Physics 208: Electricity and Magnetism. Exam 1, Secs. 801–805 7 February 2006Instructor: Dr. George R. Welch, 301D Doherty Bldg, 845-1571Print your name neatly:Last name:First name:Sign your name:Please fill in your Student ID number (UIN): – –IMPORTANTRead these directions carefully:• There are 7 problems totalling 100 points. Check your exam to make sure youhave all the pages. Work each problem on the page the problem is on. Youmay use the back. If you need extra pages, I have plenty up front.• Indicate what you are doing! We cannot give full credit for merely writingdown the answer. Neatness counts! I will give generous partial credit if Ican tell that you are on the right track. This means you must be neat andorganized.• Each problem with its associated figure is self explanatory. If you must aska question, then come to the front, being as discrete as possible so as not todisturb others.• Put your name on each page it is asked for. You will lose credit if you fail toprint your name on each page it is asked for.Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 1. 10 points.Two small charges, −Q and +Q, are separated by a distance 2d where d is a given positiveconstant. Calculate the electric field a perpendicular distance y from the midpoint of the lineconnecting the charges (this point is shown in the Figure).Use the coordinate system shown. Remember that electric field is a vector, so calculate boththe x and y components of the field.Express your answer in terms of Q, y, d, physical constants such as ²0, and numerical factors.−Qx+QyCalculate electricfield hereddBe neat. Neatness helps. Work neatly.Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 2. 20 points.Two thin rods, each of length d, carry charges −Q and +Q distributed uniformly along theirlength. The two rods are abutted end to end as shown in the Figure. Calculate the electric field aperpendicular distance y from the midpoint of the line connecting the charges (this point is shownin the Figure).Use the coordinate system shown. Remember that electric field is a vector, so calculate boththe x and y components of the field.Express your answer in terms of Q, y, d, physical constants such as ²0, and numerical factors.−Qx+QyCalculate electricfield heredd0000000000111111111100000000001111111111If you work neatly I will find more partial credit for you!Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 3. 10 points.Three small spheres shown below carry charges q1, q2, and q3respectively. Find the net electricflux through each of the five closed surfaces shown in cross section in the Figure.Express your answers in terms of q1, q2, q3, physical constants such as ²0, and numerical factors.q1qq32S 3S 2surface S 1S 4S 5(a) flux through Surface S1 =(b) flux through Surface S2 =(c) flux through Surface S3 =(d) flux through Surface S4 =(e) flux through Surface S5 =Make sure you are being neat. Working neatly will help you get it right.Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 4. (20 points)A conducting sphere of radius R1carries a net charge −Q.Surrounding this sphere is a hollow spherical shell of nonconducting material. This shell hasinner radius R2and outer radius R3. A total charge +Q is distributed uniformly throughout thespherical shell.This is shown in cross section below.R1RR23−Q+QCalculate the electric field in each of the following regions:(a) points inside the conducting sphere: r < R1.(b) points between the conducting sphere and the insulating shell: R1< r < R2.(c) points inside the insulating shell: R2< r < R3.(d) points outside the insulating shell: r > R3.If you need more room to work neatly, you can continue this problem on the next page.You need to work neatly! Don’t forget to be neat.Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 4 — continued...Work neatly! If you are neat, I can read what you did.Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 5. (10 points)Two large parallel conducting plates are separated by a distance d. A spring of force constantk is fixed to one of the plates, and on its other end is a charge Q as shown in the Figure. Ignoregravity in this problem.Suppose that the electric potential difference between the plates is increased from 0 to V . Thiswill cause the charge Q to push down on the spring.Calculate the equilibrium distance the charge pushes down the spring. Express your answer interms of Q, V , k, and d.Hint: Remember Hooke’s law for springs says that a force F will displace the spring a distanceF/k from equilibrium.dVQk−+Present your work neatly and clearly.Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 6. (20 points)A rod is bent into a quarter-circle of radius R. The rod carries a charge +Q uniformly distributedalong its length.Choose your coordinate system so that the center of the circle is at the coordinate origin andthe rod takes up the upper left quadrant. This is shown in the Figure:xy+QR(a) Calculate the electric potential V at the origin. Take the potential to be zero infinitely faraway. Express your answer in terms of Q, R, physical constants such as ²0, and numerical factors.(b) Suppose a small particle of mass m and charge q is released from rest at the center ofcurvature of the rod (that is, at the coordinate origin). When the particle is released, which waywill it initially move?(c) Calculate the speed of the particle after it has moved very far from the rod.You should be neat. You are more likely to get it right if you work neatly.Print your name:Physics 208: Electricity and Magnetism, Exam 1Problem 7. (10 points)Suppose there is some charge distribution that produced an electric potential that depends onlyon the x coordinate according to the formula:V (x) =14π²02q√a2+ x2where a is some given positive constant and q is a measure of the charge.Calculate the electric field. Express your answer in terms of q, a, ²0, numerical factors, and ofcourse, x.Working neatly will help you think about what you are doing.You may remove this sheet.If you do remove this sheet,DO NOT TURN IT IN!Derivatives:ddxaxn= an xn−1ddxsin ax = a cos axddxcos ax = −a sin axddxeax= aeaxddxln ax =1xIntegrals:Za xndx = axn+1n + 1Zdxx= ln xZsin ax dx = −1acos axZcos ax dx =1asin axZeaxdx =1aeaxZdx√a2− x2= arcsinxaZdx√x2+ a2= ln³√x2+ a2+ x´Zdxx2+ a2=1aarctanxaZdx(x2+ a2)3/2=1a2x√x2+ a2Zx