'

Gauss' Law (d) Suppose we replaced the sphere with a cube. Would the total flux still be zero? . + Q Note that it would be difficult to actually calculate the electric field flux through these surfaces (although you could certainly do it) because the electric field strength and angle of intersection vary over the surface. Applying Gauss' Law, however, made it easy. 4. Now consider a Gaussian sphere centered on + Q. (a) Draw some electric field lines. Make them long enough to intersect the sphere. (b) Is the total electric field flux through the sphere positive or negative? Does this make sense, considering the charge enclosed? Discuss. Thanks to University of Virginia Physics DepartmentGauss' Law + Q (a) Draw several electric field lines from +Q, but only ones that intersect the sphere. (For his question, omit field lines that don't intersect the sphere. This is to keep the drawing looking neat.) (b) How much charge is enclosed by the sphere? Applying Gauss' Law, what is the total electric field flux through the sphere? Justify your answer. (c) Looking back at your drawing, field lines impinge on the spherical surface from the outside heading inward (this is defined as negative flux) and eventually impinge on another part of the surface from the inside heading outward (positive flux). Does it seem reasonable that the total flux through the sphere is exactly zero? If we count the field lines leaving the sphere as positive and the field lines entering the sphere as negative, is the net number of field lines entering or leaving the sphere is zero? Thanks to University of Virginia Physics DepartmentGauss\" Law Gauss' law is always true, no matter how complicated the distribution of electric charges. In fact it's even true when the charges are in motion. However, it's rare to nd a situation with enough symmetry that applying Gauss' law becomes a convenient method to calculate the electric eld. Electric elds in the Vicinity of a conductor A point charge +0 lies at the center of an uncharged, hollow, conducting spherical shell of inner radius Rm and outer radius Rom as shown. Your ultimate goal is to nd the electric eld at all locations for this arrangement. 7. Why is it important to know whether the shell is a conductor or insulator? Is there any real difference? 8. What does the charge distribution within the shell itself (between Rm and Flour) look like (a) before the point charge +0 is inserted? (draw a picture)? (b) with the point charge +0 in place? (draw a picture)? Thanks to University of Virginia Physics Department Gauss' Law of E on the Gaussian surface. (e) Consider the following argument from a student who is trying to determine E somewhere on the previous Gaussian surface: \"The total charge enclosed by the surface is zero. According to Gauss' law this means the total electric eld flux through the surface is zero. Therefore, the electric eld is zero everywhere on the surface.\" Which, if any, of the three sentences are correct? Explain how the student came to an incorrect conclusion. (f) Refer back to 6(a). If you were asked to calculate the electric eld at that point, would you attempt to apply Gauss\" Law or would you use another method? Discuss. Thanks to University of Virginia Physics Department Gauss' Law (d) What is Eat /=0.05 m? [Your answer should be in N/C. ] (e) Apply Gauss' Law to find an expression for E when I is greater than R. (f) What is Eat /= 5.0m? (g) Make an approximate graph of E versus / on the axes shown below: Emax + 0.1 m (h) Determine the total charge inside the sphere. Outside the sphere, how does E compare to E of a point charge of that magnitude? Verify for /= 5.0m. (i) On your graph in part (g) above, use a dotted line to represent the electric field if the sphere shrank to a point charge but still contained the same total charge. Thanks to University of Virginia Physics DepartmentGauss\" Law 6. Consider two point charges of opposite sign as shown: (a) In what direction is the e ctric field at the point indicated? Draw a vector on the sketch above at the point to represent E. [Hintz Use vector addition] (b) On the sketch below, draw several (at least ten) electric eld lines for the configuration above. This should be enough to indicate E in much of the vicinity of the two charges. (c) On your drawing above, add a spherical Gaussian surface that encloses both charges, centered on the point midway between the charges. Make it fairly large, but be sure several of the electric field lines penetrate the Gaussian sphere. Add more field lines as needed. (d) Look carefully at your drawing above. In what direction does E point (inward? outward? tangent?) at various locations on the Gaussian surface? Redraw your Gaussian surface below and draw short arrows on the surface indicating the direction Thanks to University of Virginia Physics Department Gauss\" Law 5. The previous problem was mathematically fairly simple. Here's another problem requiring Gauss' Law, but this time you will have to do a bit of integration NOTE: Keep your results in symbolic form and only substitute in numbers when asked for a numerical result. Also, pay careful attention to the distinction between the radius of the sphere, R, and the distance, r, from the center of the sphere at which you are evaluating E. Consider a small sphere (an actual sphere, not a Gaussian surface) of radius R: 0.1 m that is charged throughout its interior, but not uniformly so. The charge density isp=Br , where r is the distance from the center, and B = 10'4 C/m4 is a constant. Of course, for r greater than R, the charge density is zero. H=0.l m (a) What does the charge density converge to as you approach the center of the sphere? Does it increase or decrease as we move toward the surface? Explain. (b) What does E converge to as you approach the center of the sphere? How do you know? How does this compare to the E of a point charge? [Hintz Consider the symmetry of the problem.] (c) Apply Gauss\" Law to nd an expression for E when r is less than H. [Hint the volume ofa thin spherical shell ofradius rand thickness dris dV=41Tr drl .] Thanks to University of Virginia Physics Department Gauss' Law + Q . 1 1. Use Gauss' law to find the electric field as a function of I for all three regions, and then graph it below. E r 12. Compare the electric field for I just inside Rin and just outside Rout. Are the magnitudes equal? Explain. 13. In specifying the electric field, do you need to give any other information besides the distance from the point charge (for instance, whether you're on the right side or the left side of the point charge, or something like that)? Discuss.Gauss' Law (b) Do the same for a negative charge -Q. . - Q (c) Do the same for a positive charge of +2Q. As the electric field around the charge will be twice as large as the electric field around a charge of +Q, the field line density will have to be twice as large. Make sure that you use twice as many field lines as for the field line pattern of problem 1(a). (d) Comment whether or not all field line rules provided above are obeyed. 3. Consider a "Gaussian sphere", outside of which a charge +Q lies. Remember, a Gaussian surface is just a mathematical construct to help us calculate electric fields. Nothing is actually there to interfere with any electric charges or electric fields. Thanks to University of Virginia Physics DepartmentGauss' Law (c) Are the charge distributions the same? Why or why not? 9. What is the electric field within the shell itself (between Rm and Rout, NOT including the surfaces) (a) before the point charge +0 is inserted? (b) after the point charge +0 is inserted? (c) How did the fact that the shell is a conductor help you answer the two previous questions? 10. We are still considering the conducting shell. Draw electric field lines in the region 0 Rom What are the field lines between Rm and Rout? Also show the relative amount of charge that has moved to the inner and outer surfaces of the conductor. Thanks to University of Virginia Physics Department Theoretical lab- GAUSS' LAW Purpose: Theoretical study of Gauss' law. Equipment: This is a theoretical lab so your equipment is pencil, paper, and textbook. When drawing field line pattern around charge distributions we use in general three different rules: (a) Field lines start at positive charges or in infinity and end at negative charges or in infinity. (b) Field lines never cross each other. Crossing field line would mean that at the position of the crossing field lines the direction of the electric field is no longer unambiguous determined (note that the electric field direction is tangent to the field lines) (c) The field line density is linear proportional to the electric field. So the electric field is larger in areas with a lot of field lines. Also review the lecture where we relate field lines and the magnitude of the electric field On all questions, work together as a group. 1. The statement of Gauss' Law: (a) in words: The electric flux through a closed surface is equal to the total charge enclosed by the surface divided by 60'. (b) in symbols: E . dA = Pdosed surface 2. The next few questions involve point charges. (a) Draw the electric field lines in the vicinity of a positive charge +Q. . + Q 180 = 8.85 x 10-12 C2/N.m2 Thanks to University of Virginia PhysicsGauss' Law (c) In symbols, what is the total flux through the sphere? [Use equation in step 1(b).] 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 (d) How do we know that the electric field strength does not vary over the surface? (e) Because of the simplifying conditions discussed in part (d), we can apply Gauss' Law to find the electric field due to +Q. Find the electric field for a point charge + Q. (Recall that the surface area of a sphere is 4TT/2.) (f) Graph E (the magnitude of E) versus /, using the axes given. [Both axes have linear scales.] E (g) If we had picked a cube as our Gaussian surface instead of a sphere, would it still have been easy to determine the total flux through the surface? What about calculating the electric field strength? Explain. Thanks to University of Virginia Physics Department