charge on sphere formula

Please proceed to read the whole document carefully to understand the topic completely. All rights reserved. 8-1. We will assume that the charge q of the solid sphere is positive. Ans. As the spheres are identical in material, size and shape the charge will redistribute across the two spheres so that it is shared evenly. V is the volume in m 3. 2. Cooking roast potatoes with a slow cooked roast. As an example, the below code creates a series of points (a dipole field line), shifts it away from the centre of a sphere and rotates to some angle in the xy . Everything you need to know about the sphere formula is provided below. There should be some external electric field near by to have potential energy. Therefore, the volume and total surface area of a sphere of radius 7 cm are 1437.33 cc and 616 cm2, respectively. You can photocopy, print and distribute them as often as you like. The formula of linear charge density is =q/l, such that q is the charge and l is the length of the body over which the charge is distributed. Find the total surface area and the volume of a sphere whose radius is 7cm. The inner sphere carries a charge Q1, and the spherical shell carries a charge Q2, such that Q2 = - 3 Q1 1. The first of these integrals gives you charge $Q_0$, whereas the second is zero. The next step is basically to distinguish between inside/on the sphere and outside. The greek symbol pho () typically denotes electric charge, and the subscript V indicates it is the volume charge density. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force. The same content, but different versions (branded or not) have different licenses, as explained: You are allowed and encouraged to freely copy these versions. Between R 1 and R 2?At r > R 2? Ans. The sphere enters a uniform 5.0 -tesla magnetic field that is directed into the page $(-z)$ as shown above. The formula for finding the lateral surface area or the CSA of a hemisphere is: Curved Surface Area of a Hemisphere = 2r2 Where is the constant taken as 3.142 or 22/7, and r is the radius of the hemisphere. Why does the USA not have a constitutional court? That makes sense so if I was asked to find the E field using gauss's law for r<=R would I just use E=q/(area*epsilon) but how would I know what q enclosed is? (Don't forget direction too!) The radius of the square is the line that links the centre to the boundary. (C) The charge will circle clockwise with a 4.0 m radius. Why is the federal judiciary of the United States divided into circuits? solid non-conducting sphere of radius given by p = (SQor Y (ta" has variable density of charge Where r is the radial variable distance and Qo and # are constant: Caleulate the magnitude of the electric field everywhere in space. For more information, visit Creative Commons Attribution 3.0 Unported. Download our apps to start learning, Call us and we will answer all your questions about learning on Unacademy. We shall bear in mind that the surface of the sphere is an equipotential surface, and we shall take the potential on the surface to be zero. Transcribed Image Text: A solid sphere with radius R and charge density p(r) = kr is centered at the origin and spinning with an angular velocity, w, about the z- - axis (@= @ 2). comments sorted by Best Top New Controversial Q&A Add a Comment . For example, a sphere with diameter 1 m has 52.4% the volume of a cube with edge length 1 m, or about 0.524 m 3. The formula of Electric Charge is as follows Q = I t Where, Q = Electric Charge, I = Electric Current, t = Time. Otherwise it has no other potential energy. The use of Gauss' law to examine the electric field of a charged sphere shows that the electric field environment outside the sphere is identical to that of a point charge.Therefore the potential is the same as that of a point charge:. When a solid sphere of radius $a$ and density $\sigma$ falls vertically in a viscous liquid of density $\rho(<\sigma)$ and coefficient of viscosity $\mu$, . 3. 2) An electric flux of 2 V-m goes through a sphere in vacuum space. Here, Ris the radius of the shell, is the surface charge density and !is the angular velocity, where the sphere's axis is taken to be the zaxis. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Answer: From the formula of the Gauss law, = E 4 r 2 = Q/. View formula exam 2.pdf from MATH 215 at University of Southern California. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. In this page, we are going to see how to calculate the magnitude of the electric field due to a uniformly charged solid sphere using Gauss's law. In the quick-paced environment of today, smartphone security is essential. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Is it appropriate to ignore emails from a student asking obvious questions? What happens when you connect the two spheres with a wire? A = 616 cm 2. r represents the value of the radius of the given sphere. The topic will be better understood if you use examples that are related to it. When $r > r_0$ you have to split integral into two parts: from $0$ to $r_0$ and from $r_0$ to $r$, since the integrand is represented in these two regions by different functions. The electric field inside a conducting sphere is zero, so the potential remains constant at the value it reaches at the surface: Related Articles Sample Questions Question 1: Find the volume charge density for the given charge and the volume are 8C and 12m3 respectively. From this we can calculate the magnetic eld B = A using the stan-dard formula for the curl in spherical coordinates: B= 8 <: 2 0R! 3 cos . What is the resulting motion? This boundary condition would also hold if the sphere was a conducting sphere with mobile surface charge. To learn more, see our tips on writing great answers. Deriving the volume of a sphere is the same as finding the total space available within the surface of the sphere. Find its radius. If you are given a problem to calculate volume, you will probably be given the radius. Thus, the total charge on the sphere is: q t o t a l = .4r The above equation can also be written as: E = 1 4 r q t o t a l r For the net positive charge, the direction of the electric field is from O to P, while for the negative charge, the direction of the electric field is from P to O. The potential at P from a charge +$Q$ at $Q$ and a charge $(a/R)Q$ at $I$ is (see Figure $II$.4), \[\nonumber V=\dfrac{q}{4\pi\epsilon_0}\left (\dfrac{1}{(r^2+R^2-2rR\cos \theta)^{1/2}}-\dfrac{a/R}{(r^2+a^/R^2 -2a^2r\cos \theta /R)^{1/2}}\right )\], The E field on the surface of the sphere is $V / r$ evaluated at $r = a$. Audi's show car is part of the "Podium Pass Series 4 VIP tier" will be available to gamers from December 7. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. 1. 2022 Physics Forums, All Rights Reserved, Problem with two pulleys and three masses, Newton's Laws of motion -- Bicyclist pedaling up a slope, A cylinder with cross-section area A floats with its long axis vertical, Hydrostatic pressure at a point inside a water tank that is accelerating, Forces on a rope when catching a free falling weight. To find the volume of the circle we need to utilise the formula : Ans. Name of a play about the morality of prostitution (kind of), Central limit theorem replacing radical n with n. Are there conservative socialists in the US? Find the charge on spheres A and B. The radius of a sphere is the measure from the center of the sphere to the outer edge. The volume of any circle is 2/third of the volume of any chamber with identical range and level equivalent to the distance across. Let's take a look at the concept! As a result, the formula for calculating a spheres surface area is 4 times, 12.56 times, or 88/7 times the radius square of the sphere. The formula for deriving the volume of a sphere is. If the spheres radius is r, the diameter may be calculated using the formula: A spheres circumference is equal to two times its radius. What is the electric field at r < R 1? What's the \synctex primitive? Related Topics . Step 1 - Enter the Charge Step 2 - Permittivity of Free Space (Eo) Step 3 - Enter the Radius of Charged Solid Sphere (a) Step 4 - Enter the Radius of Gaussian Sphere Step 5 - Calculate Electric field of Sphere Electric Field of Spehere Formula: E ( r ) = ( q / ( 4 * * o * a 3 ) ) * r Where, E (r) = Electric field of sphere q = Charge A spheres diameter is defined as a straight line travelling through its centre and touching two points on either side of its surface. r, rsR system difference price . Use the electroscope to determine if a charge has been transferred to the sphere. If $r r_0\end{cases}\tag{4}$. You are using an out of date browser. United States. Audi will compete in the top class of motorsport from 2026 with the power unit . Find its radius. V = q 4 0 ( 1 ( r 2 + R 2 2 r R cos ) 1 / 2 a / R ( r 2 + a / R 2 2 a 2 r cos / R) 1 / 2) The E field on the surface of the sphere is V / r evaluated at r = a. Since charge is measured in Coulombs [C], and volume is in meters^3 [m^3], the units of the electric charge density of Equation [1] are [C/m^3]. Did neanderthals need vitamin C from the diet? The formula to find the volume charge density is given by- Volume Charge density () = Charge (C)/Volume (V) = C/V where, is charge density. Be sure not to touch the edge of the hole in the sphere. 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 . Except where otherwise noted, this site is covered by a closed copyright license. MathJax reference. 2. I want to be able to quit Finder but can't edit Finder's Info.plist after disabling SIP, TypeError: unsupported operand type(s) for *: 'IntVar' and 'float', Irreducible representations of a product of two groups. Answer: The resulting current of two currents meeting at a junction is an algebraic sum, not a vector sum. By the uniqueness of . The electric field of a conducting sphere with charge Q can be obtained by a straightforward application of Gauss' law.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.The electric flux is then just the electric field times the area of the spherical surface. Better than just free, these books are also openly-licensed! We obviously have a discontinuity in the integrand at $r=r_0$ because of $\rho(r)$. The formula for the deriving surface area of a sphere is: The volume of a sphere is found using the formula: Therefore, the volume and total surface area of a sphere of radius 7 cm are 1437.33 cc and 616 cm2, respectively. b) Taking the electric potential to be zero at infinity, determine the electric potential both inside and . Fig. Mobile app security, however, is a cost associated with this power. What I mean is: I want to start at (1) and get to (4) without once arguing in a physical sense. So, the formula for the deriving volume of a sphere can be stated as 4/3 times or 4.19 times, or 88/21 times the cube of the radius of the sphere whose volume is to be determined. It only takes a minute to sign up. Thus, the total charge on the surface of the sphere is, Q = A. Is this correct? The volume of any circle is 2/third of the volume of any chamber with identical range and level equivalent to t Access free live classes and tests on the app. The mathematical formula for deriving the volume of a sphere is given as: is a constant and its value is equal to 3.14 or 22/7. For more information, visit Creative Commons Attribution-NoDerivs 3.0 Unported. Is it cheating if the proctor gives a student the answer key by mistake and the student doesn't report it? Let us take the line OQ as the $z$-axis of a coordinate system. Therefore, the radius of the sphere is 5.58 cm. Note that since electric charge can be negative or positive, the charge density . all that makes, physically, perfect sense but I don't see how we mathematically get from (3) to (4). The surface area of the sphere is A=4r 2 =4 x (0.03) 2 =0.01 m 2 Hence, the surface charge density of a sphere is = Q/A = 4C/0.01m 2 =400 C/m 2 Therefore the electric field of a charged sphere is =45.2 x 10 12 V/m is a constant with the value of 3.14 or 22/7 as its value. Download our open textbooks in different formats to use them in the way that suits you. The charge \ (Q\) that was originally on the larger sphere distributes itself onto the two spheres. Get answers to the most common queries related to the Sphere Formula. Because the charges on the large sphere can move around freely, some of them will move to the smaller sphere. The formula for calculating a spheres surface area is stated mathematically as: The total surface area of a sphere is known as TSA. This implies that. The complete surface region of some random circle is equivalent to; Ans. The complete surface region of some random circle is equivalent to; Where r is the radiusof the given circle. The D field is 0 times this, and the surface charge density is equal to D. After some patience and algebra, we obtain, for a point X on the surface of the sphere Is it correct to say "The glue on the back of the sticker is dying down so I can not stick the sticker to the wall"? Sovereign Gold Bond Scheme Everything you need to know! On the surface of a sphere, you may discover a point that is equidistant from any other point. If we have a charge q at vector position outside of a grounded sphere of radius R, the potential outside of the sphere is given by the sum of the potentials of the charge and its image charge inside the sphere. 4. Let $X$ be some point such that OX = $r$ and the angle XOQ= . PMVVY Pradhan Mantri Vaya Vandana Yojana, EPFO Employees Provident Fund Organisation. This physics video tutorial explains how to solve typical gauss law problems such as the insulating sphere which contains electric charge throughout the vol. The uniform charge per unit volume in the insulating sphere is its total charge (Q) divided by its total volume. (B) The charge will circle counterclockwise with a.0 m radius. Transcribed Image Text: A sphere with radius R centered at the origin is made out of a linear dielectric material, with electric susceptibility Xe = 2, and has a uniform (free) surface charge density, of, on its surface. Explore more about the sphere formula with solved examples. 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Making statements based on opinion; back them up with references or personal experience. The Formula for Gauss Law: As per the Gauss theorem, the total charge enclosed in any closed surface is 2proportional to the total flux enclosed by the surface. Would salt mines, lakes or flats be reasonably found in high, snowy elevations? Therefore, the diameter and circumference of the sphere are 20cm and 62.8571 cm, respectively. E = Q/ (4 r 2 ) which is the electric field due to a particle with charge Q. When would I give a checkpoint to my D&D party that they can return to if they die? The volume charge density of a sphere Charged sphere If a spherical conductor of radius R contains Q amount of charge inside its volume, then the formula for the volume charge density of the sphere is, \color {Blue}\rho=\frac {Q} {\frac {4} {3}\pi R^ {3}} = 34R3Q (2) The volume charge density of a cylinder --- not after the fact. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The volume of a sphere is found using the formula: V = (4/3) r 3. The electrostatic energy of a system of particles is the sum of the electrostatic energy of each pair. The surface charge density formula is a topic that is both significant and fascinating. It may not display this or other websites correctly. Terms and Conditions and Privacy Policy. Then the formula is: Where, . C represents charge. = Q/ [ (4/3)R 3 ] The charge enclosed by the gaussian surface is then: q enc = Qr 3 /R 3 The flux through the gaussian surface is EA = 4r 2 E Applying Gauss' Law: Net flux = E = 4r 2 E = Qr 3 /R 3 o Factors of r 2 cancel. Electric Field Due to Spherical Shell That is, the surface of our sphere. Therefore, If \phi is the total flux and \epsilon_ {0} is the electric constant and the Q is the total electric charge enclosed by the surface is. These unbranded versions of the same content are available for you to share, adapt, transform, modify or build upon in any way, with the only requirement being to give appropriate credit to Siyavula. Unacademy is Indias largest online learning platform. Where r is the fixed distance from the center point or we can call it the radius of Sphere. to personalise content to better meet the needs of our users. Volume in general terms can be defined as the total capacity of a 3D object. q = charge on the sphere 0 = 8.854 10 12 F m 1 R = Radius of the sphere. The calculation is done using spherical coordinates. $Q(r)=\int_{r'\leq r}\rho(r' ) dV' = \int_0^r dr' r'^2 \rho(r') \int_0^{2\pi}d\varphi \int_0^\pi \sin(\vartheta) d\vartheta=4\pi\int_0^r dr' r'^2 \rho(r') \tag{3}$, so far, so good. The result has to be the same as obtained calculating the field due to a solid sphere of charge using Coulomb's law. (All India, 2011, 3 Marks) The Competence Center Motorsport at the Audi Neuburg facility is being expanded for the Formula 1 project. Where does the idea of selling dragon parts come from? The $D$ field is $\epsilon_0$ times this, and the surface charge density is equal to $D$. Where is it documented? In a new building measuring around 3,000 square meters, new test benches for the development of the power unit will be installed in particular. Because there are no lateral surfaces on a sphere, its total surface area is the same as its curved surface area. Asking for help, clarification, or responding to other answers. For a better experience, please enable JavaScript in your browser before proceeding. Why is Singapore currently considered to be a dictatorial regime and a multi-party democracy by different publications? The length OI is $a^2 /R$. 1 E 1 + s = 2 E 2. Electric Field On The Surface Of The Sphere (R = r) On the surface of the conductor , where R = r , the electric field is : E = (1/4) * (q/r) Electric Field Inside Hollow Sphere If we. A uniformly charged insulating dielectric does, in general, have no surface . The electric field inside the sphere is E=0. To get there faster, you might be tempted to disregard the app's status, but doing so could be fatally mistaken. V = 1437.33 cc. I now want to properly calculate $Q(r)=\int_{r'\leq r}\rho(r' ) dV'$. Express your answer in terms of the total charge, Q, of the sphere. Thus, for purposes of calculating the potential, we can replace the metal sphere by an image of $Q$ at $I$, this image carrying a charge of $(a/R)Q$. It relates the magnitude, direction, length, and closeness of the . A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B, spheres A and B are brought in contact and then separated. Now suppose that, instead of the metal sphere, we had (in addition to the charge + $Q$ at a distance $R$ from O), a second point charge $(a/R) Q\text{ at }I$. Use MathJax to format equations. Electric Field of Uniformly Charged Solid Sphere Radius of charged solid sphere: R Electric charge on sphere: Q = rV = 4p 3 rR3. Rotate points around normal to sphere surface. Each sphere will have half of the total charge: Q = Q 1 + Q 2 2 = 9,6 10 18 + ( 4,8 10 18) 2 = 7,2 10 18 C So each sphere now has 7,2 10 18 C of charge. (2) The charge density on the surface of a conducting spherical shell is also the same as that of a conducting sphere of the same radius and the same charge. 3. A sphere is a circular solid in three dimensions. The best answers are voted up and rise to the top, Not the answer you're looking for? We think you are located in The charge in the sphere, Q=260e The radius of the sphere, r=1.85cm If Q is the total charge distributed over a volume V, then the volume charge density is given by the equation: = Q/V The volume of a sphere: V= 4/3r3 The volume charge density of the sphere is: = Q / (4/3)r3 =260e3 / 4 (1.85cm)3 =9.8ecm3 (Image to be added soon) Mobile apps are a fantastic way to keep customers interested and boost revenue. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. 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. V = (4/3) x (22/7) x (7 ) 3. and the observation point. The charge density of an electric object must also be determined using the surface area and volume of the object. Solution: Given: More answers below An object is up in the sky and so it has stored potential energy due to earth's gravitational field. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. How is the charge distributed on the spherical shell? Connect and share knowledge within a single location that is structured and easy to search. Better than just free, these books are also openly-licensed! a) Using the results of example 5.11, find the magnetic vector potential, A, inside the spinning charged sphere. Legal. (A) The charge will circle clockwise with a 2.0 m radius. The sphere diameter formula, sphere surface area formula, and sphere volume formula are the three major formulae for a sphere. b) A very thin plastic ring (radius R) has a constant linear charge density, and total charge Q. The value of r represents the radius of the provided sphere. Find out more here about the sponsorships and partnerships with others that made the production of each of the open textbooks possible. You can burn them to CD, email them around or upload them to your website. The formula for surface charge density of a conducting sphere is \small {\color {Blue} \sigma =\frac {Q} {4\pi r^ {2}}} = 4r2Q . So how do we mathematically, as pedantically as possible, get from (3) to (4)? a) Determine the electric field both inside and outside the sphere. Electric Charges and Fields - Get complete study material including notes, formulas, equations, definition, books, tips and tricks, practice questions, preparation plan prepared by subject matter experts on careers360.com. What is the formula of linear charge density? You can download them onto your mobile phone, iPad, PC or flash drive. Solution: Given, Charge (C)=8 By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. Sample Questions Question 1: An electric charge is a scalar quantity for what reason? F1 22 is the latest version of the official video game of the FIA Formula 1 World Championship, available for PlayStation, Xbox and PC - also in VR on PC. 1 answer. This page titled 2.5: A Point Charge and a Conducting Sphere is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. JavaScript is disabled. All of the current teams, drivers and tracks are included in the game. The diameter of the sphere is the longest straight line that runs through the centre of the sphere. Explanation: Some definitions: Q = Total charge on our sphere R = Radius of our sphere A = Surface area of our sphere = E = Electric Field due to a point charge = = permittivity of free space (constant) Electrons can move freely in a conductor and will move to the outside of the sphere to maximize the distance between each electron. The quantity of charge per unit area, measured in coulombs per square meter (Cm . Then $R/ = a/$, or, \[\label{2.5.1}\dfrac{1}{}-\dfrac{a/R}{}=0\]. The surface area of a sphere of radius a is 4 a 2. . The construction work on the extension began this week. Ans. As a result, a spheres circumference may be calculated as 6.28 times or 44/7 times its radius. sin2 (x) + cos2 (x) = 1, cos(2x) = cos2 (x) sin2 (x), . $$\int_0^rdr'r'^2 = \int_0^{r_0}dr'r'^2 + \int_{r_0}^rdr'r'^2.$$ Consider a sphere with radius $r_0$which is homogeneously charged with charge density, $\rho(r)=\begin{cases} \rho_0 & r\leq r_0 \\ 0 & r > r_0\end{cases} \tag{1}$, if $Q_0$ is the total charge of the sphere, we get, $\rho_0 = \frac{Q_0}{V}=\frac{4Q_0}{3 \pi r_0^3} \tag{2}$. The locus of points where the potential is zero is where, \[\label{2.5.2}\dfrac{Q}{4\pi\epsilon_0}\left ( \dfrac{1}{}-\dfrac{a/R}{}\right ) = 0\]. A rectangular glass tank is 25 cm long, 20 cm wide and 30 cm high and contains water. Error in the volume will be thrice the error in radius,as per formula-V=4/3r^3.So, error in volume=30.1=0.3% We are going to try to calculate the surface charge density induced on the surface of the sphere, as a function of position on the surface. A sphere is a fully symmetrical three-dimensional circular shaped object. The volume of a sphere is found to be 729 cc. viscosity; sphere; force; charge; 0 like 0 dislike. Coulomb's inverse-square law, or simply Coulomb's law, is an experimental law of physics that quantifies the amount of force between two stationary, electrically charged particles. After some patience and algebra, we obtain, for a point $X$ on the surface of the sphere, \[\label{2.5.3}\sigma = -\dfrac{Q}{4\pi}\dfrac{R^2-a^2}{a}\cdot \dfrac{1}{(XQ)^3}\]. In case of no surface charge, the boundary condition reduces to the continuity of the dielectric displacement. 3. Charge the sphere by touching it to the the ball . Calculate the diameter and the circumference of a sphere whose radius is 10 cm. E = 0 R 2 r 2 Notice that the total charge on the sphere is q t o t = .4 R 2 Thus we can also write the above equation as: E = 1 4 0 q t o t r 2 Notice that this is similar to the electric field due to a point charge. 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. We still have a proper integral though, so we just split it: $4\pi\int_0^r dr' r'^2 \rho(r') = 4\pi\int_0^{r_0} dr' r'^2 \underbrace{\rho(r')}_{\rho_0} + 4\pi\int_{r_0}^r dr' r'^2 \underbrace{\rho(r')}_{0}$. Answer: If the sphere is conducting, its potential will be the same as the potential of the vacuum would be at the same location if all the charge inside were located at the centre. Although the law was known earlier, it was first published in 1785 by French physicist Andrew Crane . How is the charge distributed on the sphere? One is the application of the concept of energy to electrostatic problems; the other is the evaluation of the energy in different ways. Thus, for $r>r_0$: so inside the sphere we have the constant charge density $\rho_0$ and outside we don't have any charge density. I would like to rotate an array about the axis that is normal to a given location on the surface of a sphere, but unsure how to go about it. Q = Total charge on our sphere; r = Radius of the sphere; A = Surface area of a sphere (4r 2) E = Electric Field due to a point charge Q/ 4r 2; . Let us first construct a point I such that the triangles OPI and PQO are similar, with the lengths shown in Figure $II$.3. This relation between the variables $ \text{ and }$ is in effect the equation to the sphere expressed in these variables. I do get the whole gist of it and my intuition about the problem is clear - I just noticed I'm not satisfied with how I mathematically argue. Formula sheet: Integration by parts: udv = uv vdu. To see this, first notice that the conducting sphere is necessarily an equipotential surface. 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