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# CLFds5

Module by: George Brown. E-mail the author

Summary: Page 5 of 6 of a PLTL workshop on Coulomb law fields, for undergraduate physics.

## Problems

Work on these problems with your Peer Team members. Determine analytic solutions before substituting any numerical values to find a numerical solution. Each problem is solved by use of either Equation 2 ( E [ r ] = 1 4 π ϵ 0 n = 1 N q n r n 2 r ^ n E [ r ] = 1 4 π ϵ 0 n = 1 N q n r n 2 r ^ n ) or Equation 3 ( E [ r ] = 1 4 π ϵ 0 V ρ η ^ η 2 V E [ r ] = 1 4 π ϵ 0 V ρ η ^ η 2 V , or the lower-dimension analogues involving σ := q S σ := q S or λ := q L λ := q L ). These basic relationships should form the starting point of your solutions, although other basic relationships that you have encountered before, or perhaps have to look up, may also be needed to complete the solutions.

### Problem 1

Consider a disk in the x y x y plane with radius R = 3 cm R = 3 cm and constant charge density σ = 3.4 C cm 2 σ = 3.4 C cm 2 , centered on the origin of coordinates. Find the electric field everywhere on the zz axis.

### Problem 2

Repeat Problem 1 with the charge density changed to σ = σ 0 s R σ = σ 0 s R , where s s is the radial variable in the x y x y plane: s = x 2 + y 2 s = x 2 + y 2 , restricted to the charge distribution, and σ 0 σ 0 is the charge density of Problem 1.

### Problem 3

The figure below can be taken as a very crude model of a water molecule. The red disk represents the oxygen atom and the two green disks represent the hydrogen atoms in the water molecule. Given the coordinate system shown in the diagram, find the electric field everywhere on the xx axis except the origin. In nature, and for an isolated water molecule, the bond distance d 1.0 Å = 10 - 10 m d1.0 Å = 10 - 10 m, the angle between the hydrogen atoms is 2 α 7 π 12 = 105 ° 2α 7 π 12 = 105° and the charge q 9 10 e q 9 10 e. These values can change significantly if other molecules are in the neighborhood of the water molecule. Note the coordinate system chosen: all charges lie in the x y x y plane, with the xx axis bisecting the bond angle 2α2α.

### Problem 4

Four equal point charges qq lie at the corners of a regular tetrahedron centered on the origin, as diagrammed below. The coordinates of the point charges are { a { 1 3 , 0 , 1 2 6 , } , a { - 1 2 3 , 1 2 , - 1 2 6 } , a { - 1 2 3 , 1 2 , - 1 2 6 } , a { 0 , 0 , 1 2 3 2 } } {a{ 1 3 ,0, 1 2 6 ,},a{- 1 2 3 , 1 2 ,- 1 2 6 },a{- 1 2 3 , 1 2 ,- 1 2 6 },a{0,0, 1 2 3 2 }}, where aa is the edge length of the tetrahedron, which is the distance separating any two of the charges.

(a) Demonstrate that the electric field at the origin is zero.

Now consider the red spheres to represent hydrogen atoms, and add a nitrogen atom at the center of the tetrahedron (blue sphere). This gives us a (very, very crude) model of the ammonium ion, NH 4 + NH 4 + , pictured below.

(b) For this model, consider a field point located at the center of any one of the four triangular sides of the tetrahedron. The sum for the electric field at this point reduces to contributions from only two of the five point source charges. Why? Which two charges sum to make a nonzero contribution? Is the direction of this electric field pointing out of the tetrahedron or into it? What is the value of the electric field at this field point? Justify your answers.

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