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Lecture 3. Electric Field Flux, Gauss’ Law From the concept of electric field flux – to the calculation of electric fields of complex charge distributions. 1 Iclicker Question Charged particles are fixed on grids having the same spacing. Each charge has the same magnitude Q with signs given in the figure. Rank the magnitude of the electric field (from strong to weak) at the location marked with an “x”. 1 2 3 4 A. 1 > 2 > 3 > 4 B. 2 > 3 > 1 = 4 C. 1 = 4 > 3 > 2 D. 3 > 2 > 1 > 4 2 Iclicker Question Charged particles are fixed on grids having the same spacing. Each charge has the same magnitude Q with signs given in the figure. Rank the magnitude of the electric field (from strong to weak) at the location marked with an “x”. 1 2 3 4 A. 1 > 2 > 3 > 4 B. 2 > 3 > 1 = 4 C. 1 = 4 > 3 > 2 D. 3 > 2 > 1 > 4 3 Iclicker Question A negative point charge -Q is released from rest in an electric field. At this moment, the acceleration of the point charge A. is in the direction of the electric field at the position of the point charge. B. is directly opposite the direction of the electric field at the position of the point charge. C. is perpendicular to the direction of the electric field at the position of the point charge. -Q D. is zero. E. not enough information given to decide 𝐸 4 Iclicker Question A negative point charge -Q is released from rest in an electric field. At this moment, the acceleration of the point charge A. is in the direction of the electric field at the position of the point charge. B. is directly opposite the direction of the electric field at the position of the point charge. C. is perpendicular to the direction of the electric field at the position of the point charge. -Q D. is zero. E. not enough information given to decide 𝐸 5 Iclicker Question A positive point charge +Q is released from rest in an electric field. At any later time, the velocity of the point charge A. is in the direction of the electric field at the position of the point charge. B. is directly opposite the direction of the electric field at the position of the point charge. C. is perpendicular to the direction of the electric field at the position of the point charge. 𝑣 D. is zero. E. not enough information given to decide +Q 𝐸 6 Iclicker Question A positive point charge +Q is released from rest in an electric field. At any later time, the velocity of the point charge A. is in the direction of the electric field at the position of the point charge. B. is directly opposite the direction of the electric field at the position of the point charge. C. is perpendicular to the direction of the electric field at the position of the point charge. 𝑣 D. is zero. E. not enough information given to decide +Q 𝐸 7 Flow of Fluid Flow of fluid: the speed of water, v, and the pipe cross section, A, are known. Calculate how much water you added to the bucket in 1 sec. ∆𝑉 = 𝐴 ∙ 𝑣 ∙ 1𝑠 length (water displacement in 1s) ∆𝑉 = 𝐴 ∙ 𝑣 ∙ 1𝑠 𝜙 scalar product, 𝐴 ∙ 𝑣 ∙ cos 𝜙 ∆𝑉/∆𝑉𝑚𝑎𝑥 𝜙 8 Concept of Flux small element of surface - applies to any vector field 𝑎 𝑟 (the field of velocities in a fluid flow, the E field, etc.). vector field The flux of the field 𝑎 𝑟 through a small element of surface 𝑑𝐴 : 𝐸 Φ𝑎 ≡ 𝑎 𝑟 ∙ 𝑑𝐴 = 𝑎 𝑟 𝐴 ∙ 𝑐𝑜𝑠𝜙 scalar ! 𝐸 𝐴 𝐴 (a) (b) (c) 9 Units of Flux The flux of electrostatic field: Φ𝐸 𝑬 𝒓 The flux of gravitational field: 𝑁 2 → 𝑚 𝐶 (𝐹 = 𝑞𝐸) 𝑁 2 𝑚3 Φ𝑔 → 𝑚 = 2 𝑘𝑔 𝑠 𝒈 𝒓 (𝐹 = 𝑚𝑔) 10 Iclicker Question Calculate the flux of uniform electric field through the surface shown in the Figure. 600 E= 2 N/C A. 𝜋 0.01 𝑁 𝑚2 ∙ 𝐶 B. −𝜋 0.01 𝑁 C. 𝜋 0.02 𝑁 𝑚2 ∙ 𝐶 𝑚2 ∙ 𝐶 D. −𝜋 0.02 𝑁 ∙ 𝑚2 𝐶 11 Iclicker Question Calculate the flux of uniform electric field through the surface shown in the Figure. 600 E= 2 N/C A. 𝜋 0.01 𝑁 𝑚2 ∙ 𝐶 B. −𝜋 0.01 𝑁 C. 𝜋 0.02 𝑁 𝑚2 ∙ 𝐶 𝑚2 ∙ 𝐶 D. −𝜋 0.02 𝑁 ∙ 𝑚2 𝐶 Φ𝐸 = 𝐸 𝑟 ∙ 𝐴 ∙ 𝑐𝑜𝑠𝜙 𝜙 = 1200 cos 1200 = −0.5 12 Flux through Closed Surfaces We integrate over this “Gaussian” surface The net flux of the electric field through the closed surface: Φ𝐸 = 𝐸 𝑟 ∙ 𝑑𝐴 𝑑𝐴 the integral is taken over the whole surface 𝐸 Convention: the vector normal to the closed surface points outward. 𝑑𝐴 𝑑𝐴 𝑑𝐴 13 Example: Point Charge at the Center of a Spherical Surface At each point of the surface 𝐸 ∥ 𝑑𝐴 ( cos𝜙=1 ) Φ𝐸 = 𝐸 𝑟 ∙ 𝑑𝐴 = 𝐸 𝑟 4𝜋𝑟 2 1 𝑞 𝑞 2 = 4𝜋𝑟 = 4𝜋𝜀0 𝑟 2 𝜀0 - doesn’t depend on r ! - holds for any vector field whose strength 1/𝑟 2 . (electrostatic, gravitational) The flux of the gravitational field of Earth through Earth’s surface? Φ𝑔 = − 𝐺 𝑀 2 = −4𝜋𝐺𝑀 4𝜋𝑅 𝑅2 14 Point Charge at the Center of a Spherical Surface (cont’d) What happens if we place the charge outside? Φ𝐸 = 0 ! consequence of 1/𝑟 2 field dependence = continuity of field lines The flux magnitudes are the same for these two surfaces. 15 Iclicker Question Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? A. surface A B. surface B C. surface C D. surface D E. both surface C and surface D 16 Iclicker Question Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? A. surface A B. surface B C. surface C D. surface D E. both surface C and surface D 17 Gauss’ Law The total flux of the electric field though any closed surface is proportional to the net charge inside the surface. charges the same fluxes for all three surfaces Φ𝐸 ≡ 𝐸 𝑟 ∙ 𝑑𝐴 = 1 𝜀0 𝑞𝑖 𝑖 The net charge: 𝑄= 𝑞𝑖 𝑖 𝑄= 𝜌 𝑟 𝑑𝜏 𝑣𝑜𝑙𝑢𝑚𝑒 charge density element of volume 18 Applications of Gauss Law 𝑸𝒏𝒆𝒕 𝑬 𝒓 ∙ 𝒅𝑨 = 𝜺𝟎 Gauss’ Law (GL) in its integral form (one scalar equation) is not sufficient for finding three components of a vector field E. However, GL is very useful whenever it is possible to reduce a 3D (vector) problem to a 1D (scalar) problem. The key is the proper symmetry of a problem. «Useful» symmetries: - spherical - cylindrical + translational - plane An intelligent choice of a Gaussian surface is crucial! Example of an «insufficient» symmetry (cylindrical but without translational): an electric dipole. 19 Spherical Symmetry Example: electric field of a uniformly charged ball. Radius R, total charge Q. charge density 𝑄 𝜌= 4 3 3 𝜋𝑅 R O Gaussian surface: a sphere centered at () O. Its radius r can be smaller or larger than R. enclosed charge 𝑞 𝑟′ ≤ 𝑟 2 𝐸 𝑟 ∙ 4𝜋𝑟 = = 𝜀0 r Φ𝐸 𝑅 𝑟 flux 𝑄𝑟 4𝜋𝜀0 𝑅 3 𝐸 𝑟 =𝐸 𝑟 𝑟 𝐸 𝑟 = 𝑄 4𝜋𝜀0 𝑟 2 4 𝜌 3 𝜋𝑟 3 𝜀0 𝑄 𝜀0 𝑄 𝑟3 = 𝜀0 𝑅3 𝑟<𝑅 𝑟≥𝑅 𝑟<𝑅 𝑟≥𝑅 The field outside the sphere is the same as that for a point charge Q at the sphere’s center. 20 Plane Symmetry Example: the field of uniformly charged infinite plane. Gaussian surface: a rectangular box (or cylinder)centered at the plane. enclosed charge 𝜎 𝐸 𝑥 = 2𝜀0 𝐸 𝑥 = ±𝐸 𝑥 𝑥 𝜎𝐴 𝐸 𝑥 ∙ 2𝐴 = 𝜀0 flux 𝐸 𝑥 - the surface charge density [C/m2] 𝜎 2𝜀0 𝜎 − 2𝜀0 𝑥 21 Superposition + + + - + + + - + + + + + + + + + - 𝐸 𝑥 𝐸 𝑥 𝑥 𝑥 𝑬𝟏 𝑬𝒏𝒆𝒕 = 𝑬𝟏 + 𝑬𝟐 𝑬𝟐 𝑬𝟏 𝑬𝟐 𝑬𝒏𝒆𝒕 = 𝑬𝟏 + 𝑬𝟐 22 Conclusion Gauss’ Law: works in electrodynamics, in electrostatics it is equivalent to Coulomb’s Law. Powerful tool for computing the electric fields if a problem is essentially 1D due to symmetry. Next time: Lecture 4. Applications of Gauss’ Law, Conductors in Electrostatics. §§ 22.5 Read about Metals and Dielectrics. 23 Appendix I. Cylindrical Symmetry + Translational Symmetry along the axis of symmetry Example: the field of uniformly charged infinitely long wire. Gaussian surface: a cylinder of length L centered at the wire. enclosed charge 𝐸 𝑟 ∙ 2𝜋𝑟𝑙 = - the linear density of charges (the unit length charge) 𝜌𝑙 𝜀0 flux 𝑟≥𝑅 𝜌 𝐸 𝑟 = 2𝜋𝜀0 𝑟 𝐸 𝑟 =𝐸 𝑟 𝑟 𝐸 𝑟 1/r 𝑅 𝑟 24 Appendix II: Symmetry Facilitates Flux Calculations Example 1: A charge q is at the center of a cube. What’s the flux of E through the shaded side? 𝑞 1 Φ𝐸 = × 𝜖0 6 Example 2: A charge q is at the back corner of a cube. What’s the flux of E through the shaded side? Φ𝐸 = 𝑞 1 × 𝜖0 24 25 Iclicker Question Rank the fluxes Φ𝑖 through the five surfaces (from most negative to most positive): A. Φ1 < Φ2 < Φ3 < Φ4 < Φ5 B. Φ1 < Φ4 < Φ2 < Φ3 < Φ5 C. Φ4 < Φ5 < Φ2 < Φ3 < Φ1 D. Φ2 < Φ4 < Φ1 < Φ3 < Φ5 E. Φ5 < Φ4 < Φ3 < Φ2 < Φ1 26 Iclicker Question Rank the fluxes Φ𝑖 through the five surfaces (from small to large): A. Φ1 < Φ2 < Φ3 < Φ4 < Φ5 B. Φ1 < Φ4 < Φ2 < Φ3 < Φ5 C. Φ4 < Φ5 < Φ2 < Φ3 < Φ1 D. Φ2 < Φ4 < Φ1 < Φ3 < Φ5 E. Φ5 < Φ4 < Φ3 < Φ2 < Φ1 27