University Physics: Waves and Electricity. Ch23. Finding the Electric Field – II. Lecture 8. Dr.-Ing. Erwin Sitompul. http://zitompul.wordpress.com. Homework 6: Three Particles. Three particles are fixed in place and have charges q 1 = q 2 = + p and q 3 = +2 p . Distance a = 6 μ m.
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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
University Physics: Waves and Electricity Ch23. Finding the Electric Field – II Lecture 8 Dr.-Ing. Erwin Sitompul http://zitompul.wordpress.com
Homework 6: Three Particles Three particles are fixed in place and have charges q1 = q2 = +p and q3 = +2p. Distance a = 6 μm. What are the magnitude and direction of the net electric field at point P due to the particles?
Solution of Homework 6: Three Particles • Both fields cancel one another • Magnitude • Direction
The Electric Field → • The calculation of the electric field E can be simplified by using symmetry to discard the perpendicular components of the dE vectors. • For certain charge distributions involving symmetry, we can simplify even more by using a law called Gauss’ law, developed by German mathematician and physicist Carl Friedrich Gauss (1777–1855). → → • Instead of considering dE in a given charge distribution, Gauss’ law considers a hypothetical (imaginary) closed surface enclosing the charge distribution. • Gauss’ law relates the electric fields at points on a closed Gaussian surface to the net charge enclosed by that surface.
Flux → • Suppose that a wide airstream flows with uniform velocity v flows through a small square loop of area A. • Let Φ represent the volume flow rate (volume per unit time) at which air flows through the loop. • Φ depends on the angle θ between v and the plane of the loop. → • Unit vector pointing to the normal direction of the plane
Flux → • If v is perpendicular to the plane (or parallel to the plane’s direction), the rate Φ is equal to vA. • If v is parallel to the plane (or perpendicular to the plane’s direction), no air moves through the loop, so Φ is zero. • For an intermediate angle θ, the rate of volume flow through the loop is: → • This rate of flow through an area is an example of a flux. • The flux can be interpreted as the flow of the velocity field through the loop.
Flashback: Multiplying Vectors The Scalar Product → → → → • The scalar product of the vectora andbis written asa·b and defined to be: → → • Because of the notation, a·bis also known as the dot product and is spoken as “a dot b.” • If ais perpendicular to b, means Φ = 90°, then the dot product is equal to zero. • If a is parallel tob, means Φ = 0, then the dot product is equal to ab. → → → →
Flashback: Multiplying Vectors • The dot product can be regarded as the product of the magnitude of the first vector and the projection magnitude of the second vector on the first vector
Flashback: Multiplying Vectors • When two vectors are in unit vector notation, their dot product can be written as
Flashback: Multiplying Vectors → → ^ ^ ^ ^ What is the angle Φbetweena = 3i – 4jand b = –2i + 3k ? Solution:
Flux of an Electric Field • The next figure shows an arbitrary Gaussian surface immersed in a nonuniform electric field. • The surface is divided into small squares of area ΔA, each being very small to permit us to consider the individual square to be flat. • The electric field E may now be taken as constant over any given square. • The flux of the electric field for the given Gaussian surface is: → • Φ can be positive, negative, or zero, depending on the angle θ between E and ΔA → →
Flux of an Electric Field • The exact solution of the flux of electric field through a closed surface is: • The flux is a scalar, and its Si unit is Nm2/C or Vm. • The electric flux through a Gaussian surface is proportional to the net number of field lines passing through that surface • Without any source of electric field inside the surface as in this case, the total flux through this surface is in fact equal to zero
Checkpoint The figure below shows a Gaussian cube of face area A immersed in a uniform electric field E that has the positive direction of the z axis. In terms of E and A, determine the flux flowing through: (a) the front face (xy plane) (b) the rear face (c) the top face (d) the whole cube Φ = +EA Φ = –EA Φ = 0 Φ = 0
Example: Flux of an Electric Field In a three-dimensional space, a homogenous electric field of 10 V/m is directed down to the negative z direction. Calculate the flux flowing through: (a) the square ABCD (xy plane) (b) the rectangular AEFG (xz plane) z 3 (a) G 2 1 F A B y 0 1 2 3 (b) 1 2 D C 3 x E
Homework 7 The rectangle ABCD is defined by its corner points of A(2,0,0), B(0,3,0), C(0,3,2.5), and D(2,0,2.5). Draw a sketch of the rectangular. Given an electric field of E = –2i + 6j V/m, draw the electric field on the sketch from part (a). Determine the number of flux crossing the area of the rectangular ABCD. → ^ ^
Homework 7 The triangle FGH is defined by its corner points of F(2,0,0), G(0,3,0), and H(0,0,4). Draw a sketch of the rectangular. Given an electric field of E = –2i + 6j V/m, draw the electric field on the sketch from part (a). Determine the number of flux crossing the area of the triangle FGH. → ^ ^ New