Lecture 4-1 Physics 219 Question 1 Aug.31.2016. Where (if any) is the net electric field due to the following two charges equal to zero? y Q Q a x a) at (-a,0) b) at (2a,0) c) at (a/2,0) d) at (0,a) and (0,-a) e) nowhere
Lecture 4-2 Physics 219 Question 2 Aug.31.2016. Where (if any) is the net electric potential due to the following two charges equal to zero? y Q Q a x a) at (-a,0) b) at (2a,0) c) at (a/2,0) d) at (0,a) and (0,-a) e) nowhere
Lecture 4-3 Gauss s Law: Qualitative Statement Form any closed surface around charges Count the number of electric field lines coming through the surface, those outward as positive and inward as negative. Then the net number of lines is proportional to the net charges enclosed in the surface.
Lecture 4-4 Electric flux To state Gauss s Law in a quantitative form, we first need to define Electric Flux: # of field lines N = density of field lines x area E Area Ais a vector. Ais the magnitude of the area. The direction of A is perpendicular to the surface A. The flux is : = E A E A cos E E i i i i i i i General definition of electric flux: A E A E i i i Projected = (must specify sense, i.e., which way does A points)
Lecture 4-5 Why are we interested in electric flux? E is closely related to the charge(s) which create it. Consider Point charge q A E EA q 0 To calculate the above summ over a Gaussian spherical surface We make use of the facts that are true on the surface of Gaussian sphere:
Lecture 4-6 (a) E and E are parralel thus cos =1 (b) E is constant for fixed R (c) E A E( R) A E( R)4 r q R 2 E( R) A k 4R 2 q 0 2 If we now turn to our previous discussion and use the analogy to the number of field lines, then the flux should be the same even when the surface is deformed. Thus should only depend on Q enclosed.
Lecture 4-7 Gauss s Law: Quantitative Statement E Q enclosed 0 Must use a closed surface. Take the unit normal vector always outward.
Lecture 4-8 Flux of an Electric Field The flux through a surface of arbitrary shape in an electric field can be computed by summing up fluxes through infinitesimal segments of the surface EA segment 1 makes negative contribution. segment 2 makes no contribution. segment 3 makes positive contribution.
Lecture 4-9 Electric Flux through a Closed Surface The above results to can be written a in precise mathematical form for cases of a closed surfaces in an electric field as: E S E da This represents the net flux through the closed surface. Such a closed surface is referred to as Gaussian surface. Visually, the net flux through a Gaussian surface is proportional to the net number of electric field lines passing through the surface.
Lecture 4-10 Gauss Law The net flux of an electric field through a Gaussian surface is proportional to the net charge enclosed by the surface, or quantitatively, S E da q enc 0 The equation holds only when the net charge is located in a vacuum but can be generalized to situations in which a material such as mica, oil, or glass is present. The net charge is the algebraic sum of all the enclosed positive and negative charges.
Lecture 4-11 Gauss Law Surface S 1 : the field lines are all outward going, so the flux is positive. Thus, the net charge enclosed must be positive. Surface S 2 : the field lines are all inward going, so the flux is negative. Thus, the net charge enclosed must be negative. Surface S 3 : there is no net flux, so the net charge must be zero. Surface S 4 : the net charge enclosed is also zero in this case. Thus, the net flux must also be zero.
Lecture 4-12 Calculating E from Coulomb s or Gauss s Law If we are given: (a) Location of a point charge q (x, y, z) (b) Line charge density λ (x) (c) Surface charge density σ (x, y) (d) Volume charge density ρ (x, y, z) with the help of Coulomb s law, we can always calculate the electric field E by integrating the infinitesimal electric field dq E de k r ˆ 2 r de : where dq has one of the forms: λdx, σdxdy, ρdxdydz Although Gauss s law is of fundamental validity, it is one of Maxwell s equations, it can be used to calculate E only if the charge distributions have certain symmetries:
Lecture 4-13 1. Plane Symmetry Assume an infinite plane with uniform surface charge density σ(x,y)=constant. By plane symmetry the resulting electric field E is normal to the surface of the infinite plane and it can depend only on the distance from the plane: E x, y, z E x ex
Lecture 4-14 2. Spherical Symmetry Assume: (a) point charge (b) uniform charge distribution on a spherical shell (c) spherically symmetric volume charge distribution By spherical symmetry the resulting electric field is radial and can vary only as function of r: E x, y, z E r er
Lecture 4-15 3. Cylindrical Symmetry Assume: (a) surface with uniform surface charge density (r,,z)=constant (b) cylindrically symmetric volume charge distribution (r,,z)=(r) The resulting electric field has a cylindrically symmetric distribution: E( r,, z) E( r) e r Let us make use of these symmetries!
Lecture 4-16 Infinite Non-conducting Sheet (Planar Symmetry) We choose a cylindrical Gaussian surface that is symmetric about the sheet. From symmetry, E must be perpendicular to the sheet and hence to the end caps. E however is perpendicular to the surface normal of the Barrel (circular cylinder), thus E ds 0 0 since cos 90= 0 E ds E ds E ds EdS E ds S barrel end cap end caps end caps = E ds E ds 2EA e.cap L e.cap R where E and A denote the uniform electric field and the surface of the endcap respectively
Lecture 4-17 Infinite Non-conducting Sheet-Cont d Since, there is no flux through the side wall the net flux through the Gaussian surface is then equal to that through the end caps, = EA + EA = 2EA. According to Gauss law, we have 2EA A E 2 0 0 where is the electric charge density (charge per unit area) 2 of the sheet: C/ m
Lecture 4-18 Capacitor geometry A - If E=0 on the upper side, then what is E between the plates?
Lecture 4-19 Let + and denote plus and minus charge densities on the upper and low plates. Choose the Gaussian cylinder surface to be a rectangular box with top and bottom area A. Then the flux s E ds E ds E A Eq.1 E q A Eq.2 enclosed From Gauss s Law: enclosed Substituting Eq. 1 and Eq. 2 into Gauss s Law we obtain: E 0 E q 0 is
Lecture 4-20 Infinitely long uniformly charged line ( Cylindrical Symmetry) ( end caps) ( barrel) E ( barrel) E E 2 rh ( Eq.1) Q h ( Eq.2) enclosed 0 since cos 90 = 0 charge density: C/m Gauss s Law States: E Q enclosed 0 ( Eq.3) Substituting Eq.1 and Eq. 2 into Eq.3 we obtain E 2 r 0 2k r h r
Lecture 4-21 Gauss Law Examples (Spherical Symmetry) Shell Theorem Consider a conducting sphere with a uniformly distributed charge Q. Outside: If the Gaussian spherical surface is outside the conducting sphere then E is as if Q where a point charge located at the center of the charged sphere: E Q k r 2 Inside: If however the Gaussian spherical surface is located inside the conducting sphere, E = 0 everywhere. Thus: E da Q 0 enclosed Therefore there is no charge inside the conducting sphere. Uniformly distributed charge Q on the surface 0
Lecture 4-22 Proof of the Shell Theorem by Gauss s Law (Spherical Symmetry) Electric Field Outside By symmetry, the electric field must only depend on r and is along a radial line everywhere. Apply Gauss s law to the blue surface, we get Uniformly distributed charge Q on the surface E 2 (4 r ) E 1 4 0 Q Q 2 r 0
Lecture 4-23 E = 0 inside Uniformly charged thin shell: Inside By symmetry, the electric field must only depend on r and is along a radial line everywhere. Apply Gauss s law to the blue surface, we get E = 0. Equal and opposite contributions from charges on diagonally opposite surface elements. Discontinuity in E
Lecture 4-24 Physics 219 Question 3 Aug.31.2016. A charge of Q is placed on the inner ball. What is the magnitude of the electric field on the blue shell of radius r? Q r a) 0 b) kq/r 2 c) - kq/r 2 d) kq/r e) depends on the net charge on the outer shell
Lecture 4-25 Physics 219 Question 4 Aug.31.2016. A charge of Q is placed on the inner ball. What is the magnitude of the electric field on the blue shell of radius r? Q r a) 0 b) kq/r 2 c) - kq/r 2 d) kq/r e) depends on the net charge on the outer shell