University Physics: Waves and Electricity Lecture 8 Ch23. Finding the Electric Field – II University Physics: Waves and Electricity 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 vector a and b is written as a·b and defined to be: → → Because of the notation, a·b is also known as the dot product and is spoken as “a dot b.” If a is perpendicular to b, means Φ = 90°, then the dot product is equal to zero. If a is parallel to b, 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 Φ between a = 3i – 4j and 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 Nm2/C or Vm. 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 Φ = +EA Φ = –EA Φ = 0 Φ = 0 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) x y z 1 2 3 (a) E G F A B D C (b)

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 New 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. → ^ ^