ECE341 Test 2 Your Name: Thu 4/12/2018

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5 Problem 2 One surface of an infinitely large ideal conductor plate is at the plane = 0 of the Cartesian coordinate system, as shown in the figure. The conductor plate is grounded (i.e. at a potential 0). A positive point charge Q is located at (d, 0, 0). Assuming free space (i.e. vacuum), find the following: (1) The electric field and potential anywhere with < 0, and (2) The electric field and potential at point (d, 2d, 0). Note: The filed is a vector while the potential is a scalar. You may epress a field in the form of E ˆ yˆ zˆ E y Ez. (3) The electric field and potential anywhere on the y ais. (4) (Optional; for bonus) The electric field and potential anywhere in the plane = 0. Note: For convenience, you may define your own polar coordinate in this plane. Solutions: (1) E = 0 and V = 0 for < 0. (2) Use the image method: Image charge at ( d, 0, 0). The field due to charge Q is The field due to the image is E2: Let

6 The total field is

7 Problem 3 (1) A uniform, two-dimensional (meaning zero thickness), infinitely large charge sheet is at the plane = 0. The surface charge density is σ. Epress the field E() and potential V() for all. Assuming σ is positive, schematically plot E() and V(). Note: Epress your results in the notations defined here. Copying equations from elsewhere without adapting to the notations defined here will not earn full credit. Same for below. (2) Two uniform, two-dimensional, infinitely large charge sheets are at = 0 and = d, with surface charge densities +σ and σ, respectively. Epress the field E() and potential V() for all. Assuming σ is positive, schematically plot E() and V(). E E d V V (1) (2) (1) E( ) sgn( ) ˆ. sgn. 2 0 (2) For < 0, E() = 0 and V() = 0. For 0 < < d, and. For > 0, E() = 0 and.

8 (3) As shown in the left picture below, the si two-dimensional, infinitely large charge sheets are parallel to each other, with the surface charge density of each sheet labeled. Assuming σ is positive, graphically depict the field E() and potential V() distributions. (4) Same as (3), but with surface charge densities shown in the right picture below /2 + + / E E V V Special case where all sheets are equally spaced: V (3) (4)

9 Problem 4 An infinitely large conductor sheet is at the z = 0 plane, carrying a surface current density J Jˆ, as shown in Fig. (1). (1) What is the unit of J? Find the magnetic field distribution B(z) in the entire space (for z > 0 and z < 0). Assuming J > 0, draw the magnetic field lines (indicating directions with arrows) in Fig. (1). (2) Now, another conductor sheet, carrying a surface current density J Jˆ, is placed at z = d, as shown in Fig. (2). Find the magnetic field distribution B(z) in the entire space. Assuming J > 0, draw the magnetic field lines (indicating directions with arrows) in Fig. (2). (3) Do the two sheets in Fig. (2) attract or repel each other? What is the force per unit area between the two sheets? z (1) (2) z l y y Solution: (1) The unit of J is A/m. Draw a rectangular loop with length l and width w (which is infinitesimally small) as shown. 2lB = Jl. Therefore, B = J/2. Directions shown in figure. (2) For z < 0 and z > d, B = 0. For 0 < z < d, B = J. Direction shown in Figure. (3) They repel each other. The upper sheet is in the field B = J/2, with the direction shown by red arrows, due to the lower sheet. For a rectangle in the upper sheet, with a width (perpendicular to J) w, and a length (along the direction of J) l, the force is F = IlB = JwlB = Jwl J/2 = J 2 wl/2. Therefore, the force per area is F = J 2 /2

10 Problem 5 A metal sample, with a length L, a width W, and a thickness d, is in a uniform magnetic field B = B. The electron density is n. A current I flows through it. The electrons can be regarded as moving at an average drift velocity v = v. The current is epressed as I = qnvwd, where q is the elementary charge, taken as a positive value. Since I nv, any electrical measurement alone cannot independently determine n and v. (1) A voltage, called the Hall voltage VH, is measured between the two sides of the sample, as shown in the figure. Epress VH in terms of v, B and sample dimensions. Assuming positive values for B and I, determine the polarity of VH (label the two terminal with + and signs in the figure), with the directions of B and I shown in the figure. (2) Show how to determine n from this eperiment (i.e., develop an epression of n in terms of the measured quantity VH and known parameters I, B and sample dimensions). z B L + V H I v F d y W Solution: (1) The B field eerts a force on the moving electron: Electrons pushed by this force accumulate on one side of the sample as shown, leaving positive charge on the opposite side, establishing an electric field E. At the steady state, E = vb. Polarity is shown in the Figure. (2) From (1), we have v = VH / (BW). From I = qnvwd, we get n = I / (qvwd)