A charge is near a “Gaussian sphere”. Remember that the sphere

Objective: To investigate Gauss’s law.Introduction:Gauss’s Law states that the electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the constant , called the permittivity of free space.Symbolically this statement is written: .Keep in mind that the closed surface discussed here, often called a Gaussian surface, is an imaginary surface. It is a mathematical construct used to simplify calculation and does not necessarily coincide with the surface of an actual physical object.Investigation 1: Charged Particle Outside of a Gaussian Sphere1. A charge is near a “Gaussian sphere”. Remember that the sphere is imaginary and there is really nothing to interfere with any charges or electric fields.Draw several electric field lines from , but only ones that intersect the sphere. (Do not draw the electric field lines that do not intersect the sphere to keep the drawing neat. Extend the lines through the sphere. You may want to use a ruler.)2. How much charge is enclosed by the sphere? Applying Gauss’s law, what is the total electric flux through the sphere? Justify your answer mathematically using the equation above.3. Examine your drawing. The electric field lines impinge on the spherical surface from the outside heading inward (by definition this is negative electric flux) and eventually impinge on another part of the surface from the inside heading outward (this is positive electric flux). Does it seem reasonable that the total flux through the sphere is exactly zero based on ? Explain why the total flux through the sphere is exactly zero even though the field lines exit through a larger area than they entered. (Hint: Consider the magnitude of the electric field at the points of entry and exit points.)4. Suppose the Gaussian sphere was replaced with a Gaussian cube. Would the total flux still be zero? Explain.Note that it would be difficult to actually calculate the electric flux through these surfaces using (although it can be done), because the strength of the electric field and angle of entry and exit vary over the surface. However, applying Gauss’s law makes this easy.Investigation 2: Charged Particle Inside of a Gaussian Sphere5. Now consider a Gaussian sphere centered on .Draw some electric field lines emanating from . Draw them long enough to intersect the sphere.6. Is the total electric flux through the sphere positive or negative? Does this make sense considering the charge enclosed? Explain.7. In symbols, express the total electric flux through the sphere.Notice that in this case the electric field lines intersect the sphere perpendicular to its surface and that the electric field strength is uniform over the surface.8. Explain how you know the electric field strength does not vary over the surface? (Hint: Use the diagram above.)9. Because of the simplifying conditions discussed above, Gauss’s law can be used to find the electric field due to . Find the electric field for the point charge . (The surface area of a sphere is .)10. Graph the magnitude of the electric field versus the distance from the point charge.11. If we had chosen a cube as the Gaussian surface instead of a sphere, would it have been as easy to determine the total electric flux through the surface? Explain.12. If we had chosen a cube as the Gaussian surface instead of a sphere, would it have been as easy to determine the electric field strength at each point on the surface? Explain.Investigation 3: Applying Gauss’s Law to a Non-Uniformly Charged SphereThe previous problem was mathematically simple. In the next problem you will have to do some integration. Keep your results in symbolic form and only substitute number when you are asked for a numerical result. Be sure to distinguish , the radius of the sphere, from , the distance from the center of the sphere at which you are calculating the electric field, .13. Consider a small sphere (an actual physical object, not a Gaussian surface) of radius that is charged throughout its interior, but not uniformly. The charge density in is , where is the distance from the center of the sphere, and is a constant. For (outside of the charged sphere) the charge density is zero.14. What does the charge density converge to as you approach the center of the sphere? How does the charge density change as you move from the center of the sphere toward the surface?15. What does converge to as you approach the center of the sphere? How do you know? How does this compare to the electric field of a point charge as you converge on its location?16. Apply Gauss’s law to find an expression for the magnitude of the electric field of the charged sphere, , when . Hint: draw a Gaussian sphere inside of the charged sphere with radius . The volume of a thin spherical shell is . Charge density is . Rearrange and integrate to solve for q. Remember that for this particular problem . Substitute for in the expression. You should be able to integrate this expression as it should only contain the variable . )17. Use your expression for to find the electric field at . Your answer should be in units of N/C. Use .18. Apply Gauss’s law to find an expression for the magnitude of the electric field of the charged sphere, , when . Hint: this should be easier since all of the charge is enclosed. Use the following: , , .19. What is at ?20. Sketch an approximate graph of the magnitude of the electric field versus the distance from the center of the sphere.21. Outside the sphere, how does the electric field compare to the electric field of a point charge of the same magnitude? Sketch with a dotted line on the graph above versus for a point charge with the same magnitude of the charged sphere.Investigation 4: A Gaussian Sphere around an Electric Dipole22. Consider the diagram below of two point charges with opposite signs. Draw several (at least ten) electric field lines for the configuration in the diagram. Draw enough lines to indicate the electric field in much of the vicinity of the two charges.23. On your drawing above, add a spherical Gaussian surface that encloses both charges, centered on the point midway between the charges. Be sure that several of the electric field lines you drew penetrate the Gaussian sphere. Add more field lines to your diagram if needed.24. Look carefully at your drawing above. In what direction does the electric field point (Inward? Outward? Tangent?) at various locations on the Gaussian surface? Redraw the Gaussian surface below and draw short arrows on the surface indication the direction of on the Gaussian surface.25. Consider the following argument from a student who is trying to determine somewhere on the previous Gaussian surface:“The total charge enclosed by the surface is zero. According to Gauss’s law this means the total electric flux through the surface is zero. Therefore, the electric field is zero everywhere on the surface.”Which, if any, of the three sentences are incorrect?26. Explain how you think the student came to an incorrect conclusion.Investigation 5: Applying Gauss’s Law to a Spherical ShellA point charge lies at the center of an uncharged, hollow, conducting spherical shell of inner radius and outer radius as shown in the diagram below. Your ultimate goal is to find the electric field at all locations for this arrangement.27. Why is it important to know whether the shell is a conductor or insulator when determining the electric field of the object at various locations?28. What does the charge distribution within the shell itself (between and ) look like before the point charge is placed within the hollow, conducting spherical shell.29. What does the charge distribution within the shell itself (between and ) look like after the point charge is placed within the hollow, conducting spherical shell.30. Are the charge distributions the same? Why or why not?31. What is the electric field within the shell itself (between and ) before the point charge is inserted?32. What is the electric field within the shell itself (between and ) after the point charge is inserted?33. How did the fact that the shell is a conductor help you answer the two previous questions?34. We are still considering the conducting shell. Draw electric field lines in the region ( is the distance from the point charge) and for the region . What are the field line between and ? Also show the relative amount of charge that has moved to the inner and outer surfaces of the conductor.35. Use Gauss’s law to sketch the electric field as a function of for all three regions. Be sure to label and on the horizontal axis.36. Compare the electric field for just inside and just outside . Are the magnitudes equal? Explain.37. In specifying the electric field, do you need to give any other information beside the distance from the point charge (for instance, whether you’re on the right side or left side of the point charge)? Explain.Conclusion: Write a paragraph summarizing what you learned in this lab. State important rules, relationships, and concepts you discovered or skills you acquired in performing the investigations.