Chapter 16: Electric Charge And Electric Field

A neutral atom is one that has an equal number of protons and electrons. Protons have a positive charge, while electrons have a negative charge. The charges of protons and electrons cancel each other out, resulting in a neutral overall charge for the atom. Neutrons, on the other hand, have no charge and do not affect the overall charge of the atom. Therefore, a neutral atom will always have the same number of protons as electrons.

Conductors are materials in which electrons are bound very loosely to the nuclei and can move about freely within the material. This allows for the easy flow of electric current through the material. Insulators, on the other hand, have tightly bound electrons that do not move easily, while semiconductors have properties in between conductors and insulators. Superconductors are materials that exhibit zero electrical resistance at very low temperatures.

An atom having more electrons than protons indicates that it has gained extra electrons, resulting in a negative charge. This excess of negatively charged electrons makes the atom a negative ion.

When the charges on both objects are doubled, the force between them is directly proportional to the product of the charges. Since both charges are doubled, the product of the charges is quadrupled. Therefore, the force between them also quadruples.

The correct answer is that the model of the atom shows a nucleus consisting of both protons and neutrons, surrounded by a cloud of electrons. This is based on the current understanding of atomic structure, where the nucleus contains the positively charged protons and neutral neutrons, while the negatively charged electrons orbit around the nucleus in energy levels or electron clouds.

The magnitude of the electric force between two point charges can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are 2.0 and -4.0 μC, and the distance between them is 1.5 cm. Plugging these values into the equation, we can calculate the magnitude of the electric force to be 320 N.

A proton carries a positive charge because it is one of the fundamental particles that make up an atom's nucleus. It has a charge of +1 elementary charge, which is equal in magnitude but opposite in sign to the electron's charge. This positive charge is what allows protons to attract and bind with negatively charged electrons, forming the structure of an atom.

When a second electron is placed next to the first electron, both electrons have the same charge, which is negative. Since like charges repel each other, the second electron will exert a repulsive force on the first electron. According to Newton's third law of motion, the first electron will also exert an equal and opposite force on the second electron. However, the question asks about the magnitude of the force on the proton, not the second electron. Since the proton has a positive charge, it will experience an attractive force towards the second electron. Therefore, the magnitude of the force on the proton doubles.

The charges exert forces on each other equal in magnitude and opposite in direction because according to Coulomb's law, the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Since the first charge is larger in magnitude, it will exert a larger force on the second charge. However, the second charge will also exert a force on the first charge, but in the opposite direction, resulting in forces that are equal in magnitude and opposite in direction.

The magnitude of the electric field at a distance of 1.50 m from a 50.0-nC charge is 200 N/C. The direction of the electric field is away from the charge.

The magnitude of the electric field at the location of the test charge can be calculated using the formula for electric field strength, which is given by the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge producing the electric field, and r is the distance from the charge. In this case, the charge producing the electric field is 5.0 C and the distance from the charge is 10 m. Plugging these values into the equation, we get E = (9 x 10^9 Nm^2/C^2)(5.0 C)/(10 m)^2 = 4.5 x 10^8 N/C. Therefore, the magnitude of the electric field at the location of the test charge is 4.5 x 10^8 N/C.

When a glass rod is rubbed with a piece of silk, electrons are transferred between the two materials. The glass rod has a higher affinity for electrons, so it pulls electrons from the silk, leaving the silk with a positive charge. The glass rod, on the other hand, gains these electrons and becomes negatively charged. Therefore, the correct answer is that the glass rod acquires a negative charge.

The force between two charges is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, we are given the force, the distance, and one of the charges. By rearranging the equation and substituting the known values, we can solve for the magnitude of the second charge.

When the two point charges are initially 2.0 cm apart, they experience a 1.0-N force. If they are moved to a new separation of 8.0 cm, the electric force between them can be calculated using Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Since the distance is increased by a factor of 4 (from 2.0 cm to 8.0 cm), the electric force will decrease by a factor of 16. Therefore, the electric force between them is 1/16 N.

When a positive object touches a neutral electroscope, the leaves separate because the positive charge is transferred to the electroscope, causing the leaves to repel each other. However, when a negative object is brought near the electroscope, the negative charge induces a positive charge on the leaves, attracting them towards the negative object. This attraction causes the leaves to move closer together. Therefore, the correct answer is that the leaves move closer together when a negative object is brought near the electroscope without touching it.

The correct answer is "radiate outward." This is because electric field lines always point away from positive charges. In the case of positive point charges, the electric field lines extend outward in all directions, indicating the direction in which a positive test charge would move if placed in the field.

The magnitude of the electric field at a distance from an atomic nucleus is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant, Q is the charge of the nucleus, and r is the distance from the nucleus. In this case, the charge of the nucleus is +40e, which means Q = +40 * 1.6 * 10^(-19) C. The distance from the nucleus is 1.0 m, so r = 1.0 m. Plugging these values into the equation, we get E = (9 * 10^9 N m^2/C^2)(+40 * 1.6 * 10^(-19) C)/(1.0 m)^2 = 5.8 * 10^(-8) N/C. Therefore, the correct answer is 5.8 * 10^(-8) N/C.

The given answer of 95 km is incorrect. The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the force is 1.0 N, which means the charges are relatively small. Therefore, it is unlikely that they would be 95 km apart. A more reasonable distance would be in the order of meters or millimeters.

The electrostatic force of attraction between the pennies is equal to the weight of a penny. This means that the force of attraction is equal to the gravitational force acting on the penny. The weight of a penny can be calculated using the equation weight = mass * acceleration due to gravity. Since the mass of the penny is given as 3.0 g, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the penny. Once we have the weight of the penny, we can equate it to the electrostatic force of attraction between the pennies and solve for the separation between them. The correct answer is 35 cm.

The electric field required to balance the weight of the particle can be calculated using the equation F = qE, where F is the force due to gravity (weight) and q is the charge of the particle. Since the weight is equal to the gravitational force, we can use the equation F = mg, where m is the mass of the particle and g is the acceleration due to gravity. By equating the two equations, we get mg = qE. Rearranging the equation, we find that E = (mg)/q. Plugging in the given values, we get E = (5.0 * 10^-3 kg * 9.8 m/s^2) / (4.0 * 10^-6 C) = 1.2 * 10^4 N/C.

According to the laws of electromagnetism, positive and negative charges always attract each other. This is because opposite charges have opposite electric fields, and the electric field lines between them attract each other, causing the charges to come together. Therefore, it is possible for a positive and a negative charge to attract each other.

When one charge is doubled, the force between the two charges increases by a factor of 2. When the other charge is tripled, the force between the two charges increases by a factor of 3. When the separation distance is reduced to one-fourth its original value, the force between the two charges increases by a factor of 16 (since force is inversely proportional to the square of the distance). Therefore, the final force is equal to the product of these factors: 2 x 3 x 16 = 96. Hence, the force is now equal to 96F.

When sphere B is briefly touched with a wire that is grounded, it means that any excess charge on sphere B is transferred to the ground. Since sphere A carries a net positive charge, when sphere B is touched, it will gain electrons from the ground, resulting in an excess of electrons on sphere B. This means that sphere B is now negatively charged.

Insulators are materials in which the electrons are tightly bound to the nuclei, meaning they are not free to move easily. This lack of mobility prevents the flow of electric current through the material. Conductors, on the other hand, have loosely bound electrons that can move freely, allowing for the easy flow of electric current. Semiconductors have properties that lie between those of insulators and conductors, while superconductors are materials that exhibit zero electrical resistance at very low temperatures. Therefore, the correct answer is insulators.

The electrostatic repulsion force between two charged objects is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the two spheres are identical and have the same charge. The only variable is the distance between them, which is determined by the angle between the threads. As the angle increases, the distance between the spheres increases, and therefore the force decreases. Since the force is directly proportional to the charge, the only way to maintain the same force at different distances is to adjust the charge. Therefore, the charge on each sphere must be 20 nC.

The electric field at the center of the square can be found by calculating the electric field due to each individual charge and then summing them up. The electric field due to a point charge is given by the equation E = k * q / r^2, where k is the electrostatic constant, q is the charge, and r is the distance from the charge to the point where the electric field is being calculated. Since the charges at the corners of the square are all positive, their electric fields will point away from them. The electric field due to each of the three positive charges will have the same magnitude and will be directed towards the center of the square. The electric field due to the negative charge will have the same magnitude as the positive charges but will be directed away from the center of the square. By calculating the electric field due to each charge and summing them up, we find that the total electric field at the center of the square is 4.3 * 10^5 N/C.

The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the force is given as 1.8 N. By rearranging Coulomb's law and plugging in the given charges, we can solve for the distance between the charges. The correct answer of 3.7 m indicates that the charges are 3.7 meters apart.

The electric field midway between the center and the surface of both the solid and hollow metal spheres is zero. This is because the charge on the spheres is uniformly distributed, and at this point, the electric field contributions from all points on the sphere cancel each other out. Therefore, the electric field is zero for both spheres.

The correct answer is "all of the given answers." This means that charge is both quantized, meaning it exists in discrete units, and conserved, meaning it cannot be created or destroyed but can only be transferred or redistributed. Additionally, charge is invariant, meaning it remains constant in certain physical processes. Therefore, all of these statements accurately describe the nature of charge.

When the electron is moved 0.50 m closer to the proton, the distance between them decreases by half. According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them. Therefore, when the distance is halved, the force increases by a factor of (1/0.5)^2 = 4. Hence, the size of the force on the proton increases to 4 times its original value.

not-available-via-ai

Electric charge is not a vector because it does not have both magnitude and direction. Unlike electric force, electric field, and electric line of force, electric charge is a scalar quantity that only has magnitude.

When there is a negative and positive charge along the line joining them, the electric field created by the negative charge points towards it, while the electric field created by the positive charge points away from it. These two electric fields add up to form a net electric field between the charges. Since the electric fields are in opposite directions, they cannot cancel each other out completely, meaning that a zero electric field cannot exist between the charges.

Electric field lines represent the direction and magnitude of the electric field at any given point in space. If the electric field lines were to intersect, it would mean that at that point, there are two different directions and magnitudes of the electric field, which is not possible. Electric field lines never intersect in free space.

The charge on 1 kg of protons is 9.6 * 10^7 C. This is because the charge of a proton is 1.6 * 10^-19 C, and there are approximately 6.02 * 10^23 protons in 1 kg. Therefore, the total charge on 1 kg of protons can be calculated by multiplying the charge of a single proton by the number of protons, resulting in 9.6 * 10^7 C.

The magnitude of the electric field at a point outside a charged sphere is given by the equation E = kQ/r^2, where E is the electric field, k is the electrostatic constant, Q is the charge on the sphere, and r is the distance from the center of the sphere. In this case, the charge on the sphere is +2.0 μC, the radius of the sphere is 10 cm, and the distance from the surface of the sphere is 5.0 cm. Plugging these values into the equation, we get E = (9 * 10^9 N m^2/C^2)(2.0 * 10^-6 C)/(0.05 m)^2 = 8.0 * 10^5 N/C. Therefore, the magnitude of the electric field is 8.0 * 10^5 N/C.

When the originally neutral electroscope is grounded, any excess charge is transferred to the ground, leaving the electroscope neutral. When the positively charged glass rod is brought near the electroscope, it induces a negative charge on the electroscope by repelling the electrons in the electroscope. However, once the glass rod is removed, the electroscope remains negatively charged because the excess electrons are not able to escape to the ground. Therefore, the correct answer is that the electroscope is negatively charged.

The Coulomb constant, k, is a proportionality constant that appears in Coulomb's law, which relates the force between two charged objects. The units of k can be determined by examining the units in Coulomb's law equation. The equation states that the force (F) between two charges (q1 and q2) is equal to the Coulomb constant (k) multiplied by the product of the charges (q1*q2) and divided by the square of the distance between them (r^2). By rearranging the equation, we can determine the units of k to be N*m^2/C^2, where N represents Newtons (the unit of force), m represents meters (the unit of distance), and C represents Coulombs (the unit of charge).

According to the principles of electrostatics, like charges repel each other, while opposite charges attract. Since both charges in this scenario are negative, they have the same charge and will therefore repel each other, not attract.

A third charge can be placed at x = 1.5 m on the x-axis so that the net force on it is zero. At this point, the electric forces exerted by the +3.0-C charge and the +9.0-C charge will cancel each other out due to the inverse square law of electric forces. The distance between the +3.0-C charge and the third charge is equal to the distance between the +9.0-C charge and the third charge, resulting in a net force of zero.

A negatively charged rod can charge an electroscope positively by induction. When the negatively charged rod is brought close to the electroscope, the negative charges in the rod repel the electrons in the electroscope, causing them to move away from the rod. This leaves a net positive charge on the electroscope.

The strength of the electric field is inversely proportional to the square of the distance from the point charge. This means that as the distance from the point charge doubles, the strength of the electric field decreases by a factor of four. Therefore, at twice the distance, the strength of the electric field is one-fourth of its original value.

The magnitude of the electric field at the square's center can be calculated using the formula for electric field due to point charges. The electric field due to each charge is calculated separately and then added together. In this case, the electric field due to the +4.0 μC charge at (0, 0) is directed towards the center of the square, while the electric field due to the +4.0 μC charge at (1, 1) is directed away from the center of the square. The electric fields due to the +3.0 μC charge at (1, 0) and the -3.0 μC charge at (0, 1) cancel each other out. By calculating the magnitudes of these electric fields and adding them together, the result is 1.1 * 10^5 N/C.

The force experienced by a charge in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field. In this case, the force is given as 10 N and the charge is given as 5.0 μC. Plugging these values into the equation, we get 10 N = (5.0 μC)E. Solving for E, we find that the magnitude of the electric field is 2.0 MN/C.

When a negatively charged rod is brought near one end of an uncharged metal bar, the negative charges in the metal bar will be repelled by the negative charges on the rod. This will cause the positive charges in the metal bar to move towards the end closest to the rod, leaving the end farthest from the rod with an excess of negative charges. Therefore, the end of the metal bar farthest from the charged rod will be charged negative.

When a second point charge of -Q is placed at one corner of the square, the electrostatic force between the two charges is attractive and has a magnitude of 2.0 N. However, when identical charges of -Q are placed at the other three corners of the square, the forces between the positive charge at the center and each of the negative charges cancel out. This is because the forces are equal in magnitude and opposite in direction, resulting in a net force of zero. Therefore, the magnitude of the net electrostatic force acting on the positive charge at the center of the square is zero.

When a second electron is placed next to the proton, it creates a repulsive force between the two electrons. This repulsive force cancels out the attractive force between the electron and the proton. As a result, the net force on the first electron becomes zero, causing the magnitude of the force to become zero.

The direction of the electric field is away from the 5.0 C charge. This is because electric field lines always point away from positive charges and towards negative charges. Since the 5.0 C charge is positive, the electric field will point away from it.

The net charge of +2.00 μC indicates that the piece of plastic has an excess of 2.00 μC worth of positive charge. Since protons carry a positive charge and electrons carry a negative charge, the piece of plastic must have 2.00 μC worth of extra protons compared to electrons. Therefore, the piece of plastic has 1.25 * 10^13 more protons than electrons.

The magnitude of the electrostatic force between two charges can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are identical and the distance between them is the length of one side of the equilateral triangle. By substituting the given values into the formula, the magnitude of the electrostatic force is calculated to be 1.6 N.