Problem Set 02: Electric Fields Due: 11:59pm on Monday, September 9, 2013 You will receive no credit for items you complete after the assignment is due. Grading Policy Question 1 A point charge is at the point meters, meters, and a second point charge is at the point meters, . Part A Calculate the magnitude of the net electric field at the origin due to these two point charges. Express your answer in newtons per coulomb to three significant figures. Hint 1. How to approach the problem First, draw a diagram of the charge system with both of the charges at their correct locations. Next calculate the x and y components of the electric field at the origin from each charge separately. Use vector addition to obtain the components of the combined field in each direction. Finally, use the x and y components to find the magnitude of the combined electric field using the Pythagorean theorem. Hint 2. Calculate the x component of the field created by Calculate , the x component of , the electric field at the origin created by , including its sign. Express your answer in newtons per coulomb to three significant figures. ANSWER: Correct Hint 3. Calculate the y component of Calculate , the y component of , the electric field at the origin created by , including its sign. Express your answer in newtons per coulomb to three significant figures. ANSWER: Correct Since the second charge is on the x axis, there will be no y component of . Hint 4. Calculate the magnitude of the field created by the first charge = −4.00 nCq 1 x = 0.600 y = 0.800 = +6.00 nCq 2 x = 0.600 y = 0 E q 2 E 2x E " 2 q 2 = -150 E 2x N/C E " 2 E 2y E " 2 q 2 = 0 E 2y N/C E " 2 E 1 Calculate , the magnitude of the electric field at the origin created by charge . Recall that magnitude is always nonnegative. Express your answer in newtons per coulomb to three significant figures. Hint 1. Calculate the distance from the first charge to the origin Calculate , the distance from the first charge, , to the origin. ANSWER: ANSWER: Correct Hint 5. Calculate the x component of Calculate , the x component of , the electric field at the origin created by , including its sign. Express your answer in newtons per coulomb to three significant figures. Hint 1. Splitting a vector into its components Use the x and y components of the location of charge , along with the distance from the charge to the origin, to construct a right triangle. Use this triangle to find the sine or cosine of the angle that makes with the x axis. To find the x component, use the cosine that you calculate, and use the direction in which the field points to find the appropriate sign for the component. ANSWER: Answer Requested Hint 6. Calculate the y component of Calculate the y component of , the electric field created at the origin by , including its sign. Express your answer in newtons per coulomb to three significant figures. Hint 1. Splitting a vector into its components E 1 q 1 r 1 q 1 = 1.00 r 1 m = 36.0 E 1 N/C E " 1 E 1x E " 1 q 1 q 1 E " 1 = 21.6 E 1x N/C E " 1 E " 1 q 1 Use the x and y components of the location of charge , along with the distance from the charge to the origin, to construct a right triangle. Use this triangle to find the sine or cosine of the angle that makes with the x axis. To find the y component, use the sine that you calculate, and use the direction in which the field points to find the appropriate sign for the component. ANSWER: Correct Hint 7. Putting it all together Add the components from each charge to obtain the components of the total electric field (e.g., ). Use the Pythagorean theorem to calculate the magnitude of the field: . ANSWER: Correct Part B What is the direction, relative to the negative x axis, of the net electric field at the origin due to these two point charges. Express your answer in degrees to three significant figures. Hint 1. How to approach the problem Using the components of the electric field from Part A, calculate the tangent of the angle as the ratio of the y component to the x component, then take the arctangent to find the angle. Since the angle is taken with respect to the negative x axis, use instead of for the x component. ANSWER: Correct Question 2 Two particles with positive charges and are separated by a distance . q 1 E " 1 = 28.8 E 1y N/C = +E x E 1x E 2x = +E 2 E 2 x E 2 y = 131 E N/C −E x E x = 12.6 up from the negative x axis θ # q 1 q 2 s Part A Along the line connecting the two charges, at what distance from the charge is the total electric field from the two charges zero? Express your answer in terms of some or all of the variables , , and . If your answer is difficult to enter, consider simplifying it, as it can be made relatively simple with some work. Hint 1. How to approach the problem First, draw a diagram of the system. Locate the first charge, , at the origin, and the second charge, , at a distance away. Next, define a point P, at distance from the first charge, where the field is zero. At the point in question, the electric field from each point charge will be equal but opposite. Solve for . Hint 2. Find the distance from the second charge to the zero-field point If the point where the electric field is zero is a distance away from the first charge, what is the distance of this point from the second charge? Express your answer in terms of some or all of the variables , , , and . ANSWER: Correct Hint 3. Find the electric field from the first charge At point P, what is the magnitude of the electric field due to the first charge? Express your answer in terms of some or all of the variables , , , and . ANSWER: Correct Hint 4. Find the electric field from the second charge At point P, what is the magnitude of the electric field due to the second charge? Express your answer in terms of some or all of the variables , , , and . ANSWER: q 1 s q 1 q 2 k = 1 4pi# 0 q 1 q 2 s x 1 x 1 x 1 x 2 s x 1 q 1 k = 1 4pi# 0 = x 2 s− x 1 E 1 s x 1 q 1 k = 1 4pi# 0 = E 1 kq 1 ( )x 1 2 E 2 s x 1 q 2 k = 1 4pi# 0 Correct Hint 5. Solving the quadratic equation and choosing the correct answer If you set the magnitudes of the fields due to and equal at point P, you should end up with a quadratic equation for . This equation will have two solutions, but since you know that the zero-field point must be between the two charges, you should be able to eliminate one of the results. Also, if you assume that , you can use the relation to simplify your answer. ANSWER: Correct Your answer can be reduced to . In order to arrive at this expression from its initial form, , the assumption must be made that (otherwise there's a zero in the denominator). Therefore, to solve the special (or singular) case when , it would be best to use a symmetry argument to solve this problem. Question 3 Part A For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero. If no such region exists on the horizontal axis choose the last option (nowhere). = E 2 kq 2 (s− )x 1 2 q 1 q 2 x 1 ≠q 1 q 2 a − b = ( + )( − )a √ b√ a √ b√ s +1 q 2 q 1 √ s q 1 √ +q 1 √ q 2 √ s −q 1 q 1 q 2 √ −q 1 q 2 ≠q 1 q 2 =q 1 q 2 Hint 1. Zeros of the electric field The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region. ANSWER: Correct Part B For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero. If no such region exists on the horizontal axis choose the last option (nowhere). A B C D E nowhere Hint 1. Zeros of the electric field The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region. Hint 2. Determine the regions where the electric fields could cancel In which region(s) do the electric fields from the two source charges point in opposite directions? List all the correct answers in alphabetical order. ANSWER: Correct Since the two charges produce fields that point in opposite directions in these regions, if the magnitude of the fields are equal, the net electric field will be zero. Hint 3. Consider the magnitude of the electric field For each of the three regions found in the previous hint, determine whether it is possible for the magnitudes to be equal. As an example, consider the point directly between the two charges. Which charge produces the largest magnitude field directly between the two charges? ANSWER: BCD the charge on the right the charge on the left neither, because they have the same magnitude Correct Therefore, the point directly between the two charges is not the correct answer since the right charge dominates at this point. Check the other two possible regions. ANSWER: Correct Part C For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero. If no such region exists on the horizontal axis choose the last option (nowhere). Hint 1. Zeros of the electric field The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region. ANSWER: A B C D E nowhere Correct Part D For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero. Hint 1. Zeros of the electric field The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region. ANSWER: A B C D E nowhere A B C D E Nowhere along the finite x axis Correct Question 4 Learning Goal: To understand the nature of electric fields and how to draw field lines. Electric field lines are a tool used to visualize electric fields. A field line is drawn beginning at a positive charge and ending at a negative charge. Field lines may also appear from the edge of a picture or disappear at the edge of the picture. Such lines are said to begin or end at infinity. The field lines are directed so that the electric field at any point is tangent to the field line at that point. The figure shows two different ways to visualize an electric field. On the left, vectors are drawn at various points to show the direction and magnitude of the electric field. On the right, electric field lines depict the same situation. Notice that, as stated above, the electric field lines are drawn such that their tangents point in the same direction as the electric field vectors on the left. Because of the nature of electric fields, field lines never cross. Also, the vectors shrink as you move away from the charge, and the electric field lines spread out as you move away from the charge. The spacing between electric field lines indicates the strength of the electric field, just as the length of vectors indicates the strength of the electric field. The greater the spacing between field lines, the weaker the electric field. Although the advantage of field lines over field vectors may not be apparent in the case of a single charge, electric field lines present a much less cluttered and more intuitive picture of more complicated charge arrangements. Part A Which of the following figures correctly depicts the field lines from an infinite uniformly negatively charged sheet? Note that the sheet is being viewed edge-on in all pictures. Hint 1. Description of the field Recall that the field around an infinite charged sheet is always perpendicular to the sheet and that the field strength does not change, regardless of distance from the sheet. ANSWER: Correct Part B In the diagram from part A , what is wrong with figure B? (Pick only those statements that apply to figure B.) Check all that apply. ANSWER: Correct Part C Which of the following figures shows the correct electric field lines for an electric dipole? A B C D Field lines cannot cross each other. The field lines should be parallel because of the sheet's symmetry. The field lines should spread apart as they leave the sheet to indicate the weakening of the field with distance. The field lines should always end on negative charges or at infinity. ANSWER: Correct This applet shows two charges. You can alter the charge on each independently or alter the distance between them. You should try to get a feeling for how altering the charges or the distance affects the field lines. Part D In the diagram from part C , what is wrong with figure D? (Pick only those statements that apply to figure D.) Check all that apply. ANSWER: A B C D Correct In even relatively simple setups as in the figure, electric field lines are quite helpful for understanding the field qualitatively (understanding the general direction in which a certain charge will move from a specific position, identifying locations where the field is roughly zero or where the field points a specific direction, etc.). A good figure with electric field lines can help you to organize your thoughts as well as check your calculations to see whether they make sense. Part E In the figure , the electric field lines are shown for a system of two point charges, and . Which of the following could represent the magnitudes and signs of and ? In the following, take to be a positive quantity. ANSWER: Field lines cannot cross each other. The field lines should turn sharply as you move from one charge to the other. The field lines should be smooth curves. The field lines should always end on negative charges or at infinity. Q A Q B Q A Q B q Correct Very far from the two charges, the system looks like a single charge with value . At large enough distances, the field lines will be indistinguishable from the field lines due to a single point charge . Question 5 An charge with mass and charge is emitted from the origin, . A large, flat screen is located at . There is a target on the screen at y position , where . In this problem, you will examine two different ways that the charge might hit the target. Ignore gravity in this problem. Part A Assume that the charge is emitted with velocity in the positive x direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive y direction. What should the magnitude of the electric field be if the charge is to hit the target on the screen? Express your answer in terms of , , , , and . Hint 1. How to approach the problem Once you determine the force on the charge due to the electric field, this becomes a standard two-dimensional kinematics problem. To solve the problem, first determine the equations of motion in both the x and y directions. Then use the fact that at some final time you know that the position of the charge is to obtain two equations in terms of the two unknowns and . Eliminate and solve for . Hint 2. Find the equation of motion in the x direction Find an expression for , the charge's x position as a function of time. Express your answer in terms of as well as any of the given variables and constants. Hint 1. Find the force in the x direction What net force does the charge experience in the x direction? Hint 1. Formula for the force on a charge in an electric field The formula for the force on a charge in an electric field is , , , , , = +qQ A = −qQ B = +7qQ A = −3qQ B = +3qQ A = −7qQ B = −3qQ A = +7qQ B = −7qQ A = +3qQ B + = +4qQ A Q B +4q m q (x,y) = (0,0) x = L y h > 0y h v 0 E m q y h v 0 L t final (L, )y h E t final t final E x(t) t F x F " q E " = q " " .ANSWER: Correct Hint 2. A helpful kinematic equation Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is . ANSWER: Correct Hint 3. Find the equation of motion in the y direction Find an expression for , the charge's y position as a function of time. Express your answer in terms of as well as any of the given variables and constants. Hint 1. Find the force in the y direction What is the net force acting on the charge in the y direction? Express your answer in terms of the given variables and constants. Hint 1. Formula for the force on a charge in an electric field The formula for the force on a charge in an electric field is . ANSWER: = qF " E " = 0F x s u a t s = ut + a 1 2 t 2 = x(t) tv 0 y(t) t F " q E " = qF " E " = F y qE Correct Notice that the constant field provides a constant force, which yields a constant acceleration. The kinematics of charge in a constant electric field are not very different from of those of mass in a constant gravitational field. Hint 2. A helpful kinematic equation Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is . ANSWER: Correct Hint 4. Combine Your Results At some final time , you have and . Starting with these two equations, eliminate and solve for . Hint 5. Find Use the equation for the motion of the charge in the x direction to find . Express your answer in terms of the variables and . ANSWER: Correct Now substitute into the equation and solve for . ANSWER: s u a t s = ut + a 1 2 t 2 = y(t) 1 2 qE m t 2 t final x( ) = Lt final y( ) =t final y h t final E t final t final v 0 L = t final L v 0 = L/t final v 0 y( ) =t final y h E = E 2 my h q( ) L v 0 2 Correct Part B Now assume that the charge is emitted with velocity in the positive y direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive x direction. What should the magnitude of the electric field be if the charge is to hit the target on the screen? Express your answer in terms of , , , , and . Hint 1. How to approach the problem Just as in the previous part, once you determine the force on the charge due to the electric field, this becomes a standard two-dimensional kinematics problem. To solve the problem, first determine the equations of motion in both the x and y directions. Then use the fact that at some final time you know that the position of the charge is to obtain two equations in terms of the two unknowns and . Eliminate and solve for . Hint 2. Find the equation of motion in the y direction Find an expression for , the charge's y position as a function of time. Express your answer in terms of as well as any of the given variables and constants. Hint 1. Find the force in the y direction What net force does the charge experience in the y direction? Hint 1. Formula for the force on a charge in an electric field The formula for the force on a charge in an electric field is . ANSWER: Hint 2. A helpful kinematic equation Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is . ANSWER: v 0 E m q y h v 0 L t final (L, )y h E t final t final E y(t) t F y F " q E " = qF " E " = 0 F y s u a t s = ut + a 1 2 t 2 Hint 3. Find the equation of motion in the x direction Find an expression for , the charge's x position as a function of time. Express your answer in terms of as well as any of the given variables and constants. Hint 1. Find the force in the x direction What is the net force acting on the charge in the x direction? Express your answer in terms of the given variables and constants. Hint 1. Formula for the force on a charge in an electric field The formula for the force on a charge in an electric field is . ANSWER: Hint 2. A helpful kinematic equation Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is . ANSWER: Hint 4. Combine your results At some final time , you have and . Starting with these two equations, eliminate and solve for . Hint 5. Find Use the equation for the motion of the charge in the y direction to find . Express your answer in terms of the variables and . ANSWER: = y(t) tv 0 x(t) t F " q E " = qF " E " = F x qE s u a t s = ut + a 1 2 t 2 = x(t) 1 2 qE m t 2 t final x( ) = Lt final y( ) =t final y h t final E t final t final v 0 y h ANSWER: Correct The equations of motion for this part are identical to the equations of motion for the previous part, with and interchanged. Thus it is no surprise that the answers to the two parts are also identical, with and interchanged. Question 6 Point charges 5.00 and 5.00 are separated by distance 4.00 , forming an electric dipole. Part A Find the magnitude of the electric dipole moment. Express your answer in coulomb meters to three significant figures. Hint 1. How to approach the problem Use the equation for the dipole moment (not the electric field or repulsive force on the charges). Also, check that your answer has dimensions appropriate for a dipole moment. Hint 2. Formula for dipole moment The formula for the magnitude of the dipole moment of a pair of opposite charges, each with magnitude = 5.00 , separated by a distance = 4.00 is . ANSWER: Correct Part B What is the direction of the electric dipole moment? ANSWER: = t final y h v 0 = E 2mL q( ) y h v 0 2 L y h L y h = −q 1 nC = +q 2 nC mm p q nC d mm p = qd 2.00×10 −11 C ⋅m Correct Part C The charges are in a uniform electric field whose direction makes an angle 36.7 with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10 −9 ? Express your answer in newtons per coulomb to three significant figures. Hint 1. Find an equation for the torque Which of the following is an expression for the magnitude of the torque? ANSWER: Hint 2. How to obtain Once the torque is expressed in terms of the dipole moment, the electric field, and the angle between the two, you can solve for . ANSWER: Correct Question 7 Learning Goal: To understand how the charges within a conductor respond to an externally applied electric field. To illustrate the behavior of charge inside conductors, consider a long conducting rod that is suspended by insulating strings (see the figure). Assume that the rod is initially electrically neutral, and that it remains so for this discussion. The rod is positioned along the x axis, and an external electric field that points in the positive x direction (to the right) can be applied to from to from to q 1 q 2 q 2 q 1 # N ⋅m qdE qdE sin(θ) qd E qd E sin(θ) E E 611 N/C the rod and the surrounding region. The atoms in the rod are composed of positive nuclei (indicated by plus signs) and negative electrons (indicated by minus signs). Before application of the electric field, these atoms were distributed evenly throughout the rod. Part A What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction is applied? (Ignore internal fields in the rod for the moment.) Hint 1. Formula for the force on a charge in an electric field The force on a charge in an electric field is given by . ANSWER: Correct Part B What is the motion of the negative electrons and positive atomic nuclei caused by the external field? Hint 1. How to approch this part Newton's 2nd Law tells you that an object at rest will move in the direction of the force applied on it. F " q E " = qF " E " Both electrons and nuclei experience a force to the right. The nuclei experience a force to the right and the electrons experience a force to the left. The electrons experience a force to the left but the nuclei experience no force. The electrons experience no force but the nuclei experience a force to the right. Hint 2. Masses and charges of nuclei and electrons A nucleus contains as many protons as the atom does electrons. So if the atom has N electrons, the nucleus will contain N protons. This means that the force on the nucleus will be N times as much as that on the electron. This N is of the order of 10-100. However, the mass of a nucleus is roughly 2N times the mass of a proton, since it contains both protons and neutrons. Each proton itself weighs about 1836 times as much as an electron! So a typical nucleus really does weigh a lot more than an electron. Given this information, how would the distance moved by a nucleus compare with that moved by an electron? ANSWER: Correct The nuclei of the atoms of a conducting solid remain almost in their places in the crystal lattice, while the electrons relatively move a lot. In an insulator, the electrons are constrained to stay with their atoms (or molecules), and at most, the charge distribution is displaced slightly. The motion of the electrons due to the external electric field constitutes an electric current. Since the negatively charged electrons are moving to the left, the current, which is defined as the "flow" of positive charge, moves to the right. Part C Imagine that the rightward current flows in the rod for a short time. As a result, what will the net charge on the right and left ends of the rod become? Hint 1. How to approach this part Remember that the rod as a whole must remain electrically neutral even if the charges are redistributed. This is because applying an electric field does not change the charge on the rod, only redistributes it. ANSWER: Both electrons and nuclei move to the right. The nuclei move to the right and the electrons move to the left through equal distances. The electrons move to the left and the nuclei are almost stationary. The electrons are almost stationary and the nuclei move to the right. left end negative and right end positive left end negative and right end negative left end negative and right end nearly neutral left end nearly neutral and right end positive both ends nearly neutral Correct Given that the positively charged nuclei do not move, why does the right end of the rod become positively charged? The reason is that some electrons have moved to the left end, leaving an excess of stationary nuclei at the right end. Part D The charge imbalance that results from this movement of charge will generate an additional electric field in the region within the rod. In what direction will this field point? Hint 1. Direction of the electric field The electric field point away from positive charges and towards negative ones. ANSWER: Correct An electric field that exists in an isolated conductor will cause a current flow. This flow sets up an electric field that opposes the original electric field, halting the motion of the charges on a nanosecond time scale for meter-sized conductors. For this reason, an isolated conductor will have no static electric field inside it, and will have a reduced electric field near it. This conclusion does not apply to a conductor whose ends are connected to an external circuit. In a circuit, a rod (or wire) can conduct current indefinitely. Question 8 Part A Rank in order, from largest to smallest, the electric field strengths to at points 1 to 4 in the figure. Rank from largest to smallest. To rank items as equivalent, overlap them. It will point to the right and enhance the initial applied field. It will point to the left and oppose the initial applied field. E 1 E 4 ANSWER: Correct Part B Explain. ANSWER: Submitted, grade pending Question 9 A small segment of wire in the figure contains 12 of charge. Part A The segment is shrunk to one-third of its original length. What is the ratio , where and are the initial and final linear charge densities? ANSWER: Correct Part B A proton is very far from the wire. What is the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk? ANSWER: 3642 Character(s) remaining The closer the electric field lines are the stronger the field will be. The farther the electric field lines are from one another, the weaker the field gets. nC /λ f λ i λ i λ f = 3 λ f λ i /F f F i Correct Part C Suppose the original segment of wire is stretched to 11 times its original length. How much charge must be added to the wire to keep the linear charge density unchanged? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Question 10 The irregularly shaped area of charge in the figure has surface charge density . Each dimension ( and ) of the area is reduced by a factor of 3.86. Part A What is the ratio , where is the final surface charge density? ANSWER: Correct = 1 F f F i = 120 ∆q nC η i x y /η f η i η f = 14.9/η f η i Part B An electron is very far from the area. What is the ratio of the electric force on the electron after the area is reduced to the force before the area was reduced? ANSWER: Correct Question 11 A proton orbits a long charged wire, making 1.00 revolutions per second. The radius of the orbit is 1.40 . Part A What is the wire's linear charge density? Express your answer with the appropriate units. Hint 1. How to approach this problem From the radius and frequency of the proton's orbit, we can determine its centripital acceleration. From the equation for the electric field a distance 1.40 from a long charged wire we can find the force on a proton in terms of the wire's linear charge density. Since we know the mass of a proton, we have the three components of newton's second law. Setting Force equal to Mass times Acceleration, we have an equation with one unknown, the linear charge density. ANSWER: Correct Question 12 Air "breaks down" when the electric field strength reaches , causing a spark. A parallel-plate capacitor is made from two 4.0 4.0 plates. Part A How many electrons must be transferred from one disk to the other to create a spark between the disks? Express your answer to two significant figures and include the appropriate units. /F f F i = 1.00/F f F i ×10 6 cm cm -4.49 nC m 3 × N/C10 6 cm × cm Hint 1. Electic field between two plates of capacitor Recall that the electric field between the two plates of a capacitor with charge densities of and is given by . You can use this equation to find the necessary charge on each plate, and convert this charge into number of electrons. ANSWER: Correct Question 13 Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion of infinite charged sheets. This approximation is useful when a problem deals with points whose distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you will look at the electric field from two finite sheets and compare it to the results for infinite sheets to get a better idea of when this approximation is valid. Consider two thin disks, of negligible thickness, of radius oriented perpendicular to the x axis such that the x axis runs through the center of each disk. The disk centered at has positive charge density , and the disk centered at has negative charge density , where the charge density is charge per unit area. Part A What is the magnitude of the electric field at the point on the x axis with x coordinate ? Express your answer in terms of , , , and the permittivity of free space . Hint 1. How to approach the problem When calculating the electric field from more than one charge or a continuous charge distribution, one makes use of the superposition principle, which states that the electric field from multiple charges equals the vector sum of the fields from each individual charge. To find the electric field at a point, add the field due to each disk at that point. Be careful of whether the field magnitudes should add or subtract. The easiest way to be sure is to draw a figure with the two disks and arrows for the electric field direction on each side of each disk. If the arrows for the two disks point the same way, then the magnitudes add. If they point in opposite directions, the magnitudes +σ −σ E = σ ε 0 2.7×10 11 R x = 0 η x = a −η E a/2 η R a # 0 subtract. Hint 2. The magnitude of the electric field due to a single disk The magnitude of the electric field along the x axis for a charged disk centered at is , where is the radius of the disk, is the charge density on the disk, is the permittivity of free space, and is the x coordinate. Be careful in determining the direction in which the electric field due to each disk points. Hint 3. Determine the general form of the electric field between the disks Which of the following equations represents the magnitude of the electric field between the two disks? Hint 1. Determine whether the magnitudes should add or subtract Which of the following statements properly describes the directions of the electric fields due to each disk in the region between the two disks? ANSWER: Correct Since the two fields point in the same direction, their magnitudes add. ANSWER: E disk x = 0 = [1− ]E disk η 2# 0 1 1+ /R 2 x 2 √ R η # 0 x E The electric fields due to both disks point to the left. The electric fields due to both disks point to the right. The electric field due to the positive disk points to the left, while the electric field due to the negative disk points to the right. The electric field due to the positive disk points to the right, while the electric field due to the negative disk points to the left. Correct ANSWER: Part B This question will be shown after you complete previous question(s). Question 14 The electric field is constant over each face of the cube shown in the figure . [1− ] η 2# 0 1 1+ /R 2 x 2 √ 2 [1− ] η 2# 0 1 1+ /R 2 x 2 √ [ − ] η 2# 0 1 1+ /(x−aR 2 ) 2 √ 1 1+ /R 2 x 2 √ [2− − ] η 2# 0 1 1+ /(x−aR 2 ) 2 √ 1 1+ /R 2 x 2 √ [2 + − ] η 2# 0 1 1+ /(x−aR 2 ) 2 √ 1 1+ /R 2 x 2 √ = E Part A Does the box contain positive charge, negative charge, or no charge? ANSWER: Correct Question 15 The cube in the figure contains negative charge. The electric field is constant over each face of the cube. positive charge negative charge no charge Part A Does the missing electric field vector on the front face point in or out? ANSWER: Correct Part B What strength must this field exceed? ANSWER: Correct Question 16 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: The unknown vector points into the front face of the cube. The unknown vector points out the front face of the cube. The unknown vector is zero. 5 10 15 20 N/C N/C N/C N/C = " ⋅ d = ,Φ E E " A " q encl # 0 = 8.85× /(N ⋅ )# 0 10 −12 C 2 m 2 Correct In the integral for Gauss's law, the vector represents an infinitesimal surface element. The magnitude of is the area of the surface element. The direction of is normal to the surface element, pointing out of the enclosed volume. Part B In Gauss's law, to what does refer? ANSWER: 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. Question 17 Suppose a disk with area is placed in a uniform electric field of magnitude . The disk is oriented so that the vector normal to its surface, , makes an angle with the electric field, as shown in the figure. around the perimeter of a closed loop over the surface bounded by a closed loop over a closed surface dA " dA " dA " q encl 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 A E n ^ θ Part A What is the electric flux through the surface of the disk that is facing right (the normal vector to this surface is shown in the figure)? Assume that the presence of the disk does not interfere with the electric field. Express your answer in terms of , , and Hint 1. Definition of electric flux The electric flux through a surface is given by , where is an infinitesimal element of area on the surface. Hint 2. Simplifying the integrand Note that the formula for electric flux can be expressed in the following way: . Since we are dealing with a constant electric field, which therefore does not vary across the surface of the disk, we can take outside of the integral. This gives a much simpler expression: . Hint 3. Evaluate the scalar product Find in terms of and . ANSWER: ANSWER: Φ E E A θ Φ E = ∫ ⋅ dΦ E E " A " dA " = ∫ ⋅ d = ∫ ⋅ dAΦ E E " A " E " n^ ⋅E " N ^ = ⋅ ∫ dAΦ E E " n^ ⋅E " n ^ E θ = ⋅E " n ^ Ecos(θ) Correct Question 18 Part A The electric flux through the shaded surface is ANSWER: Correct Correct! Question 19 = Φ E EAcosθ 0. 400 N /C. 200 N m/C. Flux isn't defined for an open surface. m 2 Part A Which spherical Gaussian surface has the larger electric flux? ANSWER: Correct Flux depends only on the enclosed charge, not the radius. Score Summary: Your score on this assignment is 88.7%. You received 15.09 out of a possible total of 17 points. Surface B. Not enough information to tell. They have the same flux. Surface A. Jasmin Khoo Problem Set 02: Electric Fields