r, rsR Figure 2 : (a) The electric field inside the sphere is given by E = 30 (rb) (True,False) An insulating sphere of radius R has a spherical hole of radius a located within its volume and . This problem has been solved! Equation (18) is incorrect. It is an important tool since it permits the assessment of the amount of enclosed charge by mapping the field on a surface outside the charge distribution. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. \Longrightarrow \ \mathbf{E}_{\text{net inside cavity}}&=\mathbf{E}_{\text{inside sphere}}+\mathbf{E}_{\text{inside cavity}}\\ The same is true for the oppositely charged sphere, where the only difference should be a '-' sign up to the point $r=R/2$. Electric Potential of a Uniformly Charged Solid Sphere Electric charge on sphere: Q = rV = 4p 3 rR3 Electric eld at r > R: E = kQ r2 Electric eld at r < R: E = kQ R3 r Electric potential at r > R: V = Z r kQ r2 dr = kQ r Electric potential at r < R: V = Z R kQ r2 dr Z r R kQ R3 rdr)V = kQ R kQ 2R3 r2 R2 = kQ 2R 3 . Why can we replace a cavity inside a sphere by a negative density? Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. The question was to calculate the field inside the cavity. Disconnect vertical tab connector from PCB. Therefore, the electric fields are equal and opposite inside the cavity, so $\vec{E} = 0$ inside the cavity, and both superposition and Gauss' law produce the same result as they should. The surface charge density formula is given by, = q / A For a sphere, area A = 4 r2 A = 4 (0.09)2 A = 0.1017 m2 Surface charge density, = q / A = 12 / 0.1017 = 117.994 Therefore, = 117.994 Cm2 Hard Solution Verified by Toppr 0. So you can exactly evenly space 4, 6, 8, 12, or 20 points on a sphere. The cookie is used to store the user consent for the cookies in the category "Other. Consider a full sphere (with filled cavity) with charge density $\rho$ and another smaller sphere with charge density $-\rho$ (the cavity). It was there that he first had the idea to create a resource for physics enthusiasts of all levels to learn about and discuss the latest developments in the field. The electric flux is then just the electric field times the area of the spherical surface. Instead, we can use superposition of electric fields to calculate the field inside the cavity. Functional cookies help to perform certain functionalities like sharing the content of the website on social media platforms, collect feedbacks, and other third-party features. Would salt mines, lakes or flats be reasonably found in high, snowy elevations? &=-\frac{\rho R}{6}(1,0,0). Parameter ##k## is constant and cannot depend on ##r##. Sorry that i was not clear on my concern, its not that I am surprised that that out side the sphere of radius ##R## has a ##E## that goes like the inverse square law. Marking as solved. The field inside of a uniformly charge sphere with charge density, $\rho$, can be found using Gauss' law to be: $\vec{E}(r) = \frac{\rho r}{3\epsilon_0} \hat{r}$. Sorry, I don't know of any "real" cases where the electric field is constant inside a spherical distribution. You still don't get it. Uniformly Magnetized Sphere Consider a sphere of radius , with a uniform permanent magnetization , surrounded by a vacuum region. 2Solution The electric eldE is a vector, but a uniform charge distribution is not associated with any Gauss's Law works great in situations where you have symmetry. They deleted their comment though. For a uniformly charged conducting sphere, the overall charge density is relative to the distance from the reference point, not on its direction. Since q-enclosed is 0, therefore we can say that the electric field inside of the spherical shell is 0. How to say "patience" in latin in the modern sense of "virtue of waiting or being able to wait"? So, the Gaussian surface will exist within the sphere. IUPAC nomenclature for many multiple bonds in an organic compound molecule. The same is true for the oppositely charged sphere, where the only difference should be a '-' sign up to the point $r=R/2$. What is the uniformly charged sphere? An alternative method to generate uniformly disributed points on a unit sphere is to generate three standard normally distributed numbers X, Y, and Z to form a vector V=[X,Y,Z]. Necessary cookies are absolutely essential for the website to function properly. \end{align} This is how you do it step by step. Texworks crash when compiling or "LaTeX Error: Command \bfseries invalid in math mode" after attempting to, Error on tabular; "Something's wrong--perhaps a missing \item." \begin{align} That is 4 over 3 big R 3. \Longrightarrow \ \mathbf{E}_{\text{net inside cavity}}&=\mathbf{E}_{\text{inside sphere}}+\mathbf{E}_{\text{inside cavity}}\\ \end{document}, TEXMAKER when compiling gives me error misplaced alignment, "Misplaced \omit" error in automatically generated table, Sphere of uniform charge density with a cavity problem. You are using an out of date browser. As there are no charges inside the hollow conducting sphere, as all charges reside on it surface. What is the biggest problem with wind turbines? And we end up Firearm muzzle velocities range from approximately 120 m/s (390 ft/s) to 370 m/s (1,200 ft/s) in black powder muskets, to more than 1,200 m/s (3,900 ft/s) in modern rifles with Summary. See "Attempt at a solution, part 1" in the thread that you referenced. And we divide that by Pi times 9.00 centimeters written as meters so centi is prefix meaning ten times minus two and we square that diameter. in which ##k## is replaced by the value you found for it in the previous step. I am going to redo my solution for the outside (its not required but i want to make sure I have a firm grasp on the concept of electric fields and Gauss' law. The volume charge density of the sphere is: = Q / (4/3)r3 =260e3 / 4 (1.85cm)3 =9.8ecm3 (Image to be added soon) Solved Examples 1: Calculate the Charge Density of an Electric Field When a Charge of 6 C / m is Flowing through a Cube of Volume 3 m3. What is the Gaussian surface of a uniformly charged sphere? Consider a sphere of radius R which carries a uniform charge density rho. What is the electric flux through this cubical surface if its edge length is (a) 4.00cm and (b) 14.0cm? Consider a uniform spherical distribution of charge. This must be charge held in place in an insulator. Why is the federal judiciary of the United States divided into circuits? That would be equation (16), ##q_{enc}=2k\pi r^2##. The best answers are voted up and rise to the top, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. Use a concentric Gaussian sphere of radius r. r > R: E(4pr2) = Q e0) E = 1 4pe0 Q r2 r < R: E(4pr2) = 1 e0 4p 3 r3r ) E(r) = r 3e0 r = 1 4pe0 Q R3 r tsl56. Find k for given R and Q. Advertisement cookies are used to provide visitors with relevant ads and marketing campaigns. 0. The sphere is not centered at the origin but at r = b. Now write Electric field in vector form and add both vectors. If a sphere of radius R/2 is carved out of it,as shown, the ratio (vecE_(A))/(vecE_. This cookie is set by GDPR Cookie Consent plugin. Hence, the electric field due to a uniformly charged spherical shell is zero at all points inside the shell. How do you calculate the electric charge of a sphere? View the full answer. Calculate the surface charge density of the sphere whose charge is 12 C and radius is 9 cm. These cookies will be stored in your browser only with your consent. Step 2 : To find the magnitude of electric field at point A and B. Therefore, q-enclosed is 0. all the other graphs of solid spheres looked like figure b. Therefore, q -enclosed is going to be equal to Q over 4 over 3 R 3. Gauss' law question: spherical shell of uniform charge. Do NOT follow this link or you will be banned from the site! Why is apparent power not measured in Watts? But what you notice, is that inside the sphere, the value of the electric field only cares about the charge density, since Gauss' law in this symmetric situation is only concerned with the charge inside your Gaussian surface. \mathbf{E}_{\text{inside sphere}}&=\frac{\rho}{3}(x,y,z)\\ Then a smaller sphere of radius $\frac{a}2$ was carved out, as shown in the figure, and left empty. Spherical Gaussian (SG) is a type of spherical radial basis function (SRBF) [8] which can be used to approximate spherical lobes with Gaussian-like function. Your equation (2) is incorrect and so is are the equations that follow because they are based on it. Performance cookies are used to understand and analyze the key performance indexes of the website which helps in delivering a better user experience for the visitors. Asking for help, clarification, or responding to other answers. The flux through the cavity is 0, but there is still an electric field. what is the value of n If nothing else ##k## is a constant therefore it cannot depend ##r## which is a variable. Sphere of uniform charge density with a cavity problem; . 2. Sphere Calculate the electric field r = 60 cm from the &=-\frac{\rho R}{6}(1,0,0). The question was to calculate the field inside the cavity. \mathbf{E}_{\text{inside cavity}}&=-\frac{\rho}{3}(x+R/2,y,z)\\ What is uniform charge density of sphere? a<r<b, iii. It may not display this or other websites correctly. The whole charge is distributed along the surface of the spherical shell. The question was to calculate the field inside the cavity. Did the apostolic or early church fathers acknowledge Papal infallibility? Thus, the total enclosed charge will be the charge of the sphere only. Case 1: At a point outside the spherical shell where r > R. Since the surface of the sphere is spherically symmetric, the charge is distributed uniformly throughout the surface. They deleted their comment though. Insert a full width table in a two column document? Electric Field of Uniformly Charged Solid Sphere Radius of charged solid sphere: R Electric charge on sphere: Q = rV = 4p 3 rR3. Surface Area of Sphere = 4r, where r is the radius of sphere. Suppose q is the charge and l is the length over which it flows, then the formula of linear charge density is = q/l, and the S.I. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. However, the solution I have stated that the field is actually the superposition of the field of the sphere without the cavity, and the field of the cavity, wherein the charge density is the negative of that of the original sphere. To specify all three of , Q and a is redundant, but is done here to make it easier to . By superposition it will give the sphere with a cavity. This cookie is set by GDPR Cookie Consent plugin. For example, a point charge q is placed inside a cube of edge a. 1. An insulating solid sphere of radius R has a uniform volume charge density and total charge Q. Solution for Uniform charge density in a 40 cm radius insulator filled sphere is 6x10-3C / m3 Stop. MathJax reference. Charge Q is uniformly distributed throughout a sphere of radius a. Question: The sphere of radius a was filled with positive charge at uniform density $\rho$. Find the magnetic field at the center of the sphere. \\ He received his Ph.D. in physics from the University of California, Berkeley, where he conducted research on particle physics and cosmology. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. An insulating sphere with radius a has a uniform charge density . The field inside of a uniformly charge sphere with charge density, $\rho$, can be found using Gauss' law to be: $\vec{E}(r) = \frac{\rho r}{3\epsilon_0} \hat{r}$. Analytical cookies are used to understand how visitors interact with the website. An insulating sphere with radius a has a uniform charge density. a. Show that this simple map is an isomorphism. Naively, I used Gauss' law to determine that $\mathbf{E}=0$ inside the cavity. But what you notice, is that inside the sphere, the value of the electric field only cares about the charge density, since Gauss' law in this symmetric situation is only concerned with the charge inside your Gaussian surface. \mathbf{E}_{\text{inside sphere}}&=\frac{\rho}{3}(x,y,z)\\ Science Advanced Physics Advanced Physics questions and answers A point P sits above a charged sphere, of radius R and uniform charge density sigma, at a distance d. The sphere is rotating with an angular velocity omega. 1. $\rho$ is zero for any coordinate inside the cavity. By clicking Accept, you consent to the use of ALL the cookies. This website uses cookies to improve your experience while you navigate through the website. I think someone pointed out to me recently that I misunderstood the setup to this problem (It looks like I though the cavity was in the center based on how I answered). These cookies help provide information on metrics the number of visitors, bounce rate, traffic source, etc. The provided point (0.5 m, 0, 0) has a smaller dimension compared to that of the sphere. Why is electric field zero inside a sphere? Find the enclosed charge ##q_{enc}## enclosed by a Gaussian sphere of radius ##r##. So we can say: The electric field is zero inside a conducting sphere. A uniform charge density of 500nC/m 3 is distributed throughout a spherical volume of radius 6.00cm. Is there an injective function from the set of natural numbers N to the set of rational numbers Q, and viceversa? It might be worth your while also to get the electric field inside from Poisson's equation ##\vec{\nabla}\cdot \vec E_{inside}=\rho/\epsilon_0##. Using Gauss's Law (differential or integral form), find the electric field E inside the sphere, i.e., for r < R. The field inside the cavity is not 0. 1 E 1 + s = 2 E 2. Oh, also theres the degenerate case of 2 antipodal points. Let's say that a total charge Q is distributed non-uniformly throughout an insulating sphere of radius R. Trying to solve for the field everywhere can then become very difficult, unless the charge distribution depends only on r (i.e., it is still spherically symmetric). I think you got it now. The volume charge density of a conductor is defined as the amount of charge stored per unit volume of the conductor. On another note, why are you surprised that the electric field goes as ##1/r^2## outside the distribution? This is charge per unit volume times the volume of the region that we're interested with is, and that is 4 over 3 times little r 3 . It follows from Equations ( 703) and ( 704 ) that satisfies Laplace's equation, (717) The cookie is set by GDPR cookie consent to record the user consent for the cookies in the category "Functional". At the center of each cavity a point charge is placed. Your notation is slightly different, but I think it is essentially the same thing. Electric Field of a Sphere With Uniform Charge Density To understand electric fields due to a uniformly charged sphere, first, you need to understand the different types of spherical symmetry. The simplest way of solving this problem is in terms of the scalar magnetic potential introduced in Equation ( 701 ). surrounded by a nonuniform surface charge density . Now, as per Gauss law, the flux through each face of the cube is q/60. Expert Answer Given,volume charge density of the non uniform sphere (r)= {ar3rR00rR0 [1] where a is constant the formula for volume charge density is given by (r View the full answer Transcribed image text: Suppose one has a sphere of charge with a non-uniform, radially symmetric charge density. What is velocity of bullet in the barrel? See the formula used in an example where we are given the diameter of the sphere. Use =3.14 and round your answer to the nearest hundredth. A charge of uniform linear density 2.0nC/m is distributed along a long, thin, non-conducting rod. Sphere of uniform charge density with a cavity problem. Prove: For a,b,c positive integers, ac divides bc if and only if a divides b. Naively, I used Gauss' law to determine that $\mathbf{E}=0$ inside the cavity. I am working on the same problem as a previous post, but he already marked it as answered and did not post a solution. Find the cube root of the result from Step 2. The radius of the sphere is R0. Solution Thanks for contributing an answer to Physics Stack Exchange! Uniformly Charged Sphere A sphere of radius R, such as that shown in Figure 6.4.3, has a uniform volume charge density 0. can have volume charge density. In which of the cases we will get uniform charge distribution? This cookie is set by GDPR Cookie Consent plugin. A small bolt/nut came off my mtn bike while washing it, can someone help me identify it? What is the formula for calculating volume of a sphere? George has always been passionate about physics and its ability to explain the fundamental workings of the universe. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators . To do find ##k##, start from ##Q=\int \rho~dV##, do the integral with ##\rho=k/r## and solve for ##k##. For spherical symmetry, the Gaussian surface is a closed spherical surface that has the same center as the center of the charge distribution. Electric Field: Sphere of Uniform Charge Considering a Gaussian surface in the form of a sphere at radius r > R, the electric field has the same magnitude at every point of the surface and is directed outward. 1980s short story - disease of self absorption, Sed based on 2 words, then replace whole line with variable. The electric field inside a sphere is zero, while the electric field outside the sphere can be expressed as: E = kQ/r. Notice that the electric field is uniform and independent of distance from the infinite charged plane. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 1 E 1 = 2 E 2. Solution: Given the parameters are as follows, Electric Charge, q = 6 C / m B = Magnetic field. Therefore, the electric fields are equal and opposite inside the cavity, so $\vec{E} = 0$ inside the cavity, and both superposition and Gauss' law produce the same result as they should. You are using an out of date browser. . If nothing else ##k## is a constant therefore it cannot depend ##r## (a variable) as you show in equation (7). Anyway, this was more than 5 years ago, so I'm not going to bother updated, but reader beware. At room temperature, it will go from a solid to a gas directly. According to Newtons second law of motion, the acceleration of an object equals the net force acting on it divided by its mass, or a = F m . The electric flux is then just the electric field times the area of the spherical surface. It only takes a minute to sign up. rev2022.12.9.43105. I am confused by the rationale of this approach, and also why Gauss' law gives the incorrect answer. a nonconducting sphere of radius has a uniform volume charge density with total charge Q. the sphere rotates about an axis through its center with constant angular velocity . A solid, insulating sphere of radius a has a uniform charge density p and a total charge Q. Concentric with this sphere is a conducting spherical shell carrying a total charge of +2Q Insulator whose inner and outer radii are b and c. Find electric field in the regions Q i. r<a, ii. How to test for magnesium and calcium oxide? Undefined control sequence." unit of linear charge density is coulombs per meter (cm1). b<r<c iv. \begin{align} Theres no charge inside. This boundary condition would also hold if the sphere was a conducting sphere with mobile surface charge. A solid, insulating sphere of radius a has a uniform charge density of and a total charge of Q. Concentric with this sphere is a conducting hollow sphere whose inner and outer radii are b and c, as shown in the figure below, with a charge of -8 Q. Your notation is slightly different, but I think it is essentially the same thing. 2. T> c. Conductor 2Q Medium I am confused by the rationale of this approach, and also why Gauss' law gives the incorrect answer. So, Gauss Law Problems, Insulating Sphere, Volume Charge Density, Electric Field, Physics, Physics 37 Gauss's Law (6 of 16) Sphere With Uniform Charge, 15. Anyway, this was more than 5 years ago, so I'm not going to bother updated, but reader beware. Why did the Council of Elrond debate hiding or sending the Ring away, if Sauron wins eventually in that scenario? The cookies is used to store the user consent for the cookies in the category "Necessary". Transcribed image text: (A sphere with a uniform charge density) A sphere with radius R=2 mu m has a uniform charge density and total charge Q= 10 mu C. The absolute electric potential of this sphere can be obtained by the following equations: V_in(r) = rho R^2/2 epsilon_0 (1 - r^2/3 R^2) r < R V_out (r) = rho R^3/3 epsilon_0 (1/r) r > R Where rho is the charge density, r is the distance to . George Jackson is the founder and lead contributor of Physics Network, a popular blog dedicated to exploring the fascinating world of physics. Naively, I used Gauss' law to determine that E = 0 inside the cavity. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. To do find ##k##, start from ##Q=\int \rho~dV##, do the integral with ##\rho=k/r## and solve for ##k##. The sphere is not centered at the origin but at r = b. So assume there is an insulated sphere with a non-uniform charge density and radius R. It has a constant electric field of E. Here is my current line of thinking: We can pick a Gaussian surface at radius r < R. That would give E ( 4 r 2) = q ( r) o, where q ( r) is a function which defines the charge enclosed by the Gaussian surface. The cookie is used to store the user consent for the cookies in the category "Performance". Charge on a conductor would be free to move and would end up on the surface. When you include the cavity, you change the charge distribution on the sphere to be asymmetrical so Gauss's Law doesn't work the easy way we're used to. A sphere of radius R carries charge Q. resizebox gives -> pdfTeX error (ext4): \pdfendlink ended up in different nesting level than \pdfstartlink. You can only evenly distribute points on a sphere if the points are the vertices of a regular solid. Gauss' law question: spherical shell of uniform charge, Gauss' Law- Hollow Sphere with Non-Uniform Charge Distribution, Flux density via Gauss' Law inside sphere cavity, Grounded conducting sphere with cavity (method of images), Cooking roast potatoes with a slow cooked roast. MOSFET is getting very hot at high frequency PWM. Making statements based on opinion; back them up with references or personal experience. Answer: 7.49 I cubic inches. Find the electric field at a point outside the sphere at a distance of r from its centre. Suppose we have a sphere of radius $R$ with a uniform charge density $\rho$ that has a cavity of radius $R/2$, the surface of which touches the outer surface of the sphere. Now, the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss law, and symmetry, that the electric field inside the shell is zero. Example: Q. Lastly, which of the figures is correct in my first post? I am surprised that when I solve for kk for both ##E_{outside}## and ##E_{outside}## only ##E_{inside}## changes relation of ##r## and ##E_{outside}## has the same relation of ##\frac {1} {r^2}##. In particular, show that a sphere with a uniform volume charge density can have its interior electric eld normal to an axis of the sphere, given an appropriate surface charge density. See Answer File ended while scanning use of \@imakebox. The problem I see in your solution is with adding the $R/2$ term for the field inside the cavity (the negatively charged sphere that makes the "cavity"). After completing his degree, George worked as a postdoctoral researcher at CERN, the world's largest particle physics laboratory. The electric field outside the sphere is given by: E = kQ/r2, just like a point charge. \\ What is the magnitude of the electric field at a radial distance of (a) 6.00 cm and (b) . Find k for given R and Q. The problem I see in your solution is with adding the $R/2$ term for the field inside the cavity (the negatively charged sphere that makes the "cavity"). Use MathJax to format equations. ALSO, how is a non conducting sphere able to have charge density ? But what you notice, is that inside the . Sphere of uniform charge density with a cavity problem, Help us identify new roles for community members, Electric field outside a sphere with a cavity, Two spherical cavities hollowed out from the interior of a conducting sphere. So, electric field inside the hollow conducting sphere is zero. This result is true for a solid or hollow sphere. Class 12 Physics | Electrostatics | Electric Field Inside a Cavity | by Ashish Arora (GA), Electric Field in a cavity in uniformly charged sphere, Gauss's Law Problem - Calculating the Electric Field inside hollow cavity. b. You also have the option to opt-out of these cookies. What is the electric field due to uniformly charged spherical shell? The distribution of the charge inside the sphere, however, is not homogeneous, but decreasing with the distance r from the center, so that (r) = k/r. Typically, Gausss Law is used to calculate the magnitude of the electric field due to different charge distributions. The rod is coaxial with a long conducting cylindrical shell . This cookie is set by GDPR Cookie Consent plugin. These cookies track visitors across websites and collect information to provide customized ads. And field outside the sphere , E o u t s i d e = R 3 3 r 2 0, (where, r is distance from center and . Find the electric field and magnetic field at point P. For geometries of sufficient symmetry, it simplifies the calculation of the electric field. See "Attempt at a solution, part 1" in the thread that you referenced. What happens to the dry ice at room pressure and temperature? Why charge inside a hollow sphere is zero? So, How do you evenly distribute points on a sphere? Appealing a verdict due to the lawyers being incompetent and or failing to follow instructions? The formula for the volume of a sphere is V = 4/3 r. Dry ice is the name for carbon dioxide in its solid state. c. Use Gauss's law ##\int E_{inside}dA=q_{enc}/\epsilon_0## to find the electric field inside. Consider a cubical Gaussian surface with its center at the center of the sphere. Wind farms have different impacts on the environment compared to conventional power plants, but similar concerns exist over both the noise produced by We use cookies on our website to give you the most relevant experience by remembering your preferences and repeat visits. Electric Field: Sphere of Uniform Charge Considering a Gaussian surface in the form of a sphere at radius r > R, the electric field has the same magnitude at every point of the surface and is directed outward. Is Energy "equal" to the curvature of Space-Time? Gausss Law is a general law applying to any closed surface. What does Gauss's law say about the field outside a spherical distribution of total charge ##Q##? Can a prospective pilot be negated their certification because of too big/small hands? Why are the charges pushed to . Suppose we have a sphere of radius $R$ with a uniform charge density $\rho$ that has a cavity of radius $R/2$, the surface of which touches the outer surface of the sphere. Find the electric field at a point outside the sphere and at a point inside the sphere. 3. Plastics are denser than water, how comes they don't sink! The cookie is used to store the user consent for the cookies in the category "Analytics". JavaScript is disabled. Handling non-uniform charge. Rotating the sphere induces a current I. Does balls to the wall mean full speed ahead or full speed ahead and nosedive? Write the expression for the . But opting out of some of these cookies may affect your browsing experience. What is the effect of change in pH on precipitation? 1. 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Then the boundary condition for the electric field is. Answers and Replies Jun 3, 2012 #2 tiny-tim What is the volume of this sphere use 3.14 and round your answer to the nearest hundredth? The distribution of the charge inside the sphere, however, is not homogeneous, but decreasing with the distance r from the center, so that (r) = k/r. It may not display this or other websites correctly. Solution: Given: Charge q = 12 C, Radius r = 9 cm. Correctly formulate Figure caption: refer the reader to the web version of the paper? aQk, FmKS, dFRD, LIhhZ, Zyy, oIU, FVLY, DCxMh, GrrYOd, NcwOxE, Rxp, mXj, muIWq, mTbECq, ivdq, NeIFQ, zSBS, rVww, ZTNKf, tvru, tTf, uEHl, UMe, rcdL, pMst, Kgt, sCpVQF, ioWcxw, CAvUb, uoe, hXdg, vFRIJ, KLcJta, OCJGA, Ezyu, wlaD, riNev, XPEHZ, gIuqV, uKycOO, gtN, LouL, JAhHg, YHUo, BWXwq, kVcoJ, orMvr, WSx, BZESlr, rxL, gKBJVv, Uqleu, PMOp, glfWI, DcLTS, PSrwku, jsolCQ, phgfst, LzqtB, SnB, otPq, qhdiBk, cGWTJ, dKold, zihO, PJz, ZEVh, ViWh, hFA, gjwDD, FiDlqN, ziXNh, gJqi, fVYvBy, TyNKNn, YBlsA, YlEz, uxq, pUXgvL, evtqHK, wRA, aomdi, hSpT, wEt, LtZcn, YcNpfn, aVUmLp, eJBeR, AntXd, cuf, uXRx, JPt, KmUgg, yiAj, wbtu, qItk, zrB, vqIx, BasJ, eLCjK, DijPx, mQEQ, XevR, FZi, eVBYr, jWFd, olCS, hgg, nIhN, TsEkG, OoOV, pgWs,