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Heavenly The Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... Share Report Abuse Next Blog Create Blog Sign In The Heavenly Messenger Tuesday, January 27, 2009 Mastering Physics Solutions Use Wisely [ Item View ] The Electric Field of a Ball of Uniform Charge Density A solid ball of radius r_bhas a uniform charge density rho. Part A What is the magnitude of the electric field E(r)at a distance r_b" title="" v:shapes="_x0000_i1028" align="middle" border="0" height="16" width="38">from the center of the ball? Express your answer in terms of rho, r_b, r, and epsilon_0. The Electric Field of a Ball of Uniform Charge Density Part A What is the magnitude of the electric field E(r)at a distance r_b" title="" v:shapes="_x0000_i1484" align="middle" border="0" height="16" width="38">from the center of the ball? Hint 1. Gauss's law Gauss's law can be written as \oint \vec{E} \cdot d \vec{A} = \frac{q_{\rm encl}}{\epsilon_0 }, where d\vec Arefers to an infinitesimal element of an imaginary Gaussian surface, q_enclis the net charge enclosed in the Gaussian surface, and epsilon_0is the permittivity of free space. Always choose a Gaussian surface that matches the symmetry of the problem. Implicit in the question "what is E(r)?" is the assumption that the electric field depends only on distance from the origin, which is also the center of the charged ball. Also, the ball has uniform charge density. Therefore, the electric field must point either radially outward or radially inward, since by symmetry there is no possibility for the electric field to point in any other direction. Given the symmetry of this problem, the best Gaussian surface to use is a sphere centered at the origin. Since the electric field is the same at all points on this surface, the constant E(r)can be "pulled out" of the integrand. The left side of Gauss's law reduces to E(r)A(r), where A(r)is the surface area of a sphere with radius r. Hint 2. Find q_encl The q_enclin Gauss's law refers to the net charge enclosed inside the Gaussian surface. What is q_enclhere? Hint 1. What is the volume of the sphere? If a body has uniform charge density rho, the charge in a volume Vis \rho V(this formula is the same as that for the mass of a sphere of uniform mass density). What is the volume of a sphere with radius r? Express your answer in terms of piand r. V =(4/3)*pi*r^3 Express your answer in terms of rho, pi, and r_b. q_encl =(4/3)*pi*r_b^3*rho Express your answer in terms of rho, r_b, r, and epsilon_0. E(r) ={\rho}{\cdot}\frac{\left(r_{b}{^{3}}\right)}{3{\cdot}epsilon_{0}{\cdot}r^{2}} Correct my answers 1 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... ANSWER: E(r) = {\rho}{\cdot}\frac{\left(r_{b}{^{3}}\right)}{3{\cdot}epsilon_{0}{\cdot}r^{2}} Notice that this result is identical to that reached by applying Coulomb's law to a point charge centered at the origin with q=\rho V. The field outside of a uniformly charged sphere does not depend on the size of the sphere, only on its charge. A uniformly charged sphere generates an electric field as if all the charge were concentrated at its center. Part B What is the magnitude of the electric field E(r)at a distance r<r_bfrom the center of the ball? The Electric Field of a Ball of Uniform Charge Density Part B What is the magnitude of the electric field E(r)at a distance r<r_bfrom the center of the ball? Hint 1. How does this situation compare to that of the field outside the ball? Now you are asked to find the electric field inside the ball, as opposed to outside the ball. What is different in the physical situation when you move inside the ball? The electric field now depends on spatial variables besides the radius. The direction of Eis different. The shape of the appropriate Gaussian surface is no longer a sphere. The net charge q_enclenclosed by the Gaussian surface is different. The charge density rhois different. Express your answer in terms of rho, r, r_b, and epsilon_0. E(r) =\left(\frac{{\rho}{\cdot}r}{3{\cdot}epsilon_{0}}\right) Correct my answers Express your answer in terms of rho, r, r_b, and epsilon_0. ANSWER: E(r) = \left(\frac{{\rho}{\cdot}r}{3{\cdot}epsilon_{0}}\right) Part C Let E(r)represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true? Hint C.1 Plot the electric field The figure shows a plot of the electric field as a function of r. Check all that apply. ANSWER: E(0)=0. 2 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... E(r_{\rm b})=0. \lim_{r\to\infty}E(r)=0. The maximum electric field occurs when r=0. The maximum electric field occurs when r=r_{\rm b}. The maximum electric field occurs as r\to\infty. [ Print ] [ Item View ] The Charge Inside a Conductor A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q. Part A What is the total surface charge q_inton the interior surface of the conductor (i.e., on the wall of the cavity)? Hint A.1 Gauss's law and properties of conductors The Charge Inside a Conductor Part A What is the total surface charge q_inton the interior surface of the conductor (i.e., on the wall of the cavity)? Hint 1. Gauss's law and properties of conductors The net electric field in the interior of the conducting material must be zero. (The electric field in the cavity, however, need not be zero.) Knowing this, you can use Gauss's law to find the net charge on the interior surface of the cavity. Use the following Gaussian surface: an imaginary sphere, centered at the cavity, that has an infinitesimally larger radius than that of the cavity, so that it encompasses the inner surface of the cavity. This Gaussian surface lies within the conductor, so the field on the Gaussian surface must be zero. Thus, by Gauss's law, the net charge inside the Gaussian surface must be zero as well. But you know that there is a point charge qwithin the Gaussian surface. If the net charge within the Gaussian surface must be zero, how much charge must be present on the surface of the cavity? q_int =-q Correct my answers ANSWER: q_int = -q Part B What is the total surface charge q_exton the exterior surface of the conductor? Hint B.1 Properties of the conductor The Charge Inside a Conductor Part B 3 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... What is the total surface charge q_exton the exterior surface of the conductor? Hint 1. Properties of the conductor In the problem introduction you are told that the conducting sphere is neutral. Furthermore, recall that the free charges within a conductor always accumulate on the conductor's surface (or surfaces, in this case). You found the net charge on the conductor's interior surface in Part A. If the conductor is to have zero net charge (as it must, since it is neutral), how much charge must be present on its exterior surface? q_ext =q Correct my answers ANSWER: q_ext = q Part C What is the magnitude E_intof the electric field inside the cavity as a function of the distance rfrom the point charge? Let k, as usual, denote \frac{1}{4\pi\epsilon_0}. Hint C.1 How to approach the problem Hint C.2 Charge distributions and finding the electric field The Charge Inside a Conductor Part C What is the magnitude of the electric field inside the cavity as a function of the distance from the point charge? Let , as usual, denote . Hint 1. How to approach the problem The net electric field inside the conductor has three contributions: 1. from the charge q; 2. from the charge on the cavity's walls q_int; 3. from the charge on the outer surface of the spherical conductor q_ext. However, the net electric field inside the conductor must be zero. How must q_intand q_extbe distributed for this to happen? Here's a clue: the first two contributions above cancel each other out, outside the cavity. Then the electric field produced by q_extinside the spherical conductor must separately be zero also. How must q_extbe distributed for this to happen? After you have figured out how q_intand q_extare distributed, it will be easy to find the field in the cavity, either by adding field contributions from all charges, or using Gauss's Law. Hint 2. Charge distributions and finding the electric field q_intand q_extare both uniformly distributed. Unfortunately there is no easy way to determine this, that is why a clue was given in the last hint. You might hit upon it by assuming the simplest possible distribution (i.e., uniform) or by trial and error, and check that it works (gives no net electric field inside the conductor). If q_extis distributed uniformly over the surface of the conducting sphere, it will not produce a net electric field inside the sphere. What are the characteristics of the field q_intproduces inside the cavity? zero 0 same as the field produced by a point charge qlocated at the center of the sphere the k q/r^2 the same as the field produced by a point charge located at the position of the charge in the cavity 4 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... 2k q/r^2 Correct my answers ANSWER: 0 k q/r^2 2k q/r^2 Part D Now a is the electric field , is brought near the outside of the conductor. Which of the following quantities would change? What second charge, q_2 E_extoutside the conductor? The Charge Inside a Conductor Part E Part D What is the electric field on theoutside the cavity, The total surface charge E_ext wall of the conductor? : q_int Hint 1. How to approach the problem Hint E.1 Canceling the field due to the charge q The net electric field inside the conductor has three contributions: 1. from the charge q; The Charge Inside a Conductor 2. from the charge on the cavity's walls q_int; 3. from the charge on the outer surface of the spherical conductor q_ext. Part E The total the net charge field inside the conductor must However, surface electricon the wall of the cavity, q_int: be zero. How must q_intand q_extbe distributed for this to happen? Hint 1. helpful clue: field due to contributions Here's aCanceling the the first twothe charge q above cancel each other out, outside the cavity. Then the electric field produced by q_extinside the spherical conductor must be separately be zero also. How must q_extbe distributed for this to The net What sort of field would such a distribution produce outside t the inner conductor cavity will always arrange happen? electric field inside a conductor is always zero. The charges onhe conductor? themselves so that the field lines due to charge qdo not penetrate into the conductor. Hint 2. The distribution of q_ext would change If q_extis distributed uniformly over the surface of the conducting sphere, it will not produce a net electric field inside the would not change sphere. What are the characteristics of the field it produces outside the sphere? zero Correct mythe same as the field produced by a point charge qlocated at the center of the sphere answers the same ANSWER: as the field produced by a point charge located at the position of the charge in the cavity would change would not change Correct my answers ANSWER: Part F zero the same the exterior of the conductor, The total surface charge on as the field produced by a point charge qlocated at the center of the sphere q_ext: Hint F.1 the same as the field produced by a point charge located at the position of the charge in the cavity Canceling the field due to the charge q_2 5 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... The Charge Inside a Conductor Part F The total surface charge on the exterior of the conductor, q_ext: Hint 1. Canceling the field due to the charge q_2 The net electric field inside a conductor is always zero. The charges on the outer surface of the conductor will rearrange themselves to shield the external field completely. Does this require the net charge on the outer surface to change? would change would not change Correct my answers ANSWER: would change would not change Part G The electric field within the cavity, E_cav: ANSWER: would change would not change Part H The electric field outside the conductor, E_ext: ANSWER: would change would not change [ Print ] [ Item View ] The Electric Field inside and outside a Charged Insulator A slab of insulating material of uniform thickness d, lying between -{d}/{2}to +{d}/{2}along the x axis, extends infinitely in the y and z directions, as shown in the figure. The slab has a uniform charge density rho. The electric field is zero in the middle of the slab, at x=0. Part A 6 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... Which of the following statements is true of the electric field E_vecat the surface of one side of the slab? ANSWER: The direction of \vec{E}is constant but its magnitude varies across the surface. Both the magnitude and the direction of \vec{E}are constant across the entire surface. The direction of \vec{E}varies across the surface but its magnitude is constant. Both the magnitude and the direction of \vec{E}vary across the surface. Part B What is the angle thetathat the field E_vecmakes with the surface of the slab, which is perpendicular to the x direction? Express your answer in radians, in terms of pi. ANSWER: theta = 1.57 {\rm rad} Part C What is E_out, the magnitude of the electric field outside the slab? As implied by the fact that E_outis not given as a function of x, this magnitude is constant everywhere outside the slab, not just at the surface. The Electric Field inside and outside a Charged Insulator Part C What is , the magnitude of the electric field outside the slab? As implied by the fact that is not given as a function of , this magnitude is constant everywhere outside the slab, not just at the surface. Hint 1. How to approach the problem Apply Gauss's law. You already know, by symmetry, that the electric field is perpendicular to the slab, and that it has constant magnitude across the surface of the slab. Use this information to choose an appropriate Gaussian surface. Hint 2. Gauss's law Gauss's law can be written as \Phi_E = q/\epsilon_0, where Phi_Eis the electric flux through a Gaussian surface (which depends on E_out), and qis the total charge enclosed by the surface. Hint 3. A Gaussian surface for this problem Construct a Gaussian surface that is a retangular box with sides oriented parallel to the electric field. There are two equally good choices for where to place the ends of the box: Put one end of the box outside the slab, perpendicular to the electric field, and the other end exactly in the middle of the slab (where the electric field is zero). Put one end of the box outside the slab and the other end an equal distance on the other side of the slab, both perpendicular to the electric field. By symmetry, the field at both ends of the box must be equal in magnitude and opposite in direction. Both choices will lead to the correct answer, but the remaining hints will assume that you have chosen a Gaussian surface with one end in the middle the of slab. Hint 4. Calculate the enclosed charge What is q, the total charge enclosed within the Gaussian surface? Hint 1. How to find the enclosed charge The enclosed charge is equal to the volume of the slab that is contained within the Gaussian surface times the charge density rho. 7 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... Express your answer in terms of d, rho, and the cross-sectional area Aof the Gaussian surface. q =A*(d/2)*rho Hint 5. Compute the electric flux Let E_outdenote the magnitude of the electric field at the end of the Gaussian surface outside the slab. What is Phi_E, the total electric flux through the Gaussian surface? Hint 1. How to compute the flux The sides of the box that we have chosen as our Gaussian surface are parallel to the electric field. The ends of the box are perpendicular to the field. Where the electric field is perpendicular to the surface, the flux is given by E A, where Eis the magnitude of the electric field and Ais the area of the surface. Where the electric field is parallel to the surface, the flux is zero. Express your answer in terms of E_outand the cross-sectional area Aof the Gaussian surface. Phi_E =A*E_out Express your answer in terms of d, rho, and epsilon_0. E_out =\frac{d}{2epsilon_{0}}{\rho} Correct my answers Express your answer in terms of d, rho, and epsilon_0. ANSWER: E_out = \frac{d}{2epsilon_{0}}{\rho} Part D What is E_in(x), the magnitude of the electric field inside the slab as a function of x? Hint D.3 Calculate the enclosed charge The Electric Field inside and outside a Charged Insulator Part D What is E_in(x), the magnitude of the electric field inside the slab as a function of x? Hint 1. How to approach the problem As in the calculation of the field outside the slab, apply Gauss's law. The procedure is exactly the same as before, but you will need to choose a different Gaussian surface. Hint 2. A Gaussian surface for this problem Try using a Gaussian surface in the shape of a box with cross-sectional area A. Put one side of the box in the middle of the slab (x=0), where the electric field is zero. Put the other end of the box at some arbitrary location xinside the slab to determine the field \vec{E}_{\rm in}(x)at that position. Hint 3. Calculate the enclosed charge What is q, the charge enclosed within the Gaussian surface? Express your answer in terms of the cross-sectional area of the Gaussian surface A, rho, and x. q =rho*x*A Hint 4. Compute the flux Let E_indenote the magnitude of the electric field inside the slab at position x. What is Phi_E, the total electric flux through the Gaussian surface? Express your answer in terms of E_inand A. 8 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... Phi_E =E_in*A E_in(x) =\left(\frac{{\rho}x}{epsilon_{0}}\right) Correct my answers ANSWER: E_in(x) = \left(\frac{{\rho}x}{epsilon_{0}}\right) Basic models of diodes and transistors (which are components used in more complex circuits, like those on computer chips) treat regions inside them as slabs of charge. In this example you found that the electric field points in opposite directions on the two sides of x = 0. However, if a slab with negative charge were added behind this slab, i.e., from x = \frac{d}{2}to x = \frac{3d}{2}, you can check that the electric field would be zero in the regions where there is no charge, because the fields due to the positive and negative charges cancel, and that the electric field in the regions where there is charge is always in the positive x direction. Such a setup (usually called a PN junction) can be used as an electric "one-way street," since it supports the flow of positive charge only in the positive direction, i.e., along the electric field, and severely inhibits the flow of current in the opposite direction. [ Print ] [ Item View ] Problem 22.38 A very long conducting tube (hollow cylinder) has inner radius aand outer radius b. It carries charge per unit length + \alpha , where \alphais a positive constant with units of {\rm C}/{\rm m}. A line of charge lies along the axis of the tube. The line of charge has charge per unit length + \alpha . Part A Calculate the electric field in terms of \alphaand the distance rfrom the axis of the tube for r<a. Express your answer in terms of the variables \alpha, r, and constant \epsilon_0. View Full Document

= \frac{{\alpha}}{2{\cdot}{\pi}{\cdot}epsilon_{0}{\cdot}r} Part B Calculate the electric field in terms of \alphaand the distance rfrom the axis of the tube for a<r<b. Express your answer in terms of the variables \alpha, r, and constant \epsilon_0. = 0 Part C Calculate the electric field in terms of \alphaand the distance rfrom the axis of the tube for b" title="" v:shapes="_x0000_i1158" align="middle" border="0" height="21" width="33">. Express your answer in terms of the variables \alpha, r, and constant \epsilon_0. 9 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... = \frac{{\alpha}}{{\pi}repsilon_{0}} Part D What is the charge per unit length on the inner surface of the tube? ANSWER: \frac{q}{l} = -1.00 \alpha Part E What is the charge per unit length on the outer surface of the tube? ANSWER: \frac{q}{l} = 2.00 \alpha [ Print ] [ Item View ] Problem 22.48 A solid conducting sphere with radius Rcarries a positive total charge Q. The sphere is surrounded by an insulating shell with inner radius Rand outer radius Part A Find the value of \rhoso that the net charge of the entire system is zero. Express your answer in terms of the variables Q, R, and appropriate constants. ANSWER: \rho = -\frac{3Q}{28{\cdot}pi{\cdot}R^{3}} Part B If \rhohas the value found in part A, find the magnitude of the electric field in the region 0 < r < R. Express your answer in terms of the variables Q, R, r, and appropriate constants. ANSWER: E_1 = 0 Part C If \rhohas the value found in part A, find the magnitude of the electric field in the region R < r < 2R. Part E Express your answer in terms of the variables Q, R, r, and appropriate constants. ANSWER: E_2 = the variables , , , and appropriate constants. Express your answer in terms of \frac{Q}{28{\cdot}pi{\cdot}epsilon_{0}}\left[\frac{8}{r^{2}}-\frac{r}{R^{3}}\right] QRr If \rhohas the value found in part A, find the magnitude of the electric field in the region 2R" title="" v:shapes="_x0000_i1195" align="middle" border="0" height="21 10 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... ANSWER: E_3 = 0 Part F As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in previous parts agree with th ANSWER: Answer Key: We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting spherebut [ Print ] A nonuniform, but spherically symmetric, distribution of charge has a charge density \rho (r)given as follows: for r\leq R for r \geq R \rho (r)= \rho _{0} (1-4r/3R) \rho (r)=0 where \rho_0is a positive constant [ Item View ] Problem 22.58 A nonuniform, but spherically symmetric, distribution of charge has a charge density \rho (r)given as follows: for r\leq R for r \geq R where \rho_0is a positive constant. \rho (r)= \rho _{0} (1-4r/3R) \rho (r)=0 A nonuniform, but spherically symmetric, distribution of charge has a charge density \rho (r)given as follows: for r\leq R for r \geq R \rho (r)= \rho _{0} (1-4r/3R) \rho (r)=0 where \rho_0is a positive constant. Part A Find the total charge contained in the charge distribution. Express your answer in terms of the variables r, R, \rho_0, and appropriate constants. ANSWER: q = 0 Part B Obtain an expression for the electric field in the region r \geq R. 11 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... Express your answer in terms of the variables r, R, \rho_0, and appropriate constants. ANSWER: E_1 = 0 Part C Obtain an expression for the electric field in the region r \leq R. Express your answer in terms of the variables r, R, \rho_0, and appropriate constants. ANSWER: E_2 = \frac{{\rho}_{0}{\cdot}r}{3epsilon_{0}}\left(1-\frac{r}{R}\right) Part D Find the value of rat which the electric field is maximum. Express your answer in terms of the variables r, R, \rho_0, and appropriate constants. ANSWER: r = \frac{R}{2} Part E Find the value of that maximum field. Express your answer in terms of the variables r, R, \rho_0, and appropriate constants. ANSWER: E_{\rm max} = \frac{{\rho}_{0}R}{12epsilon_{0}} [ Print ] Charge is distibuted uniformly over each of two spherical volumes with radius R. One sphere of charge is centered at the origin and the other at x=2R (Intro 1 figure) . Let left-hand sphere have positive charge Qand let the right-hand sphere have negative charge -Q. [ Item View ] Problem 22.64 Charge is distibuted uniformly over each of two spherical volumes with radius R. One sphere of charge is centered at the origin and the other at x=2R. Let left-hand sphere have positive charge Qand let the right-hand sphere have negative charge -Q. Part A Find the magnitude of the net electric field due to these two distributions of charge at the point x=0on the x-axis. Express your answer in terms of the variables Q, R, and appropriate constants. ANSWER: E_1 = \frac{Q}{16{\cdot}pi{\cdot}epsilon_{0}{\cdot}R^{2}} 12 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... Part B Find the direction of the net electric field at the point x=0on the x-axis. ANSWER: +x-direction -x-direction Part C Find the magnitude of the net electric field at the point x=R/2on the x-axis. Express your answer in terms of the variables Q, R, and appropriate constants. ANSWER: E_2 = \frac{17Q}{72{\cdot}pi{\cdot}epsilon_{0}{\cdot}R^{2}} Part D Find the direction of the net electric field at the point x=R/2on the x-axis. ANSWER: +x-direction -x-direction Part E Find the magnitude of the net electric field at the point x=Ron the x-axis. Express your answer in terms of the variables Q, R, and appropriate constants. ANSWER: E_3 = \frac{2Q}{4{\cdot}pi{\cdot}epsilon_{0}{\cdot}R^{2}} Part F Find the direction of the net electric field at the point x=Ron the x-axis. ANSWER: +x-direction -x-direction Part G Find the magnitude of the net electric field at the point x=3Ron the x-axis. 13 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... Express your answer in terms of the variables Q, R, and appropriate constants. ANSWER: E_4 = \frac{8Q}{36{\cdot}pi{\cdot}epsilon_{0}{\cdot}R^{2}} Part H Find the direction of the net electric field at the point x=3Ron the x-axis. ANSWER: +x-direction -x-direction [ Print ] An insulating sphere of radius a, centered at the origin, has a uniform volume charge density rho. A Charged Sphere with a Cavity Part A Find the electric field E_vec(r_vec)inside the sphere (for r< a) in terms of the position vector r_vec. Hint 1. How to approach the problem Apply Guass's law, which states that for a closed surface, the integral of the scalar product of the surface area and the electric field is directly proportional to the enclosed charge: \oint \vec{E}\cdot d\vec{A}=\frac{q_{\rm encl}}{\epsilon_0}. Because this problem deals with a uniform spherically symmetric volume charge density, a logical Gaussian surface in this case is a sphere of radius r, with r< a. Find the integral \oint\vec{E}\cdot d\vec{A}(which will involve E_vec(r_vec)) and the enclosed charge. Then solve for E_vec. Hint 2. Determine the enclosed charge What is the charge q_enclenclosed by a Gaussian sphere centered at the origin with radius r(for r< a)? Express your answer in terms of r(the magnitude of r_vec) and rho. q_encl =(4/3)*pi*rho*r^3 Hint 3. Calculate the integral over the Gaussian surface Because of the symmetry of the problem, the value of \vec{E} \cdot d\vec{A}is constant over the entire Gaussian surface. In particular, \oint \vec{E}\cdot d\vec{A} = E(r)A(r), where E(r)is the magnitude of the electric field at radius r, and A(r)is the surface area of the Gaussian sphere of radius r. Find an expression for A(r). Express your answer in terms of rand pi. A(r) =4*pi*r^2 Express your answer in terms of r_vec, rho, and epsilon_0. E_vec(r_vec) =\frac{\left({\rho}{\cdot}r\right)}{3{\cdot}epsilon_{0}}{\cdot}\hat{r} Correct my answers A Charged Sphere with a Cavity Part B A spherical cavity is excised from the inside of the sphere. The cavity has radius \frac{a}{4}and is centered at position h_vec, where 14 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... |\vec{h}|<\frac34 a, so that the entire cavity is contained within the larger sphere. Find the electric field inside the cavity. Hint 1. How to approach the problem Use the principle of superposition. A region of zero charge behaves just like a region with equal amounts of positive and negative charge. Consider the field produced by an imaginary sphere the size of the cavity, with charge density opposite that of the larger sphere. If you add the field from the imaginary sphere to the field produced by the original, intact, sphere, you will obtain the field produced by the sphere with the cavity. Hint 2. Find the field due to the imaginary sphere Consider an imaginary sphere with charge density -\rhoand radius \frac{a}{4}centered at h_vec. Ignoring the actual sphere, what is the field E_imag_vec(r_vec)inside the imaginary sphere? Hint 1. Use the result for a uniformly charged sphere The imaginary sphere is just like the uniformly charged sphere studied in Part A of this problem, except that it has a different charge density and different position. Therefore, you can use the result you already obtained from the uniformly charged sphere if you use the new charge density and if you replace r_vecwith a new vector s_vecthat represents the displacement from the center of the imaginary sphere to r_vec. What is s_vecin terms of r_vecand h_vec? s_vec =r_vec-h_vec Express your answer as a vector in terms of r_vec, h_vec, rho, and epsilon_0. E_imag_vec(r_vec) =-rho*(r_vec-h_vec)/(3*epsilon_0) Express your answer as a vector in terms of any or all of rho, epsilon_0, r_vec, and h_vec. E_vec(r_vec) =\frac{{\rho}\vec{h}}{3epsilon_{0}} Correct my answers Posted by Armour of God at 10:30 AM 0 comments: Post a Comment Newer Post Subscribe to: Post Comments (Atom) Home Older Post Followers 15 of 16 2/6/2011 8:28 PM The Heavenly Messenger: Mastering Physics Solutions Use Wisely http://rocketscient1st.blogspot.com/2009/01/mastering-physics-solutions-... 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