Spring 2011[GAUSS’ LAW]1 | P a g eSTUDIO Unit 04PHY 2054 STUDIO College Physics IIELECTRIC FLUX AND GAUSS’ LAWObjectives- to understand that the area of a flat surface can be represented by a vector to understandthe concept of electric flux qualitatively and quantitatively- to understand Gauss’ Law qualitativelyEquipment:1 sheet of graph paper1 index cardExcel softwareAn Applet on Gauss’s LawA. Area Vectorsa. Obtain a sheet of graph paper and an index card. Using a ruler, determine the area of the graph paper and the index card. Is there a unique way torepresent the area of a flat surface by a vector? Is there a unique direction in which it would itpoint? How long would it be? Explain if you can.2 | P a g eb. Consider a closed box three dimensional box. Is it possible to represent each surface of thebox unambiguously by a vector? What is different from the previous question? Explain yourreasoning. Again, is it possible to represent a flat single surface unambiguously by a vector?Draw a diagram and explain. We will discuss this in class so be prepared to participate.SKETCH:EXPLAIN:3 | P a g ec. Write a rule for representing the area of a surface by a vector if it is isolated and if it is aportion of a closed volume.Prepare to discuss your rule.d. Roll the graph paper into a cylinder. Is it now possible to represent the area of this surface by avector? Explain your reasoning.4 | P a g ee. Could you represent the area of a small box on the graph paper by a vector when it is rolledinto a cylinder? Explain.The area of a flat surface can be represented by a single vector perpendicular to the plane of thesurface. The length of the vector is proportional to the area but its direction is not clearlydetermined. Why?B. Electric FluxIt is useful to define a quantity called electric flux. Electric flux is defined as a measure of thenumber of electric field lines passing through a surface (whatever that means).a. We can picture an arrangement of nails to represent the electric field lines in the center of aparallel plate capacitor. A capacitor is constructed by creating two charged surfaces that areparallel to each other with each surface having the same charge but each is of the opposite sign.5 | P a g eThe Parallel Plate CapacitorIn such a configuration, the electric field is perpendicular to the surfaces and configured asshown. (We will prove this later.) The field is the same throughout the entire volume (except atthe ends where we will ignore the differences shown in the figure to the above right). We will represent a surface as the area surrounded by a perimeter that we will call a “loop”.The loop defines the boundary of the area. The boundary of the shaded area below is the loopdefined above.6 | P a g eImagine two flat areas, one 3 x 4 units and the other 8 x 8 units. The first one has 24 electricfield lines passing through it while the second has 28. The electric field lines are perpendicular tothe areas. (See the drawing on the cover page). Which area has the stronger electric fieldassociated with it? Explain.b. Consider a 10 x 10 unit surface and 100 evenly spaced electric field lines perpendicular to thesurface. Consider a square loop 5 x 5 units in size held above this surface. What does thenumber of field lines passing through the surface depend on? Explain. c. Imagine the 5 x 5 “loop” at different orientations in each of the “fields”. Which orientation hasthe least number of field lines through it? What is the direction of the vector that represents the area of the surface in this case? How does it compare to the direction of the electric field? Which orientation has the greatest number of field lines through it? What is the direction of the vector that represents the area of the surface in this case? How does it compare to the direction of the 7 | P a g eelectric field?d. For orientations in between those that have the maximum and minimum number of field linespassing through the loop, qualitatively, how does the number of field lines passing through theloop depend on the angle between the vector that represents the area of the surface and thedirection of the electric field? Explain.You can verify your conclusions and investigate them further by going to the following link: http://webphysics.davidson.edu/physlet_resources/bu_semester2/index.html and click on Electric Fluxwhich you can download from the class website (bindell) or type the address into a browser. This is a great site for “Simulations in Physics” and you can refer to it for other subjects as well. When you get to the site, scroll down on the left until you find “Electric Flux” and click on it. You will find the following:8 | P a g eIn this simulation you can rotate the “frame” and count the number of lines that pass through it asa function of angle. This provides you with a number that sort of represents the “amount” ofelectric field that passes through the surface. We call this the Electric Flux but this is not yet theproper definition for it. Fill in the following table:Angle Cos(angle) Number of Dots251550Show from this simple example that the number of dots is proportional to the cosine of the anglethat the normal to the surface makes with the Electric Field Lines.e. Refer to the following figure again:9 | P a g eWhat is the component of the electric field perpendicular to the surface? How does thecomponent of the electric field perpendicular to the surface change (increases or decreases orstays the same) as the angle - is increased?f. Summarize what the number of field lines passing through a surface depends on. Does thisrelate to the simulation that you just completed?The electric flux through a surface is defined as the magnitude of the electric field times the areaof the surface times the cosine of the angle between the direction of the electric field and the areavector of surfacecosEA10 | P a g eThe units of electric flux is CmN2are Nm2/C. Qualitatively; flux is the number of field linespassing through a surface. When the angle between the area vector of the surface and thedirection of the electric field is greater than 90-, the flux is negative. C. SOME PROBLEMSa. Consider the situations shown below. Each picture contains some charge and an imaginarybox. In each case, determine if there is net flux through the box. The