Chapter Gauss Law.1 Electric Flu. Gauss Law. A charged Isolated Conductor.4 Applying Gauss Law: Cylindrical Symmetry.5 Applying Gauss Law: Planar Symmetry.6 Applying Gauss Law: Spherical Symmetry You will learn: 1. Recognize and use symmetry to determine the shape of electric fields.. Calculate the electric flu through a surface. Use Gauss law to calculate the electric field of symmetric charge distributions 4....1 Electric Flu Electric flu through a surface: the amount of electric field passing through a surface (i.e., Gaussian surface) Flu: flow E E Dot product : Φ= E A= EAcosθ θ = Φ = EA Electric flu is a dot product of E and A, where both E and A are vectors. How to define the direction of A?? When we use Gauss law related to flu and charge, we need a closed surface 1
.1 Electric Flu For a closed surface make up of small elements ΔA, the total flu is: Φ = E Δ A For a closed (Gaussian) surface, in the limit ΔA, Φ= E Δ A Φ = E da E : electric field A: area of the surface Φ: electric flu (Unit: N m /C) Vector A: It is perpendicular to the Gaussian surface and directed away from the interior of the surface. Skim: zero flu Left cap: A direction? Right cap: A direction? Cylindrical surface: A?.1 Electric Flu Gaussian surface: A closed surface enclosing the charge distribution. It can be any shape, but the shape that minimizes our calculations of the electric field is the one that mimics the symmetry of the charge distribution. Vector A: It is perpendicular to the Gaussian surface and directed away from the interior of the surface. Symmetry: For eample: spherical, cylindrical 1. The use of symmetry may provide us a simple way to solve comple problems.. Symmetry determines the geometry of electric field. Gauss s law (in net section): it allows us to find E (for a charge distribution) that would be very difficult to find using Coulomb law.
Sample problems : Fig. -4 shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform electric field E, with the cylinder ais parallel to the field. What is the flu Φ of the electric field through this closed surface? 1. How many surfaces in this closed surface? surfaces left cap a; cylindrical surface b; right cap c 1. Draw A directions for surfaces. (note: directed away from the interior of the surface) A a = A b ; Αngle θ (Direction of A relative to E): θ a = 18, θ b = 9, θ c =,. Calculate Φ 1, Φ, Φ and sum the three terms as the flu of the E through this closed surface. Here, E is a uniform field, we may use: Φ= E A= EAcosθ Φ=Φ +Φ +Φ = E A = EA cos18 + EA cos9 + EA cos = a b c a b c.1 Electric Flu Sample problems.: A non-uniform electric filed given by E =.i + 4. j pierces the Gaussian cube shown in Fig. -5. (E is in newtons per coulomb and is in meters) What is the electric flu through the right face, the left face, and the top face? Unit vector: i i = j j = k k = 1, i j = j k = k i = y Right face: d A = dydzi, Φ = E da =? z y z
.1 Electric Flu Left face: d A = dydzi, Φ= E da =? y y z z Top face: da= ddzj, Φ = E da =?. Gauss Law Gauss law: it relates the net flu Φ of an electric field through a closed surface (a Gaussian surface) to the net charge q enc that is enclosed by that surface. ε Φ = qenc ε E da = q enc - q enc is the algebraic sum of all the enclosed charges - The eact form and location of the charges inside the Gaussian surface are of no concern; the only things that matter are the magnitude and sign of the net enclosed charges - If q enc is negative, then net flu is inward - If q enc is positive, then net flu is outward 4
. Gauss Law z Problems -6: At each point on the surface of the cube in Fig. -1, the electric field is parallel to the z ais. The length of each edge of the cube is. m. On the top face of the cube E = 4k N/C, and on the bottom face E =+ k N/C. Determine the net charge contained within the cube. y. Gauss Law Coulomb s law can be derived from Gauss law with some symmetry considerations. 1. Draw the situation. Choose a Gaussian surface appropriate to the symmetry. Apply Gauss Law How do we choose Gaussian surface? To make the integration as simple as possible, we choose the shape that mimics the symmetry of the charge distribution. For a point charge, choose a concentric sphere as a Gaussian surface. E is constant over the surface and directed radially outwards Gauss Law ε E da = q enc Coulomb s Law: q E = 4πε r 5
. Gauss Law Sample Problems.: Figure -1a shows, in cross section, a plastic, spherical shell with uniform charge Q = -16e and radius R = 1 cm. A particle with charge q = +5e is at the center, what is the electric filed (magnitude and direction) at (1) point P 1 at radial distance r 1 = 6. cm and (b) point P at radial distance r = 1. cm? Problems -14: Flu and nonconducting shells. A charged particle is suspended at the center of two concentric spherical shells that are very thin and made of nonconducting material. Figure - 1a shows a cross section. Figure -9b gives the net flu Φ through a Gaussian sphere centered on the particles, as a function of the radius r of the sphere. The scale of the vertical ais is set by Φ s = 5.1 5 Nm /C. (a) what is the charge of the central particles? What are the net charges of (b) shell A and (c) shell B? 6
. A Charged Isolated Conductor An isolated conductor: If an ecess charge is placed on an isolated conductor, that ecess charge will move entirely to the surface of the conductor. An isolated conductor with a cavity: No net charge on the cavity walls; all the ecess charge remains on the outer surface of the conductor. Electric field inside a conductor: The charges are in electrostatic equilibrium. Electric field just outside conductor (conductor free space interface): E t = E n E = Tangential direction Normal direction σ = ε o There is flu only through the eternal end face. A Charged Isolated Conductor Sample problem.5: Figure -1a shows a cross section of a spherical metal shell of inner radius R. A point charge of 5.μC is located at a distance R/ from the center of the shell. If the shell is electrically neutral, what are the (induced) charges on its inner and outer surfaces? Are those charges uniformly distributed? What is the field pattern inside and outside the shell? For conducting shell, due to the inside point charge, there are induced charges on inner wall of shell. The reason is: electric field inside a conductor has to be zero. Within a point charge of -5μC within the shell, a charge of +5μC must lie on the inner of the shell The conducting shell is electrically neutral. Inner wall can have a charge of +5μC, only if electrons with charge of -5μC move to outer wall. 7
.4 Applying Gauss Law: Cylindrical Symmetry Infinitely long cylindrical plastic rod with a uniform linear charge density λ How can we find E at a distance r from the ais of the rod, for the infinitely long cylindrical plastic rod, by using Gauss law? 1. Choose a circular cylinder of radius r and length h with two end caps as Gaussian surface. Calculate net flu F: - E direction: radial outward (or inward) up to the sign of the charge - surfaces: Cylindrical surface (Φ = Ε π r h), top cap (Φ = ), and end cap (Φ = ). Find E by using Gauss law. There is flu only through the curved surface ε Φ= εeπrh = λh qenc λ E = πε r Problem -9: Fig.-4 is a section of a conducting rod of radius R 1 =1. mm and length L = 11. m inside a thin-walled coaial conducting cylindrical shell of radius R = 1.R 1 and the (same) length L. The net charge on the rod is Q 1 =.41-1 C ; that on the shell is Q = -.Q 1. What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r =.R? What are (c) E and (d) the direction at r = 5.R 1? What is the charge on the (e) interior and (f) eterior surface of the shell? 8
.5 Applying Gauss Law: Planar Symmetry Thin, infinite, non conducting sheet with a uniform surface charge density σ How can we find E at a distance r in front of the sheet, by using Gauss law? 1. Choose a closed cylinder with end caps of area A as a Gaussian surface. Calculate net flu Φ: - E direction: directed away (or toward) up to the sign of the charge - surfaces: Cylindrical surface (Φ = ), top cap (Φ = ΕΑ), and end cap (Φ = ΕΑ). Find E by using Gauss law : ε Φ= ε EA qenc There is flu only through the two end surface = σ A E = σ /ε.5 Applying Gauss Law: Planar Symmetry Two conducting plates: (a) & (b) single plate: Ecess charges lies on surface ε Φ= ε EA σ 1A qenc E = σ / ε = 1 (c) Two plates: Ecess charges lies on inner surface ε Φ= ε EA σ 1A qenc E = σ / ε = 1 9
.5 Applying Gauss Law: Planar Symmetry Problem -5: Fig.-46a shows three plastic sheets that are large, parallel, and uniformly charged. Figure -46b gives the component of the net electric field along an ais through the sheets. The scale of the vertical ais is set by Es = 6.1 5 N/C. What is the ratio of the charge density on sheet to that on sheet? 1 σ 1 σ σ.6 Applying Gauss Law: Spherical Symmetry Spherical shell with total charge of q and radius of R How about E outside the spherical shell and inside the shell? Inside shell: r < R 1. Choose a spherical Gaussian surface S 1. Calculate net flu Φ: Φ = Ε 4π r. Find E by using Gauss law: q enc = E =. Outside shell: r R 1. Choose a spherical Gaussian surface S. Calculate net flu Φ: Φ = Ε 4π r. Find E by using Gauss law: Fig.- charge q ε Φ= qenc εe4π r = q E = 4πε r - E direction: radial outward (or inward) up to the sign of the charge Recall Shell theorems in chapter 1: If a charged particle is located inside a shell of uniform charge, there is no net electrostatic force on the particle from the shell. A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell s charge were concentrated at its center. 1
.6 Applying Gauss Law: Spherical Symmetry Sphere with total charge of q How about E outside the spherical shell and inside the shell? Outside sphere: r R 1. Choose a spherical Gaussian surface as shown in Fig.-19(a). Calculate net flu Φ: Φ = Ε 4π r. Find E by using Gauss law: q ε Φ= qenc εe4π r = q E = 4πε r - E direction: radial outward (or inward) up to the sign of the charge Inside sphere: r < R 1. Choose a spherical Gaussian surface as shown in Fig.-19(b). Calculate net flu Φ: Φ = Ε 4π r. Find charge q enc enclosed in Gaussian surface: 4π r q r qenc = ρ = 4 πr /= q 4 π R / R 4. Find E by using Gauss law: ε Φ=qenc r εe4π r = q R q r E = 4πε R The flu through the surface depends on only the enclosed charge Problem -49: In Fig.-54, a solid sphere of radius a =. cm is concentric with a spherical conducting shell of inner radius b = a and outer radius c =.4a. The sphere has a net uniform charge q 1 = 5 fc; the shell has a net charge q = - q 1. What is the magnitude of the electric field at radial distance (a) r =, (b) r = a/, (c) r = a, (d) r =1.5a, (e) r =.a, and (f) r =.5a? What is the net charge on the (g) inner and (h) outer surface of the shell? 11