 C Then, plugging in the numbers gives 𝑁 = 50.0 C / (1.60 × 10^−19 C) = 3.13 × 10^20
When a certain wire is connected across a voltage of 1 V, the resulting current through the wire is 1 A. If the same wire is connected across a voltage of 2 V, the current will be
2A Explanation: Ohm's Law: V = iR
When a second identical bulb is added in series to a circuit with a single bulb, the resistance of the circuit:
Doubles Explanation: When resistors are in series with one another, their resistances add directly to one another. Therefore, adding a second identical bulb in series will double the resistance.
On which variable(s) does the magnetic flux depend?
All of these: -the magnetic field -the area of a region through which magnetic field passes -the orientation of the field with respect to the region through which it passes Explanation: ΦB=𝐵𝐴cos(𝜃)
Consider a rubber rod that has been rubbed with fur to give the rod a net negative charge, and a glass rod that has been rubbed with silk to give it a net positive charge. After being charged by contact by the fur and silk:
The rubber rod has more mass and the glass rod has less mass. Explanation: Both electrons and protons have mass, but only electrons are transferred when an object is charged by contact. A negatively charged object has an excess of electrons, whereas a positively charged object has a deficit of electrons. The rubber rod acquired electrons to become negatively charged, so it gained mass; the glass rod shed electrons to become positively charged, so it lost mass.
Consider the two cases shown. In both cases, a central test charge 𝑞 has two source charges of equal magnitude fixed at equal distances above and below it. In Case 1, the two outer charges are oppositely charged. In Case 2, both charges are positive. The sign of the test charge 𝑞 in the center is not given. In which case is the magnitude of the net force on the center charge bigger?
Case 1 Explanation: The symmetric placement of the source charges to either side of the test charge means that, regardless of the sign of the test charge, Case 1 produces the larger net force on it. Consider Case 2. Because both source charges are positive but are placed to either side of the test charge, they produce forces on the test charge that are equal in magnitude but opposite in direction. The net force on the test charge 𝑞q in Case 2 is zero. For Case 1, the magnitude of the net force on the test charge is always greater than zero. If the test charge is negative, it is attracted to the positive charge and repelled by the negative charge; if the test charge is positive, the opposite occurs. Therefore, the individual forces add to produce a non-zero net force.
The figures show snapshots of a rectangular loop as it passes from left to right through a region of constant magnetic field. The field points into the screen and is perpendicular to the plane of the loop. In which of the sequences is the magnetic flux through the loop decreasing?
D, leaving the magnetic field Explanation: The magnetic flux is proportional to the number of magnetic field lines passing through the loop. In sequence D, the magnetic flux is decreasing, since the number of magnetic field lines is also decreasing.
A parallel‑plate capacitor is connected to a battery until it is fully charged. Then, the charged capacitor is disconnected from the battery and connected to an uncharged parallel‑plate capacitor, as shown. When connected to the uncharged capacitor, the potential between the plates of the initially charged capacitor will:
Decrease Explanation: Connecting the charged capacitor to the uncharged capacitor creates a potential difference between connected conducting plates. Charge flows until electrostatic equilibrium is reestablished, which occurs when each pair of connected plates is at a single potential (i.e., when the same voltage exists between both capacitors). The voltage 𝑉 across the plates of a capacitor is related to the magnitude of charge 𝑞 on each plate and the capacitance 𝐶 of the capacitor. 𝑉=𝑞/𝐶 Capacitance depends upon the geometry of the capacitor, which remains constant. Because the initial capacitor loses some of its initial charge, the voltage between its plates decreases.
Two charges, 𝑞=+1 μC and 𝑄=+10 μC, are placed near each other. Which of the diagrams best depicts the forces acting on the charges?
Equal repelling forces Explanation: The charges exert forces on each other, and so form a Newton's third law pair. According to Newton's third law, the forces the charges exert on each other must be equal and opposite. Because both charges share the same sign, the forces are repulsive.
Three charges are fixed in place as shown. The distance between charges +𝑞 and −𝑞 is equal to the distance between charges −3𝑞 and −𝑞. Identify which of the statements best describes the magnitudes of the forces exerted on charge −𝑞 by charges +𝑞 and −3𝑞.
Explanation: Coulomb's law states that the magnitude of the force is proportional to the product of the charges and inversely proportional to the square of the distance between the charges. In the problem, both source charges +𝑞+q and −3𝑞−3q are the same distance from the test charge −𝑞−q , so the difference in the force magnitudes depends only on the size of the charges. Because ∣−3𝑞∣>∣+𝑞∣∣−3q∣>∣+q∣ , the −3𝑞−3q charge exerts a force of larger magnitude on charge −𝑞−q .
Metal sphere A has a charge of −𝑄. An identical metal sphere B has a charge of +2𝑄. The magnitude of the electric force on sphere B due to sphere A is 𝐹. The magnitude of the electric force on sphere A due to sphere B must be:
F Explanation: Newton's third law of motion states that, in any interaction between two objects, the forces exerted on each object by the other are equal in magnitude and opposite in direction. So, the force exerted on sphere A by sphere B is the same as the force exerted on sphere B by sphere A. If the force on sphere B due to sphere A is 𝐹,F, then the force on sphere A due to sphere B must also be 𝐹.F. This can also be seen by examining Coulomb's law, which states that the force between two charges varies directly with the product of the charges. Since the electric force on each charge is proportional to the product of the two charges, the magnitude of the force must be the same for both charges (because 𝑞1 × 𝑞2 = 𝑞2 × 𝑞1)
How is it possible for an object that experiences no net magnetic force to experience a net magnetic torque?
If equal but opposite magnetic forces act at separate points on an object, and do not lie along the same line, then the forces will sum to zero but the torques generated by them will not. Explanation: Any object made to rotate without translating has experienced a net torque without a net force. If an object is free to rotate, and equal and opposite forces that do not lie along the same line act at separate points on the object, then there will be zero net force but nonzero net torque.
A parallel‑plate capacitor is made of two conducting plates of area 𝐴 separated by a distance 𝑑. The capacitor is initially connected to a battery until a charge 𝑄 is established on the capacitor. The battery is then disconnected from the plates, and the separation between the plates is doubled. The electric potential energy stored in the capacitor will
Increase Explanation: The capacitance C of a capacitor is determined by the geometrical properties of the capacitor's plates and the material between them. For a parallel-plate capacitor without a dielectric, the capacitance is inversely proportional to the plate separation distance d . Thus, increasing the plate separation decreases the capacitance. The potential energy 𝑈electric stored in a capacitor is related to the magnitude of charge 𝑄 on each plate and the capacitance 𝐶. 𝑈electric=𝑄22𝐶 The charge 𝑄 remains constant while 𝑑 is increased and 𝐶 decreases, therefore the energy stored in the capacitor 𝑈electric increases. This result is understood in the context of work. Because each plate is oppositely charged and electrically attracted to the other plate, work is required to further separate the charges. The energy supplied during the work done on the capacitor is then stored in the capacitor in the form of electric potential energy.
Suppose the distance between the plates of a parallel‑plate capacitor is increased without changing the amount of charge stored on the plates. What will happen to the energy stored in the capacitor?
It increases Explanation: This result is understood in the context of work. Because each plate is oppositely charged and electrically attracted to the other plate, work is required to further separate the charges. The energy supplied during the work done on the capacitor is then stored in the capacitor in the form of electric potential energy.
If a magnetic field exerts a force on moving, charged particles, is it capable of doing work on the particles? Explain your answer.
No, the magnetic force always acts perpendicularly to the displacement of each moving, charged particle, so the work done on each particle is always zero. (when force and displacement are perpendicular, no work is done.)
A spherical Gaussian surface is drawn inside a spherical object, as shown. Upon inspection, no net charge exists within the Gaussian surface. What can you conclude from this result?
None of the above - correct answer Choices: The object is a conductor. The object contains no net charge. The charge density throughout the object is uniform. The object is hollow with a vacuum interior. There is no E-field within the Gaussian sphere. The E-field is constant within the Gaussian sphere. None of the choices listed are correct.
An electroscope is an apparatus used to detect electric charge. It consists of a flat plate attached to a stationary, vertical support; attached to the vertical support is a freely‑rotating needle. The plate, the support, and the needle are all made of metal, and are affixed to an insulating container as shown in the figure. After a balloon is rubbed on a sleeve, the balloon is brought near to the metallic plate of an electroscope without touching it. As a result, the rotating needle of the electroscope deflects away from the stationary, vertical support. Identify the sign of the charge 𝑄p on the plate of the electroscope, and the sign of the charge 𝑄n on the rotating needle.
Plate is positive, needle is negative Explanation: When the negatively charged balloon is brought close to the electroscope plate, it repels the free electrons in the plate, leaving the plate positively charged, 𝑄p>0Qp>0 . The repelled electrons flow down into the rotating needle apparatus, causing the entire appratus to become negatively charged. Therefore, the needle acquires a charge 𝑄n<0Qn<0 and is repelled away from the negatively charged support.
A positively charged rod is brought near one end of an uncharged metal bar. The end of the metal bar farthest from the charged rod will be charged:
Positively
Which color of light, red or blue, travels faster in crown glass?
Red Explanation: The different colors of light correspond to different wavelengths, and the speed of light in a medium is inversely proportional to the medium's index of refraction. The index of refraction itself depends on the light's wavelength, being typically inversely proportional to the wavelength. Red light, which has the longer wavelength, has a lower refractive index in crown glass and so travels faster through the glass.
A solid sphere of radius 𝑅 is made of a metallic conductor. A hollow spherical shell of the same radius 𝑅 is made of the same conducting material. An excess charge 𝑄 is deposited on each. Which object has the greatest surface charge density?
Same in both cases Explanation: One property of conductors is that charge distributes itself on the surface of the conductor in such a way that the electric field inside the conductor is zero. In the case of a symmetric, closed surface like a sphere, that property requires that the charge be uniformly distributed on the outer surface. Therefore, the interior structure of a spherical conductor has no effect on the charge distribution, and both spheres have the same surface charge density.
How does a person become "charged" as he or she shuffles across a carpet with bare feet on a dry winter day?
Shuffling across the carpet physically transfers electrons from the person to the carpet, producing a net positive charge on the person. Explanation: Human skin gives up electrons much more readily than carpet. Shuffling across the floor results in a transfer of negative charge from the person's skin to the carpeted floor, leaving the person with a net positive charge. Because the air is dry there are few airborne water molecules, making the air less conductive and therefore less able to neutralize the static charge buildup.
You are given three iron rods. Two of the rods are magnets but the third one is not. How could you use the three rods to definitively identify the unmagnetized rod?
Test the rods one pair at a time in each possible end‑to‑end orientation until finding a pair that repels; the leftover rod is unmagnetized.
The change in potential energy and the change in electric potential will only be opposite for _____________ charges.
The change in potential energy and the change in electric potential will only be opposite for negative charges.
A parallel‑plate capacitor is made of two conducting plates of area 𝐴A separated by a distance 𝑑. The capacitor is initially connected to a battery until a charge 𝑄 is established on the capacitor. The battery is then disconnected from the plates, and the separation between the plates is doubled. Which of the following remains constant?
The charge on the capacitor Explanation: Charge gathers on the plates of the capacitor when it is initially connected to the battery. Charge conservation requires that any charge flowing off the capacitor's plates must flow somewhere else. Disconnecting the battery removes the path through which charge can flow. Therefore, the charge on the capacitor remains constant.
Why is the gravitational force usually ignored in problems involving particles such as electrons and protons?
The electric force between charged particles is much greater than the gravitational force.
A current‑carrying wire lies in a region where there is an external magnetic field, but there is no magnetic force acting on the wire. How can this be?
The length of the wire is oriented either parallel or antiparallel to the magnetic field lines at the location of the wire. Explanation: Use the equation that describes the magnetic force F acting on a wire of length ℓ carrying a current 𝑖 in a region of magnetic field of magnitude 𝐵. 𝐹=𝑖ℓ𝐵sin𝜃 In the equation, 𝜃 is the angle between the direction of current flow and the direction of the magnetic field. There are three scenarios in which the magnetic force on the wire equals zero. There is no current in the wire. There is no magnetic field. The direction of current flow is parallel or antiparallel to the direction of the magnetic field. The first two scenarios are trivial. The final scenario describes the only way a current‑carrying wire can lie in a region of nonzero magnetic field and not experience a magnetic force.
A conducting loop moves at a constant speed parallel to a long, straight, current‑carrying wire, as shown in the figure. Which statement is true?
There will be no induced current in the loop. Explanation: Current is induced in a wire loop as a result of a change in magnetic flux through the loop. In this problem, there is a long, straight wire, so the magnetic field is uniform along a path parallel to the wire. There is also a constant current in the wire. Therefore, as the loop moves parallel to the wire, the magnetic flux through the loop remains constant. A change in magnetic flux is necessary to induce a current in a wire loop. However, in this case, the loop has the same magnetic flux through it at all points in its motion as it travels parallel to the wire. This constant flux induces no current.
In what way or ways would the physical universe be different if protons were negatively charged and electrons were positively charged?
There would be no difference other than the sign on the charged particles.
An electron is moving northward in a magnetic field. The magnetic force on the electron is toward the northeast. What is the direction of the magnetic field?
This situation cannot exist because of the relative orientations of the velocity and force vectors. Explanation: The direction of the magnetic force on a moving charge is perpendicular both to the velocity of the charge and to the magnetic field. The right‑hand rule is a useful mnemonic device for keeping track of the relative directions. In the situation described, however, the magnetic force is not perpendicular to the velocity of the charge. Therefore, no matter what direction the field is pointing, this scenario is impossible.
A free positive charge released in an electric field will
accelerate in the direction in which the electric field is pointing.
A free negative charge released in an electric field will
accelerate in the direction opposite the electric field.
If two uncharged objects are rubbed together and one of them acquires a negative charge, then the other one
acquires a positive charge
A balloon can be charged by rubbing it with the sleeve of your sweater while holding it in your hand. You can conclude from this that a balloon is:
an insulator Explanation: Any excess charge on the balloon will repel the other excess charges on the balloon. The fact that the balloon remains charged for a significant period of time indicates that charges on the surface of the balloon cannot move around freely. If charge could move around freely on the surface of the balloon, any excess charge would be repelled onto your body via your hand and would quickly be neutralized by the opposite charge accumulated on your sweater. An object through which charges cannot move freely is classified as an insulator. You can tell that the balloon is an insulator because any excess charge on the balloon tends to stay where it has been put rather than spreading out.
The copper ring of radius 𝑅R in the figure lies in a magnetic field pointing into the screen. The field is uniformly decreasing in magnitude. The induced current in the ring is
clockwise Explanation: Use Lenz's law to determine the direction in which the induced emf drives the current. Remember that Lenz's law states that induced current flows in the direction that generates a magnetic field that opposes the change in flux. In this case, the magnetic field is into the screen and decreasing, so the induced emf will drive a current that produces a secondary magnetic field that also points into the screen in order to resist the decrease in flux. The right‑hand rule for current loops indicates that a clockwise current produces a magnetic field into the screen.
Two metal rings with a common axis are placed near each other, as shown in the figure. If current 𝑖A is suddenly set up and is increasing in ring A as shown, the current in ring B will be
antiparallel to 𝑖A. Explanation: The changing current in ring A results in a change in the magnetic flux through ring B, which results in an induced emf (ℰ) according to Faraday's law. The direction of this induced emf can be most easily determined by applying Lenz's law, which states that the induced current will be in such a direction as to produce a secondary magnetic field that opposes the change in flux that induced it. First, determine what is happening at ring B due to ring A. As the current in ring A increases, the magnetic flux through ring B is in the direction pointing from right to left and increasing. This change in flux results in a generated emf, which drives a current to counter this change of flux. Ring B will therefore have an induced current in the opposite direction of that in ring A, or antiparallel to 𝑖A.
Electric charges of opposite sign
attract each other
Positively charged particle trajectories ____________________ follow electric field lines, because _________________________________.
can but do not have to; the particle velocities may or may not be in the same direction as the electric field lines.
When a bulb is added in parallel to a circuit with a single bulb, the resistance of the circuit
decreases Explanation: 1/R12 = 1/R1 + 1/R2
An isolated, parallel‑plate capacitor carries a charge Q. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be
doubled Explanation: Since the capacitor is isolated, the charge 𝑄 remains constant. Since the plate separation is doubled, the capacitance is halved. Because the capacitance is halved and the capacitor is isolated, the voltage 𝑉 across the plates is doubled.
Which wave types refract when crossing from one medium to another?
electromagnetic, sound, and water waves Explanation: Refraction is a change in the direction of wave propagation that occurs at the boundary between two media when the speed of the wave in the two media is different. Since all of the listed wave types have speeds that depend on the material through which they propagate, they all undergo refraction.
The electric potential, when measured at a point equidistant from two particles that have charges equal in magnitude but of opposite sign, is
equal to zero. Explanation: The potential at any point is the sum of all the potentials at that point. Since the two potentials are equal in magnitude but opposite in sign, the sum will be zero.
If a negative charge is released in a uniform electric field, it will move:
from low potential to high potential. Explanation: A negative charge moves opposite of the direction of an electric field. Electric fields point in the direction of decreasing potential. Therefore, a negative charge will move in the direction of increasing potential.
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be
halved Explanation: C is inversely proportional to d. When the separation between the plates is doubled, the capacitance is halved. U = 1/2 CV^2 so halving C will cut the U in half too.
A parallel‑plate capacitor is connected to a battery until it is fully charged. Then, the capacitor is disconnected from the battery. Which action will reduce the magnitude of the electric field between the plates of the capacitor?
inserting a dielectric between the plates of the capacitor Explanation: Disconnecting the charged capacitor keeps its charge fixed. Therefore, decreasing the area of the plates increases their surface charge density. Since the electric field magnitude is proportional to the surface charge density, decreasing the area increases the electric field magnitude. Decreasing the distance between the plates of the capacitor does not change the area of the plates. Therefore, the surface charge density does not change and the electric field remains constant. Inserting a dielectric with a dielectric constant 𝜅κ between the plates of the capacitor does change the electric field to a new value 𝐸. A dielectric's molecules are polar, or are polarized by an external electric field. Consequently, the dielectric is polarized when inserted between the charged plates. The effect of the polarization of the dielectric is modeled by assuming that the dielectric has a net negative charge on the surface adjacent to the positively charged plate, and has a net positive charge on the surface adjacent to the negatively charged plate. The two layers of induced charge give rise to an electric field inside the dielectric that opposes the electric field of the capacitor plates. Thus, the net electric field is reduced. The size of the reduction depends on the value of the dielectric constant 𝜅, where 𝜅>1 for all physical materials.
The magnetic force on a charged moving particle
is described by all of these statements: depends on the sign of the charge on the particle. depends on the magnetic field at the particle's instantaneous position. is in the direction which is mutually perpendicular to the direction of motion of the charge and the direction of the magnetic field. is proportional both to the charge and to the magnitude of the magnetic field.
A positive charge is moved from one point to another point along an equipotential surface. The work required to move the charge
is zero Explanation: An equipotential surface is a surface that has the same potential at every point. The work required to move any charge from one point on an equipotential surface to any other point on the same equipotential surface is zero because work is equal to the change in potential energy of the charge. Since there is no change in potential on an equipotential surface, the potential energy of the charge also remains constant, and therefore the work must be zero.
For a positive charge moving in the direction of the electric field,
its potential energy decreases and its electric potential decreases. Explanation: When a positive charge moves in the direction of the electric field, the field does work on the charge. In other words, potential energy decreases as a positive charge moves in the direction of an electric field and increases as it moves against the direction of an electric field. The electric potential at some location is the potential energy per unit charge at that location. When the charge is positive, the electric potential has the same sign as the potential energy, so when the potential energy increases, the electric potential increases. When the potential energy decreases, the electric potential decreases. If the charge were negative, the electric potential would increase when the potential energy decreased and the electric potential would decrease when the potential energy increased.
A proton with a velocity in the positive 𝑥‑direction enters a region where there is a uniform magnetic field 𝐵 in the positive 𝑦‑direction. You want to balance the magnetic force with an electric force so that the proton will continue along a straight line. The electric field should be in the:
negative 𝑧‑direction. Explanation: The right hand rule tells us that the force from the magnetic field would send the proton upward (in the positive z direction), so the electric field would need to counteract that force to keep it moving straight on its path. - A positive charge experiences an electric force along the direction of the electric field, so an electric field in the -z direction would counteract it.
A parallel‑plate capacitor is made of two circular conducting plates separated by a distance d . The capacitor carries a charge Q. If the plates are reshaped into identical squares while keeping the thickness of the plates and the separation distance constant, how does the capacitance change?
no change in capacitance Explanation: The expression for capacitance is valid as long as the area of the plates is much larger than the plate separation distance, regardless of the plate geometry. Reshaping the plates from a circular shape into a square shape without changing the plate thickness implies that the area of the plates remains constant. Therefore, there is no change in capacitance.
An equipotential surface must be
perpendicular to the electric field at every point. Explanation: An equipotential surface is a surface that has the same potential at every point. In order for the potential to be the same at every point on the surface, there must be no work required to move a charge between any two points on the surface. The only possible way to move a charge in an electric field without doing work is to move the charge perpendicularly to the field. Therefore, the equipotential surface must be perpendicular to the electric field at every point.
A long solenoid carries a current. If the radius of the solenoid were doubled, and all other quantities remained the same, the magnetic field inside the solenoid would
remain the same Explanation: Magnetic field of solenoid: B = u0ni It doesn't depend on the radius.
Suppose that part of a wire in a circuit is only half the diameter of the rest of the wire. Compared to the rest of the wire, the current in the thinner part is:
the same Explanation: Current is the amount of charge moving past any point in the circuit per unit of time. Much like fluid flowing in a pipe, the amount of charge moving through the circuit is constant. Moving charges do not "pile up" or accumulate at any point in the circuit because electrostatic forces ensure that current remains constant throughout the wire in the steady state. Nor can charges be added or removed from the flow of charge since the circuit is a closed system. Therefore, the value of the current is the same everywhere in the circuit.
A solid sphere of radius 𝑅 is made of a metallic conductor. Another solid sphere of radius 𝑅 is made of an insulating material. Each sphere holds an excess charge 𝑄. Assume the insulating sphere has a uniform charge distribution. Which object has the greatest surface charge density?
the solid metal sphere Explanation: A property of conductors is that excess charges distribute themselves on the surface in such a way that the electric field inside the conductor is zero. A property of insulators is that excess charge may be distributed throughout the volume of the insulator, and the excess charges do not move. Because all of the charge on the conducting sphere is concentrated on the surface, whereas the charge on the insulator is distributed uniformly throughout its volume, the conducting sphere has the greater surface charge density.
When a certain charged particle is placed at the center of a sphere, the net flux through the surface of the sphere is Φ0. If the radius of the sphere is doubled, what will be the new net flux through the sphere?
Φ0 Explanation: According to Gauss' law, the electric flux Φ through a closed surface depends only on the charge enclosed by the surface, 𝑞enc, and the constant ε0. The electric flux is independent of the geometry of the enclosing surface, therefore the flux remains Φ0.
