IMP 13 First Hour Exam Solutions

Transcription

1 IMP 13 First Hour Exam Solutions 1. Two charges are arranged along the x-axis as shown, with q 1 = 4 µc at x = -4 m and q 2 = -9 µc at x = +2 m. (Remember 1 µc = 10-6 C). y * q 1 * q 2 x a. What is the force on q 1? F = 1 q 1 q 2 4πε o r i 2 = / (6) 2 i = 0.009N i b. Find the electric field (vector) at the origin. E = 1 q 1 4πε o q i = i = N / C i c. Where should a 3 µc charge be placed so that there is no force on it? Show all work. E = 0 at a distance x to the left of q 1 where q 1 x 2 = q 2 (x + 6) 2 or 4(x + 6) 2 = 9x 2 or 2(x + 6) = 3x or x = 12m

2 2. Suppose that I am standing on a mountain whose height, as a function of position in the xy plane, is given by h(x,y) = xy 2 + 3x - y. a. Determine the (xy) direction of the steepest downhill at the point where x=1 and y= -2. h(x,y) = (y 2 +3)i + (2xy-1)j Hence, direction of steepest downhill at (1,-2) = - h(1,-2) = - (7i -5j) = - 7i + 5j b. What is the slope (i.e., the rate of change of vertical with respect to horizontal) in the direction of the steepest downhill? rate of change of vertical with respect to horizontal in the direction of the steepest downhill = - h(1,-2) = - 7i - 5j = = - 74 c. If I start walking northeast (where the positive x direction is east and the positive y direction is north) from the point where x=1 and y= -2, what is my instantaneous rate of change of height with respect to horizontal distance traveled? We are asked to compute a certain directional derivative. A unit vector in the desired direction is u = 1 i + 1 j, and h(1,-2) = 7i -5j. Hence, the desired directional 2 2 derivative is given by D u h(1, 2) = u h(1, 2) = ( ) 1 2 i j = 2 2 = 2 (7i 5j) =

3 d. Notice that since h(1,-2) = 9, the point (1,-2,9) is on the surface. Find an equation for the plane that is tangent to the mountain at this point. The surface is given by z = xy 2 + 3x - y. Let ϕ(x,y,z) = xy 2 + 3x - y - z. Then, the surface is the level surface ϕ(x,y,z) = 0 and the gradient of this function at the point (1,-2,9) is a vector normal to the surface at this point. ϕ(x,y,z) = (y 2 +3)i + (2xy-1)j - k and therefore ϕ(1,-2,9) = 7i - 5j - k Hence, an equation for the tangent plane is given by 7(x-1) - 5(y+2) - (z-9) = 0 7x - 5y - z = 8 3. Two equal positive charges of magnitude 2q lie along the x-axis at the corners of an equilateral triangle of side L. a. Find the electric field at the third corner which lies on the y-axis. Write your answer in vector form as E r = E xˆ i + E yˆ j. E lies along y axis and is given by E 2q = 2 4πε o L 2 cos30 j == 3 2 q πε o L j 2 b. Find the electric potential at the third corner of the triangle. V = 1 4πε o 2q L 2 = q πε o L c. How much work is required to bring a charge Q from very far away and place it at the third corner of the triangle? W = QV = Qq πε o L

4 4. Consider the vector field E(x,y) = <3x 2 y+6x, x 3 +4>. a. Show that E is conservative by finding a scalar potential function for E. We wish to find ϕ so that ϕ = E. Thus, we wish to find ϕ so that the following two equalities hold: ϕ x = 3x2 y+6x ϕ y = x3 +4 The first of these statements implies that ϕ = x 3 y + 3x 2 + f(y). The second of these statements implies that ϕ = x 3 y + 4y + g(x). We observe that ϕ(x,y) = x 3 y + 3x 2 + 4y satisfies each of these requirements. Thus, ϕ is a potential function for E. b. Now suppose that E is the electric field due to some distribution of electric charge. Find V(x,y), the electric potential function. We know from part a that E = ϕ. We also know that E = - V. Hence, V(x,y) = -ϕ(x,y) = - x 3 y - 3x 2-4y. c. Use your work above to compute the work done by the electric field on a particle with charge q as the particle moves from (1,3) to (-2,1). Work = q ( 2,1) E ds = (1,3) q[ϕ( 2,1) ϕ(1,3)] = q[( ) ( )] = q(8 18) = 10q

5 5. a. Rank the magnitude of the electric field at points A, B, and C shown in the figure with greatest magnitude first. 1. A 2. B 3. C b. If you stretch a cylindrical wire and it remains cylindrical, does the resistance of the wire (measured end-to-end along its length) increase, decrease, or remain the same? Increase

6 6. Suppose that the electrical potential function in some region of the xy plane is given by V(x,y) = y+x 2. a. Draw the equipotential curve that contains the point (2,1). Label the point (2,1) in your sketch. V(2,1) = 5. Thus, we wish to draw the graph of y+x 2 = 5. b. Compute - V(2,1) and add this vector to your sketch from part a. (Draw this vector with initial point (2,1).) - V(x,y) = - (2xi + j) = -2xi - j. - V(2,1) = -4i - j. c. What is the geometric relationship between your equipotential curve from part a and your vector from part b? They are perpendicular. d. What is the physical significance of your answer to part b (i.e., to what physical quantity does your answer to part b correspond)? The vector in part b is the electrical field vector at the point (2,1).

7 7. A thin rod of length L and uniform charge per unit length λ lies on the y-axis as shown. y L dy y P * x x a. First, please write down an expression for the magnitude of the electric field at point P, a distance x from the center of the rod along the x-axis, due to a length dy of the charged rod at location y with a charge dq, in terms of x, y, λ, and dy de = 1 dq 4πε o (x 2 + y 2 ) = λdy 4πε o (x 2 + y 2 ) b. Explain in a sentence or two why the electric field at point P has no y-component. (Don t just use the word symmetry!) For every length dy at y, there will be a length dy at y such that the y component of E at P cancels. c. Now, please write a complete integral expression for the net (vector) electric field at point P in terms of x, y, λ, dy, and L but do not solve the integral. E = 1 4πε o L /2 L/2 λdy (x 2 + y 2 ) x x 2 + y 2 i = λx 4πε o L/2 L/2 dy (x 2 + y 2 ) 3/2 i d. Write a complete expression for the electric potential at point P due to the rod set it up in the same way as in parts (a) and (c) using the same variables and again do not solve the integral. V = 1 4πε o L/2 L /2 λdy x 2 + y 2

8 8. Match each of the following vector fields with its graph. vector field -.5xi -.5yj 2i - j.3xj -yi + xj graph C D B A A B C D

9 9. Consider the surface S consisting of the portion of the cone z = 4 2 x 2 + y 2 which is above the xy-plane. S is shown below. Compute the mass of S if its density is given by ρ(x,y,z)=3 x 2 + y 2 kg/m 2. We parameterize S as follows: x = u cos v y = u sin v z = 4 2u 0 u 2 0 v 2π (Note that the base of the cone is a circle, center at the origin, radius 2.) (Other parameterizations are possible.) Then, r(u,v) = u cosv i + u sinv j + (4-2u) k and r u x r v = i j k cos v sin v 2 usin v u cosv 0 = 2u cosv i +2u sin v j+(ucos 2 v + usin 2 v)k = 2u cosv i +2u sin v j+uk.

10 Hence, r u x r v = 2ucos vi +2usin v j+ uk = 4u 2 cos 2 v + 4u 2 sin 2 v + u 2 = 4u 2 + u 2 = 5u 2 = 5 u Finally, we compute the mass as follows: M = = ρ(x,y, z)ds S = R ρ(x(u, v),y(u, v),z(u,v)) r u x r v da v=2π u=2 (3u)( 5 u)dudv = v=0 v=2π u=0 u=2 3 5 u 2 dudv = v=0 v=2π u=0 u=2 dv 5 u 3 = v=0 u=0 v=2π 8 5dv = v=0 8 5 v v=0 v=2π = 16 5 π kg