Ch.2 Gauss' Law

7/23/2021 Ch 22 HW Ch 22 HW Due: 11:59pm on Sunday, June 6, 2021 Gauss's Law Learning Goal: To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written where is the permittivity of vacuum. Part A How should the integral in Gauss's law be evaluated? ANSWER: around the perimeter of a closed loop over the surface bounded by a closed loop over a closed surface Correct In the integral for Gauss's law, the vector represents an infinitesimal surface element. The magnitude of the area of the surface element. The direction of volume. is is normal to the surface element, pointing out of the enclosed Part B In Gauss's law, to what does refer? ANSWER: the net charge inside the closed surface the charge residing on insulators inside the closed surface all the charge in the physical system any charge inside the closed surface that is arranged symmetrically Correct The major use of Gauss's law is to determine an electric field when the charge distribution, both inside and outside the Gaussian surface, is symmetric. Of course, the electric field can always be found by adding up (or integrating) the contributions of all the charge in the problem. In highly symmetric situations, however, Gauss's law is much simpler computationally than dealing with all such contributions, and it provides better physical insight, too. 1/30 7/23/2021 Ch 22 HW Gauss's Law in 3, 2, and 1 Dimension Gauss's law relates the electric flux through a closed surface to the total charge enclosed by the surface: . You can use Gauss's law to determine the charge enclosed inside a closed surface on which the electric field is known. However, Gauss's law is most frequently used to determine the electric field from a symmetric charge distribution. The simplest case in which Gauss's law can be used to determine the electric field is that in which the charge is localized at a point, a line, or a plane. When the charge is localized at a point, so that the electric field radiates in three-dimensional space, the Gaussian surface is a sphere, and computations can be done in spherical coordinates. Now consider extending all elements of the problem (charge, Gaussian surface, boundary conditions) infinitely along some direction, say along the z axis. In this case, the point has been extended to a line, namely, the z axis, and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions, since the field varies only with x and y, or with and in cylindrical coordinates. A onedimensional problem may be achieved by extending the problem uniformly in two directions. In this case, the point is extended to a plane, and consequently, it has planar symmetry. Three dimensions Consider a point charge in three-dimensional space. Symmetry requires the electric field to point directly away from the charge in all directions. To find , the magnitude of the field at distance from the charge, the logical Gaussian surface is a sphere centered at the charge. The electric field is normal to this surface, so the dot product of the electric field and an infinitesimal surface element involves . The flux integral is therefore reduced to the magnitude of the electric field on the Gaussian surface, and , where is is the area of the surface. Part A Determine the magnitude Express by applying Gauss's law. in terms of some or all of the variables/constants , , and . You did not open hints for this part. ANSWER: = Two dimensions 2/30 7/23/2021 Ch 22 HW Now consider the case that the charge has been extended along the z axis. This is generally called a line charge. The usual variable for a line charge density (charge per unit length) is , and it has units (in the SI system) of coulombs per meter. Part B By symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie in planes perpendicular to the wire. In solving for the magnitude of the radial electric field produced by a line charge with charge density , one should use a cylindrical Gaussian surface whose axis is the line charge. The length of the cylindrical surface should cancel out of the expression for . Apply Gauss's law to this situation to find an expression for . Express in terms of some or all of the variables , , and any needed constants. You did not open hints for this part. ANSWER: = One dimension Now consider the case with one effective direction. In order to make a problem effectively one-dimensional, it is necessary to extend a charge to infinity along two orthogonal axes, conventionally taken to be x and y. When the charge is extended to infinity in the xy plane (so that by symmetry, the electric field will be directed in the z direction and depend only on z), the charge distribution is sometimes called a sheet charge. The symbol usually used for two-dimensional charge density is either , or . In this problem we will use . has units of coulombs per meter squared. Part C In solving for the magnitude of the electric field produced by a sheet charge with charge density , use the planar symmetry since the charge distribution doesn't change if you slide it in any direction of xy plane parallel to the sheet. Therefore at each point, the electric field is perpendicular to the sheet and must have the same magnitude at any given distance on either side of the sheet. To take advantage of these symmetry properties, use a Gaussian surface in the shape of a cylinder with its axis perpendicular to the sheet of charge, with ends of area which will cancel out of the expression for in the end. The result of applying Gauss's law to this situation then gives an expression for for both and . Express for in terms of some or all of the variables/constants , , and . 3/30 7/23/2021 Ch 22 HW You did not open hints for this part. ANSWER: = A Charged Sphere with a Cavity An insulating sphere of radius , centered at the origin, has a uniform volume charge density . Part A Find the electric field inside the sphere (for < ) in terms of the position vector . Express your answer in terms of , (Greek letter rho), and . You did not open hints for this part. ANSWER: = Part B Complete previous part(s) Flux out of a Cube A point charge of magnitude is at the center of a cube with sides of length . Part A What is the electric flux Use through each of the six faces of the cube? for the permittivity of free space (not the EMF symbol ). 4/30 7/23/2021 Ch 22 HW Hint 1. How to approach the problem Since the magnitude and direction of changes over the entire surface, you cannot use the expression find the electric flux. Integration is possible, of course—but who likes to integrate if there is an easier way? to In this case, there is a way indeed. Just use Gauss's law to find the total electric flux through the cube and then exploit the symmetry of the problem. Hint 2. Calculate the total electric flux Calculate the total electric flux coming out of the cube. Hint 1. Calculate the flux for a sphere If the charge were in the center of a sphere of radius be the total flux out of the sphere? instead of a cube with sides of length , what would ANSWER: = Hint 2. Flux and surface shape The total flux through any closed surface depends solely on the total amount of charge enclosed by that surface, not on its shape. ANSWER: = Correct Hint 3. Flux through a face Since the charge is at the center of the cube, by symmetry the flux through each face is the same. ANSWER: = Correct The shape of the surface enclosing a charge, in this case a cube, does not affect the total electric flux through the surface. The flux depends only on the total enclosed charge. Part B What would be the flux Use through a face of the cube if its sides were of length ? for the permittivity of free space. 5/30 7/23/2021 Ch 22 HW Hint 1. How to approach the problem Gauss's law states that the flux through a closed surface depends only on the charge enclosed by that surface. Does changing the size of the cube affect the total charge enclosed by the cube? ANSWER: = Correct Just as the shape of the surface does not affect the total electric flux coming out of that surface, its size does not make any difference in the total electric flux either. The only relevant quantity is the total enclosed charge. PSS 22.1: Gauss's Law Learning Goal: To practice Problem-Solving Strategy 22.1: Gauss's Law. An infinite cylindrical rod has a uniform volume charge density (where ). The cross section of the rod has radius the magnitude of the electric field at a distance from the axis of the rod. Assume that . . Find Problem-Solving Strategy 22.1: Gauss's Law IDENTIFY the relevant concepts: Gauss’s law is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over a plane. In these situations we determine the direction of from the symmetry of the charge distribution. If we are given the charge distribution, we can use Gauss’s law to find the magnitude of . Alternatively, if we are given the field, we can use Gauss’s law to determine the details of the charge distribution. In either case, begin your analysis by asking the question: What is the symmetry? SET UP the problem using the following steps: 1. Select the surface that you will use with Gauss’s law. We often call it a Gaussian surface. If you are trying to find the field at a particular point, then that point must lie on your Gaussian surface. 2. The Gaussian surface does not have to be a real physical surface. Often the appropriate surface is an imaginary geometric surface. 3. If the charge distribution has cylindrical or spherical symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere, respectively. EXECUTE the solution as follows: 1. Carry out the integral The symmetry of the charge distribution and your careful choice of a Gaussian surface make the integration straightforward. 2. Often you can think of the closed Gaussian surface as being made up of several separate surfaces. The integral over the entire closed surface is always equal to the sum of the integrals over all the separate surfaces. 3. If is perpendicular (normal) at every point to a surface with area , if it points outward from the interior of the surface, and if it also has the same magnitude at every point on the surface, then , and over that surface is equal to . If instead is perpendicular and inward, then and . 4. If is tangent to a surface at every point, then and the integral over that surface is zero. 6/30 7/23/2021 Ch 22 HW 5. If at every point on a surface, the integral is zero. 6. In , is always the perpendicular component of the total electric field at each point on the closed Gaussian surface. In general, this field may be caused partly by charges within the surface and partly by charges outside it. Even when there is no charge within the surface, the field at points on the Gaussian surface is not necessarily zero. In that case, however, the integral over the Gaussian surface—that is, the total electric flux through the Gaussian surface—is always zero. 7. Once you have evaluated the integral, use the equation variable. to solve for your target EVALUATE your answer: Often your result will be a function that describes how the magnitude of the electric field varies with position. Examine this function with a critical eye to see whether it makes sense. IDENTIFY the relevant concepts The charge distribution described in this problem has cylindrical symmetry. The electric field must point away from the positive charges distributed on the rod. Since you are given information on the charge distribution, use Gauss’s law to determine the magnitude of at the required position. SET UP the problem using the following steps Part A Considering the symmetry of the charge distribution, choose one of the following options as the most appropriate choice of Gaussian surface to use in this problem. ANSWER: a finite closed cylinder whose axis coincides with the axis of the rod and whose cross section has radius a finite closed cylinder whose axis coincides with the axis of the rod and whose cross section has radius a sphere of radius whose center lies on the axis of the rod a finite closed cylinder whose axis coincides with the axis of the rod and whose cross section has radius an infinite cylinder whose axis coincides with the axis of the rod and whose cross section has radius Part B Complete previous part(s) EXECUTE the solution as follows Part C Complete previous part(s) EVALUATE your answer Part D Complete previous part(s) ± Calculating Flux for Hemispheres of Different Radii Learning Goal: 7/30 7/23/2021 Ch 22 HW To understand the definition of electric flux, and how to calculate it. Flux is the amount of a vector field that "flows" through a surface. We now discuss the electric flux through a surface (a quantity needed in Gauss's law): , where is the flux through a surface with differential area element the electric field in which the surface lies. There are several important points to consider in this expression: , and is 1. It is an integral over a surface, involving the electric field at the surface. 2. is a vector with magnitude equal to the area of an infinitesmal surface element and pointing in a direction normal (and usually outward) to the infinitesmal surface element. 3. The scalar (dot) product implies that only the component of integral. That is, , where normal to the surface contributes to the is the angle between When you compute flux, try to pick a surface that is either parallel or perpendicular to compute. and . , so that the dot product is easy to Two hemispherical surfaces, 1 and 2, of respective radii and , are centered at a point charge and are facing each other so that their edges define an annular ring (surface 3), as shown. The field at position due to the point charge is: where is a constant proportional to the charge, is the unit vector in the radial direction. , and Part A What is the electric flux through the annular ring, surface 3? Express your answer in terms of , , , and any constants. Hint 1. Apply the definition of electric flux The integrand in the equation defining flux (in the problem introduction) can be calculated by noting the following: . Since the electric field is everywhere parallel to the surface of the annular ring, the element electric field, and thus . Therefore, . is normal to the ANSWER: = 0 Correct Part B 8/30 7/23/2021 Ch 22 HW What is the electric flux Express in terms of through surface 1? , , , and any needed constants. Hint 1. Apply the definition of electric flux The integrand in the equation defining flux (in the problem introduction) can be calculated by noting the following: Since the electric field is everywhere perpendicular to surface 2, . Therefore, the integral simply becomes the magnitude of the electric field multiplied by the surface area of surface 2. Hint 2. Find the area of surface 1 Find the area Express of the hemisphere that is surface 1. in terms of and other given or known quantities. ANSWER: = ANSWER: = Correct Part C What is the electric flux Express in terms of passing outward through surface 2? , , , and any constants or other known quantities. Hint 1. Apply the definition of electric flux The integrand in the equation defining flux (in the problem introduction) can be calculated by noting the following: Since the electric field is everywhere perpendicular to surface 2, . Therefore, the integral simply becomes the magnitude of the electric field multiplied by the surface area of surface 2. Hint 2. Find the area of surface 2 Find the area of the hemisphere that is surface 2. Express your answer in terms of , and any needed constants. ANSWER: = 9/30 7/23/2021 Ch 22 HW ANSWER: = Correct Observe that the electric flux through surface 1 is the same as that through surface 2, despite the fact that surface 2 has a larger area. If you think in terms of field lines, this means that there is the same number of field lines passing through both surfaces. This is because of the inverse square, , behavior of the electric field surrounding a point particle. A good rule of thumb is that the flux through a surface is proportional to the number of field lines that pass through the surface. Exercise 22.9 - Enhanced - with Feedback A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 of -59.0 . , giving it a charge Part A Find the electric field just inside the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed toward the center of the sphere. ANSWER: = 0 Correct Part B Find the electric field just outside the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed toward the center of the sphere. ANSWER: = −6.55×107 Correct Part C Find the electric field 6.00 outside the surface of the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed toward the center of the sphere. 10/30 7/23/2021 Ch 22 HW ANSWER: = −2.36×107 Correct Exercise 22.10 - Enhanced - with Feedback A point charge = 1.30 . = 3.45 is located on the x-axis at = 2.00 , and a second point charge = -5.55 is on the y-axis at Part A What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius = 0.480 ? Express your answer in newton times meters squared per coulomb. ANSWER: = 0 Correct Part B What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius = 1.50 ? Express your answer in newton times meters squared per coulomb. ANSWER: = -627 Correct Part C What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius = 2.45 ? Express your answer in newton times meters squared per coulomb. ANSWER: = -237 11/30 7/23/2021 Ch 22 HW Correct Exercise 22.12 - Enhanced - with Feedback The nuclei of large atoms, such as uranium, with radius of the uranium nucleus is approximately protons, can be modeled as spherically symmetric spheres of charge. The . Part A What is the electric field this nucleus produces just outside its surface? Express your answer in newtons per coulomb. ANSWER: = 2.4×1021 Correct Part B What magnitude of electric field does it produce at the distance of the electrons, which is about 1.9×10−10 ? Express your answer in newtons per coulomb. ANSWER: = 3.7×1012 Correct Part C The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus? Express your answer in newtons per coulomb. ANSWER: = 0 Correct ± The Charge on a Thundercloud In a thunderstorm, charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considered to be distributed uniformly throughout the cloud. For the purposes of this problem, take the cloud to be a sphere of diameter 1.00 12/30 7/23/2021 Ch 22 HW kilometer. The point of this problem is to estimate the maximum amount of charge that this cloud can contain, assuming that the charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air breaks down. This breakdown means that the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. The ionized air will then emit light due to the recombination of the electrons and atoms to form excited molecules that radiate light. In addition, the large current will heat up the air, resulting in its rapid expansion. These two phenomena account for the appearance of lightning and the sound of thunder. Take the breakdown electric field of air to be . Part A Estimate the total charge on the cloud when the breakdown of the surrounding air is reached. Express your answer numerically, to three significant figures, using . Hint 1. Use Gauss's law This problem calls for the use of Gauss' law: . Choose a Gaussian surface that lies just on the surface of the cloud and apply Gauss's law to find the charge enclosed. Hint 2. Evaluate Gauss's law Solve Gauss's law for the enclosed charge . Keep in mind that the electric field produced by a uniform spherical charge distribution is in the same direction as the electric field of a point charge. Therefore, the integral should be very easy to evaluate. Express your answer in terms of the magnitude of the electric field, surface, ; and the permittivity constant, . ; the surface area of the Gaussian ANSWER: = ANSWER: = 83.4 Coulombs Correct Charge Distribution on a Conductor with a Cavity A positive charge is brought close to a fixed neutral conductor that has a cavity. The cavity is neutral; that is, there is no net charge inside the cavity. Part A Which of the figures best represents the charge distribution on the inner and outer walls of the conductor? 13/30 7/23/2021 Ch 22 HW Hint 1. Conductors have no internal field At steady state, conductors have no internal electric field (otherwise, charge would flow). Therefore, the arrangement of charges on the surfaces of the conductor must exactly cancel out any external electric field to ensure that the internal field is zero. Hint 2. Charges on the cavity walls Think about what the answer would be for a conductor without a cavity. Would there be net charges on the surface of some imaginary sphere drawn inside of the conductor? Would this change if you removed all of the material inside of that sphere? ANSWER: 1 2 3 Correct Exercise 22.16 - Enhanced - with Feedback Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of at the planet's surface. Part A Calculate the total electric charge on the planet. Express your answer with the appropriate units. ANSWER: 14/30 7/23/2021 Ch 22 HW = Part B Calculate the electric field at the planet's surface (refer to the astronomical data in the textbook). Express your answer with the appropriate units. ANSWER: = Part C Calculate the charge density on Mars, assuming all the charge is uniformly distributed over the planet's surface. Express your answer with the appropriate units. ANSWER: = The Electric Field and Surface Charge at a Conductor Learning Goal: To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge. Part A Which of the following describes the electric field inside this conductor? ANSWER: It is in the same direction as the original external field. It is in the opposite direction from that of the original external field. It has a direction determined entirely by the charge on its surface. It is always zero. Correct The net electric field inside a conductor is always zero. If the net electric field were not zero, a current would flow inside the conductor. This would build up charge on the exterior of the conductor. This charge would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero. 15/30 7/23/2021 Ch 22 HW Part B The charge density inside the conductor is: ANSWER: 0 non-zero; but uniform non-zero; non-uniform infinite Correct You already know that there is a zero net electric field inside a conductor; therefore, if you surround any internal point with a Gaussian surface, there will be no flux at any point on this surface, and hence the surface will enclose zero net charge. This surface can be imagined around any point inside the conductor with the same result, so the charge density must be zero everywhere inside the conductor. This argument breaks down at the surface of the conductor, because in that case, part of the Gaussian surface must lie outside the conducting object, where there is an electric field. Part C Assume that at some point just outside the surface of the conductor, the electric field has magnitude and is directed toward the surface of the conductor. What is the charge density on the surface of the conductor at that point? Express your answer in terms of and . Hint 1. How to approach the problem Which of the following is the best way to solve this problem? ANSWER: Use Coulomb's law for the electric field from each charge element. Use Gauss's law with a short and flat cylindrical surface (picture a coin) with one end just below (inside) and the other just above (outside) the surface of the conductor. Use Gauss's law with a long and thin cylindrical surface (picture a straw capped at both ends) with one end far below (but inside the object) and the other far above the surface of the conductor. None of these: You need to know the surface charge distribution everywhere on the surface. None of these: You need to know the electric field at all points on the surface. Hint 2. Calculate the flux through the top of the cylinder Using a flat cylinder with large top and bottom each of area just above and just below the surface of the conductor, find the flux generated through the top surface of the cylinder by the electric field of magnitude that points into the surface. Express your answer in terms of , , and any needed constants. ANSWER: 16/30 7/23/2021 Ch 22 HW = Hint 3. Calculate the flux through the bottom of the box Using a flat cylinder with large top and bottom of area just above and just below the surface of the conductor, find the flux generated through the bottom surface of the cylinder by the electric field inside the conductor (keep in mind that positive flux is outward through the cylinder's surface, which is downward into the conductor). Answer in terms of , , and any needed constants. ANSWER: = Hint 4. What is the charge inside the Gaussian surface? Find the net charge inside this Gaussian surface. Express your answer in terms of the charge density and other given quantities. ANSWER: = Hint 5. Apply Gauss's law Now apply Gauss's law, neglecting any contribution to the flux due to the very short sides of the cylinder. Gauss's law states that . The area should cancel out of your result. ANSWER: = Correct Problem 22.37 - Enhanced - with Feedback A long coaxial cable consists of an inner cylindrical conductor with radius and an outer coaxial cylinder with inner radius and outer radius . The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length . Part A Calculate the magnitude of the electric field at any point between the cylinders a distance Express your answer in terms of some or all of the variables , , , , and constants from the axis. and . ANSWER: = 17/30 7/23/2021 Ch 22 HW Part B Find the direction of the electric field at any point between the cylinders a distance from the axis. ANSWER: parallel cylinders' axis radially outward radially inward Part C Calculate the magnitude of the electric field at any point outside the outer cylinder a distance Express your answer in terms of some or all of the variables , , , , and constants from the axis. and . ANSWER: = Part D Find the direction of the electric field at any point outside the outer cylinder a distance from the axis. ANSWER: parallel cylinders' axis radially outward radially inward Part E Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder. Express your answers in terms of some or all of the variables , , , , answers separated by a comma. and constants and . Enter your ANSWER: = Problem 22.55 - Enhanced - with Feedback In one experiment the electric field is measured for points at distances from a uniform line of charge that has charge per unit length and length , where . In a second experiment the electric field is measured for points at distances from the 18/30 7/23/2021 Ch 22 HW center of a uniformly charged insulating sphere that has volume charge density and radius = 7.60 , where results of the two measurements are listed in the table, but you aren't told which set of data applies to which experiment. ( ) . The 1.00 1.50 2.00 2.50 3.00 3.50 4.00 2.72 1.79 1.34 1.07 0.902 0.770 0.677 5.45 2.42 1.34 0.861 0.605 0.443 0.335 Measurement A (10 ) Measurement B (10 ) Part A For the measurement , select the correct graph for versus . ANSWER: 19/30 7/23/2021 Ch 22 HW Part B For the measurement , select the correct graph for versus . ANSWER: 20/30 7/23/2021 Ch 22 HW 21/30 7/23/2021 Ch 22 HW Part C For the measurement , select the correct graph for versus . ANSWER: 22/30 7/23/2021 Ch 22 HW Part D For the measurement , select the correct graph for versus . ANSWER: 23/30 7/23/2021 Ch 22 HW Part E Complete previous part(s) Part F Complete previous part(s) Part G Complete previous part(s) Video Tutor: Electroscope in Conducting Shell 24/30 7/23/2021 Ch 22 HW First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. Part A As in the video, we apply a charge + to the half-shell that carries the electroscope. This time, we also apply a charge – to the other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again? Hint 1. How to approach the problem. The half-shell with a charge of – equal deficit of electrons. has an excess of electrons, and the half-shell with a charge of + has an exactly What happens when these two charged, conducting half-shells are brought together? What is the net charge of the resulting whole sphere? (Recall that electrons can move through a conductor.) Will the half-shells have a net charge when you separate them again? If the half-shell with the electroscope needle carries a net charge, the needle will deflect. ANSWER: It deflects more than it did at the end of the video. It does not deflect at all. It deflects the same amount as at end of the video. It deflects less than it did at the end of the video. Correct The spherical surface has zero net charge after the two halves are brought together. The two half-spheres remain electrically neutral after they are separated. Exercise 22.22 - Enhanced - with Feedback A point charge of -3.00 is located in the center of a spherical cavity of radius 6.90 solid. The charge density in the solid is 7.35 10 . inside an insulating spherical charged 25/30 7/23/2021 Ch 22 HW Part A Calculate the magnitude of the electric field inside the solid at a distance of 9.50 from the center of the cavity. Express your answer with the appropriate units. ANSWER: = Part B Find the direction of this electric field. ANSWER: radially outward radially inward Problem 22.41 - Enhanced - with Feedback A solid conducting sphere with radius that also carries charge . The charge that carries positive charge is concentric with a very thin insulating shell of radius 2 is distributed uniformly over the insulating shell. Part A Find the magnitude of the electric field in the region Express your answer in terms of some or all of the variables . , , , and constants and . , and constants and . ANSWER: = Part B Complete previous part(s) Part C Find the magnitude of the electric field in the region Express your answer in terms of some or all of the variables . , , ANSWER: = Part D Complete previous part(s) 26/30 7/23/2021 Ch 22 HW Part E Find the magnitude of the electric field in the region . Express your answer in terms of some or all of the variables , , , and constants and . ANSWER: = Part F Complete previous part(s) Problem 22.35 Negative charge is distributed uniformly over the surface of a thin spherical insulating shell with radius . Part A Calculate the magnitude of the force that the shell exerts on a positive point charge center of the shell (outside the shell). Express your answer in terms of the variables , , , , and constants and located a distance from the . ANSWER: = Correct Part B Find the direction of the force that the shell exerts on a positive point charge the shell (outside the shell). located a distance from the center of ANSWER: from the center of the shell toward the center of the shell the force is zero Correct Part C Calculate the magnitude of the force that the shell exerts on a positive point charge center of the shell (inside the shell). located a distance from the 27/30 7/23/2021 Ch 22 HW Express your answer in terms of some ot all of the variables , , , , and constants and . ANSWER: = 0 Correct Part D Find the direction of the force that the shell exerts on a positive point charge the shell (inside the shell). located a distance from the center of ANSWER: from the center of the shell toward the center of the shell the force is zero Correct Exercise 22.2 - Enhanced - with Feedback A flat sheet is in the shape of a rectangle with sides of lengths 0.400 and 0.600 . The sheet is immersed in a uniform electric field of magnitude 95.0 that is directed at 20 from the plane of the sheet . Part A Find the magnitude of the electric flux through the sheet. Express your answer with the appropriate units. ANSWER: 28/30 7/23/2021 Ch 22 HW = 7.8 Correct Exercise 22.4 - Enhanced - with Feedback It was shown in Example 21.11 (Section 21.5) in the textbook that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude . Consider an imaginary cylinder with a radius of = 0.240 and a length of = 0.465 that has an infinite line of positive charge running along its axis. The charge per unit length on the line is = 6.00 . Part A What is the electric flux through the cylinder due to this infinite line of charge? Express your answer in newton times meters squared per coulomb. ANSWER: = 3.15×105 Correct Part B What is the flux through the cylinder if its radius is increased to 0.580 ? 0.905 ? Express your answer in newton times meters squared per coulomb. ANSWER: = 3.15×105 Correct Part C What is the flux through the cylinder if its length is increased to Express your answer in newton times meters squared per coulomb. ANSWER: = 6.13×105 Correct Score Summary: 29/30 7/23/2021 Ch 22 HW Your score on this assignment is 101%. You received 86.09 out of a possible total of 85 points. 30/30