exam 2 fall 2009 solution
20 Pages
exam 2 fall 2009 solution

Course Number: ECE 2317, Fall 2013

College/University: U. Houston

Word Count: 2948

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on problem: This problem actually has a mathematical inconsistency in it, in that we dont get the same result if we use Stokess theorem or if we calculate the integral directly by integrating on the circle (which is much harder to do). This is because the function V blows up on the z axis, and hence violates the assumptions inherent in Stokess theorem. 13 Problem 3 (30 pts.) A power line has a radius a, and the...

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problem: on This problem actually has a mathematical inconsistency in it, in that we dont get the same result if we use Stokess theorem or if we calculate the integral directly by integrating on the circle (which is much harder to do). This is because the function V blows up on the z axis, and hence violates the assumptions inherent in Stokess theorem. 13 Problem 3 (30 pts.) A power line has a radius a, and the center of the line is at a height h above the ground, which may be modeled as a perfect conductor. The air surrounding the line has a critical breakdown field strength that is denoted as Ec. (a) Find an expression for the line charge density l0 on the line when the air at the surface of the line is at the threshold of breaking down. (b) Find an expression for the maximum voltage Vmax that can be placed on the line (assuming that the earth is at zero volts) before the air breaks down at the surface of the line and a corona forms. (Feel free to use any formula that you find on the formula sheet to help you do this calculation as simply as possible). (c) Find an expression for the electric field vector at a point directly below the line (at x = 0), at a height d above the surface of the earth, when the maximum voltage Vmax is on the line. y 2a h x Earth 14 ROOM FOR WORK The image picture looks like this: y z l0 h original line q r h d x x h h -l0 image line Part (a) Breakdown will occur on the line at the bottom surface (y = h a). Along the vertical dashed line we have 1 1 E = y l 0 + . 2 0 h y h + y We set E ( h a ) = Ec . This gives us l 0 1 1 + = Ec . 2 0 a 2h a 15 We then have l 0 = 2 0 Ec 1 1 + a 2h a [C / m] . Part the (b) Using formula on the formula sheet, we have l0 2 0 V0 2h ln . a Hence, 2h Vmax = l0 ln . 2 0 a Using the answer from part (a), we then have Vmax = 2h ln [ V ] 1 a 1 + a 2h a Ec Part (c) Using the formula from part (a), and setting y = d, we have 1 1 E = y l 0 + [ V / m] . 2 0 h d h + d Hence, we have 1 1 hd + h+d E = y Ec 1+ 1 a 2h a [ V / m] . 16 17 Problem 4 (20 pts.) A capacitor consists of two metal spheres in free space that are separated by a distance h between their centers, as shown below. Each sphere has a radius a. You may assume that the centers of the spheres lie on the z axis, at z = h / 2 and z = -h / 2. Derive an approximate formula for the capacitance of this system. Assume that the spheres are far enough apart that the charge density on each sphere may be assumed to be constant. z h x 18 ROOM FOR WORK The capacitance formula is C= Q V Lets assume a charge +Q on the top sphere and a charge Q on the bottom sphere. The voltage V then means VAB, where A is a point on the top sphere and B is a point on the bottom sphere. For convenience, we choose the points located on the spheres along the line x = 0. We then have ( h /2 a ) V= Ez dz . ( h /2 a ) Along the line x = 0, the electric field is given by Q 1 1 E = z + 2 2. ( h / 2 + z) 4 0 ( h / 2 z ) Hence, we have ( h /2 a ) Q V = 4 0 ( h /2 a ) 1 1 dz + ( h / 2 z) 2 ( h / 2 + z) 2 . Integrating, we have Q 1 1 V = ( h / 2 z) ( h / 2 + z) 4 0 ( h /2 a ) . ( h /2 a ) Hence, Q 1 1 1 1 Q 1 1 V = + = . 4 0 a h a h a a 2 0 a h a We then have 19 C= 2 0 1 1 a ha [F] 20

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