Exam1 - Mid-Term Exam: Graduate Electromagnetics I, 95.657...

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Mid-Term Exam: Graduate Electromagnetics I, 95.657 Fall 2008, UMass Lowell, Dr. Baird This in-class examination lasts 80 minutes. No calculators, electronic devices, books or notes allowed. The later problems become increasingly harder and are correspondingly worth more points. It is in your interest to leave more time for the later problems than for the first few. PROBLEM 1. Consider two conducting planes that are joined at one end to form a deep corner. (a) What is the general behavior of the surface charge distribution for varying corner angles and why? (Explain conceptually, no numbers required.) SOLUTION: Near the corner, the charge on one plane is very close to the charge on the other plane and they experience a strong repulsion. Farther from the corner, the charges on the plane are farther apart and do not exert as much force. As a result, most of the charge is repulsed away from the corner. As the corner angle becomes smaller, the charge is more strongly repulsed away from the corner. (b) As the corner angle becomes zero, what is the behavior of the surface-charge density, the electric field and the electrostatic potential energy density inside the deep corner? SOLUTION: As the corner angle becomes zero, the surface-charge becomes completely repulsed out of the deep corner. There is no charge and therefore no electric field in the deep corner. Because of the definition, w = ( ε 0 /2)| E | 2 , the electrostatic potential energy density also becomes zero. Physically, when the corner angle is zero, the deep corner is gone and there is only a solid mass of conducting material. This is one way to show that the charges, fields, and energy density in a conductor are zero. (c) Considering the previous parts, plot the surface charge density and electric field lines corresponding to the top of the periodic conducting object shown here. This object extends uniformly out of the page. SOLUTION: . . . . . . σ σ = 0 . . . . . . E
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PROBLEM 2. Two concentric spherical shells are centered at the origin. The inner sphere has radius a and the outer sphere has radius b so that b > a . The inner sphere is held at zero potential, and the outer sphere is held at a potential V  ,  . What is the potential everywhere between the two spheres? SOLUTION: We use the general solution to the Laplace equation in spherical coordinates when a valid solution is required on the full azimuthal sweep and full polar sweep:  r , , = l = 0 m =− l l A l , m r l B l ,m r l 1 Y lm  ,  We cannot eliminate A l or B l because we do not seek a valid solution at the origin or at infinity. Apply the boundary condition  r = a = 0 0 = l = 0 m =− l l A l ,m a l B l , m a l 1 Y lm  ,  This must hold for all values of the independent variables so that the the coefficient of each sum must vanish separately.
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Exam1 - Mid-Term Exam: Graduate Electromagnetics I, 95.657...

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