1 / 43

Lecture 5 Electric Flux and Gauss ’ s Law

Lecture 5 Electric Flux and Gauss ’ s Law. E. A. q. Electric Flux. Electric flux quantifies the notion “number of field lines crossing a surface.”

afya
Download Presentation

Lecture 5 Electric Flux and Gauss ’ s Law

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 5Electric Flux and Gauss’s Law General Physics II, Lec 5, By/ T.A. Eleyan

  2. E A q Electric Flux • Electric flux quantifies the notion “number of field lines crossing a surface.” • The electric flux  through a flat surface in a uniform electric field depends on the field strength E, the surface area A, and the angle  between the field and the normal to the surface. • Mathematically, the flux is given by • Here A is a vector whose magnitude is the surface area A and whose orientation is normal to the surface. General Physics II, Lec 5, By/ T.A. Eleyan

  3. When q < 90˚, the flux is positive (out of the surface), and when q > 90˚, the flux is negative. • Units: Nm2/C in SI units, the electric flux is a SCALAR quantity • Find the electric flux through the area A = 2 m2, which is perpendicular to an electric field E=22 N/C Answer: F = 44 Nm2/C. General Physics II, Lec 5, By/ T.A. Eleyan

  4. Example: Calculate the flux of a constant E field (along x) through a cube of side “L”. y 2 1 Solution E x z General Physics II, Lec 5, By/ T.A. Eleyan

  5. Question: The flux through side B of the cube in the figure is the same as the flux through side C. What is a correct expression for the flux through each of these sides? General Physics II, Lec 5, By/ T.A. Eleyan

  6. Question: What is the electric flux through a cylindrical surface? The electric field, E, is uniform and perpendicular to the surface. The cylinder has radius r and length L • A) E 4/3 p r3 L • B) E r L • C) E p r2 L • D) E 2 p rL • E) 0 General Physics II, Lec 5, By/ T.A. Eleyan

  7. When we have a complicated surface, we can divide it up into tiny elemental areas: General Physics II, Lec 5, By/ T.A. Eleyan

  8. Example: • What’s the total flux on a closed surface with a charge inside? • The shape and size don’t matter! • Just use a sphere General Physics II, Lec 5, By/ T.A. Eleyan

  9. What is Gauss’s Law? • Gauss’s Law does not tell us anything new, it is NOT a new law of physics, but another way of expressing Coulomb’s Law • Gauss’s law makes it possible to find the electric field easily in highly symmetric situations. General Physics II, Lec 5, By/ T.A. Eleyan

  10. Gauss’ Law • The precise relation between flux and the enclosed charge is given by Gauss’ Law • e0 is the permittivity of free space in the Coulomb’s law • The symbol has a little circle to indicate that the integral is over a closed surface. The flux through a closed surface is equal to the total charge contained divided by permittivity of free space General Physics II, Lec 5, By/ T.A. Eleyan

  11. A few important points on Gauss’ Law • The integral is over the value of E on a closed surface of our choice in any given situation • The charge Qencl is the net charge enclosed by the arbitrary close surface of our choice. • It does NOT matter where or how much charge is distributed inside the surface • The charge outside the surface does not contribute. Why? General Physics II, Lec 5, By/ T.A. Eleyan

  12. Question • What’s the total flux with the charge outside? Why? Solution: • Zero. • Because the surface surrounds no charge General Physics II, Lec 5, By/ T.A. Eleyan

  13. Gauss Coulomb • Calculate E of point like (+) charge Q • Consider sphere radius r centered at the charge • Spherical symmetry: E is the same everywhere on the sphere, perpendicular to the sphere General Physics II, Lec 5, By/ T.A. Eleyan

  14. Application of Gauss’s lawGauss’s law can be used to calculate the electric field if the symmetry of the charge distribution is high. Here we concentrate in three different ways of charge distribution • A linear charge distribution • A surface charge distribution • A volume charge distribution General Physics II, Lec 5, By/ T.A. Eleyan

  15. A linear charge distribution • Let’s calculate the electric field from a conducting wire with charge per unit length  using Gauss’ Law • We start by assuming a Gaussian surface in the form of a right cylinder with radius r and length L placed around the wire such that the wire is along the axis of the cylinder General Physics II, Lec 5, By/ T.A. Eleyan

  16. From symmetry we can see that the electric field will extend radially from the wire. • How? • If we rotate the wire along its axis, the electric field must look the same • Cylindrical symmetry • If we imagine a very long wire, the electric field cannot be different anywhere along the length of the wire • Translational symmetry • Thus our assumption of a right cylinder as a Gaussian surface is perfectly suited for the calculation of the electric field using Gauss’ Law. General Physics II, Lec 5, By/ T.A. Eleyan

  17. The electric flux through the ends of the cylinder is zero because the electric field is always parallel to the ends. • The electric field is always perpendicular to the wall of the cylinder so • … and now solve for the electric field General Physics II, Lec 5, By/ T.A. Eleyan

  18. A surface charge distribution • Assume that we have a thin, infinite non-conducting sheet of positive charge The charge density in this case is the charge per unit area,  From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet General Physics II, Lec 5, By/ T.A. Eleyan

  19. To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly. • Because the electric field is perpendicular to the planeeverywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder. • Using Gauss’ Law we get • … so the electric field from an infinitenon-conducting sheet with charge density  General Physics II, Lec 5, By/ T.A. Eleyan

  20. Assume that we have a thin, infinite conductor (metal plate) with positive charge • The “charge density” in this case is also the charge per unit area, , on either surface; there is equal surface charge on both sides. • From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet • To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height r, chosen to cut through one side of the plane perpendicularly. General Physics II, Lec 5, By/ T.A. Eleyan

  21. The field inside the conductor is zero so the end inside the conductor does not contribute to the integral. • Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the end of the cylinder outside the conductor. • Using Gauss’ Law we get • … so the electric field from an infiniteconducting sheet with surface charge density  is General Physics II, Lec 5, By/ T.A. Eleyan

  22. A volume charge distribution let’s calculate the electric field from charge distributed uniformly throughout charged sphere. • Assume that we have insolating a solid sphere of charge Q with radius r with constant charge density per unit volume . • We will assume two different spherical Gaussian surfaces • r2 > r (outside) • r1 < r (inside) General Physics II, Lec 5, By/ T.A. Eleyan

  23. Let’s start with a Gaussian surface with r1 < r. • From spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. • Gauss’ Law gives us • Solving for E we find inside General Physics II, Lec 5, By/ T.A. Eleyan

  24. In terms of the total charge Q … inside General Physics II, Lec 5, By/ T.A. Eleyan

  25. Now consider a Gaussian surface with radius r2 > r. • Again by spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. • Gauss’ Law gives us • Solving for E we find outside same as a point charge! General Physics II, Lec 5, By/ T.A. Eleyan

  26. Electric field vs. radius for a conducting sphere General Physics II, Lec 5, By/ T.A. Eleyan

  27. Properties of Conductors E is zero within conductor If there is a field in the conductor, then the free electrons would feel a force and be accelerated. They would then move and since there are charges moving the conductor would not be in electrostatic equilibrium. Thus E=0 net charge within the surface is zero How ? General Physics II, Lec 5, By/ T.A. Eleyan

  28. Lecture 6 Application (Gauss’s Law) General Physics II, Lec 5, By/ T.A. Eleyan

  29. [1] A solid conducting sphere of radius a has a net charge +2Q.  A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge –Q as shown in figure.  Using Gauss’s law find the electric field in the regions labeled 1, 2, 3, 4 and find the charge distribution on the spherical shell. Region (1) r < a To find the E inside the solid sphere of radius a we construct a Gaussian surface of radius r < a E = 0 since no charge inside the Gaussian surface Region (3) b > r < c E=0 How? General Physics II, Lec 5, By/ T.A. Eleyan

  30. Region (2) a < r < b we construct a spherical Gaussian surface of radius r Region (4) r > c we construct a spherical Gaussian surface of radius r > c, the total net charge inside the Gaussian surface is q = 2Q + (-Q) = +Q Therefore Gauss’s law gives General Physics II, Lec 5, By/ T.A. Eleyan

  31. [2] A long straight wire is surrounded by a hollow cylinder whose axis coincides with that wire as shown in figure.  The solid wire has a charge per unit length of +λ , and the hollow cylinder has a net charge per unit length of +2λ .  Use Gauss law to find (a) the charge per unit length on the inner and outer surfaces of the hollow cylinder and (b) the electric field outside the hollow cylinder, a distance r from the axis. (a) Use a cylindrical Gaussian surface S1 within the conducting cylinder where E=0 General Physics II, Lec 5, By/ T.A. Eleyan

  32. (b) For a Gaussian surface S2 outside the conducting cylinder General Physics II, Lec 5, By/ T.A. Eleyan

  33. [3] Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ.  Find the electric field at distance r from the axis where r < R. If we choose a cylindrical Gaussian surface of length L and radius r, Its volume is πr²L, and it encloses a charge ρπr²L. By applying Gauss’s law we get, radially outward from the cylinder axis Thus Notice that the electric field will increase as r increases, and also the electric field is proportional to r for r<R. For the region outside the cylinder (r>R), the electric field will decrease as r increases. General Physics II, Lec 5, By/ T.A. Eleyan

  34. Two Parallel Conducting Plates • When we have the situation shown in the left two panels (a positively charged plate and another negatively charged plate with the same magnitude of charge), both in isolation, they each have equal amounts of charge (surface charge density s) on both faces. • But when we bring them close together, the charges on the far sides move to the near sides, so on that inner surface the charge density is now 2s. • A Gaussian surface shows that the net charge is zero (no flux through sides — dA perpendicular to E, or ends — E = 0). E = 0 outside, too, due to shielding, in just the same way we saw for the sphere. General Physics II, Lec 5, By/ T.A. Eleyan

  35. Two Parallel Nonconducting Sheets • The situation is different if you bring two nonconducting sheets of charge close to each other. • In this case, the charges cannot move, so there is no shielding, but now we can use the principle of superposition. • In this case, the electric field on the left due to the positively charged sheet is canceled by the electric field on the left of the negatively charged sheet, so the field there is zero. • Likewise, the electric field on the right due to the negatively charged sheet is canceled by the electric field on the right of the positively charged sheet. • The result is much the same as before, with the electric field in between being twice what it was previously. General Physics II, Lec 5, By/ T.A. Eleyan

  36. [4] Two large non-conducting sheets of +ve charge face each other as shown in figure.  What is E at points (i) to the left of the sheets (ii) between them and (iii) to the right of the sheets? We know previously that for each sheet, the magnitude of the field at any point is a) At point to the left of the two parallel sheets General Physics II, Lec 5, By/ T.A. Eleyan

  37. b) At point between the two sheets (c) At point to the right of the two parallel sheets General Physics II, Lec 5, By/ T.A. Eleyan

  38. [5] A square plate of copper of sides 50cm is placed in an extended electric field of 8*104N/C directed perpendicular to the plate.  Find (a) the charge density of each face of the plate General Physics II, Lec 5, By/ T.A. Eleyan

  39. [6] An electric field of intensity 3.5*103N/C is applied the x axis.  Calculate the electric flux through a rectangular plane 0.35m wide and 0.70m long if (a) the plane is parallel to the yz plane, (b) the plane is parallel to the xy plane, and (c) the plane contains the y axis and its normal makes an angle of 40o with the x axis. (a) the plane is parallel to the yz plane (b) the plane is parallel to the xy plane The angel 90 c) the plane is parallel to the xy plane The angel 40 General Physics II, Lec 5, By/ T.A. Eleyan

  40. [7] A long, straight metal rod has a radius of 5cm and a charge per unit length of 30nC/m.  Find the electric field at the following distances from the axis of the rod: (a) 3cm, (b) 10cm, (c) 100cm. General Physics II, Lec 5, By/ T.A. Eleyan

  41. [8] The electric field everywhere on the surface of a conducting hollow sphere of radius 0.75m is measured to be equal to 8.90*102N/C and points radially toward the center of the sphere. What is the net charge within the surface? General Physics II, Lec 5, By/ T.A. Eleyan

  42. [9] A point charge of +5mC is located at the center of a sphere with a radius of 12cm.  What is the electric flux through the surface of this sphere? [10] (a) Two charges of 8mC and -5mC are inside a cube of sides 0.45m.  What is the total electric flux through the cube? (b) Repeat (a) if the same two charges are inside a spherical shell of radius 0. 45 m. General Physics II, Lec 5, By/ T.A. Eleyan

  43. (a) At 12 cm the charge in side the Gaussian surface is zero so the electric field E=0 [12] A solid copper sphere 15cm in radius has a total charge of 40nC.  Find the electric field at the following distances measured from the center of the sphere: (a) 12cm, (b) 17cm, (c) 75cm. (b) General Physics II, Lec 5, By/ T.A. Eleyan

More Related