Aribtary Point - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Aribtary Point - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! Part I 1 Use coordinate system centered at the midpoint of the wires, y-axis parallel to the wires. An aribtary point (x, 0). Field due to wire A: ~BA = µ0I 4π ∮ d~l × r̂ r2 = µ0I 4π ∫ dy cos θ r2 = µ0I 4π ∫ (a− x) sec2 θ cos θdθ (a− x)2 sec2 θ = µ0I 2π(a− x) Similarly, field due to wire B ~BB = µ0I 2π(a + x) , so, Bz = BA + BB = µ0I 2π 2a (a− x)(a + x) 2 Let sphere 1 and sphere 2 have (variable) charge Q1 and Q2 respectively, and fixed radius R1 and R2 respectively. Sum of charges must be conserved, therefore Q0 = Q1 + Q2 is a constant. Potential energy of sphere 1,W1 = 1 2 ρV = Q2 8π0R1 sphere 2,W2 = Q22 8π0R2 E = W1 + W2 = 1 8π0 ( Q21 R1 + (Q0 −Q1)2 R2 ) = 1 8π0 [ (R1 + R2)Q 2 1 − 2R1Q0Q1 + R1Q20 R1R2 ] minimum energy, dE dQ1 = 1 8π0 [ 2(R2 + R1)Q1 − 2R1Q0 R1R2 ] = 0 Q1 = R1 R1 + R2 Q2 Q1 R1 = Q2 R2 Q1 4π0R1 = Q2 4π0R2 V1 = V2 3 The only relavant Maxwell equation is Gauss’ Law, ∇ · ~E = ρt 0∮ ~E · d~a = Q 0 1 For a stationary point charge, choose Gaussian surface of radius r centered at the point charge: ∫ EdA = Q 0 E = Q 4π0r2 Field always points away form point charge, ~E = Qr̂ 4π0r2 Boundary condition: ~E = 0 at r = ∞ 4 The general equation of capacitance for coaxial cables:- Gauss’ Law EA = Q  E · 2πrL = λL  , λ=charge per unit length E = λ 2πr potential difference, V = − ∫ Edr = ∫ b a λ 2πr dr = λ 2π ln b a capacitance formula, Q = CV λl = C λ 2π ln b a l=length of cable C = 2π ln b a l Now, the left part of the cable is empty, while the right part is filled with material of permittivity r. Therefore Cleft = 2π0 ln b a (L− x), Cright = 2π0r ln b a x Together, they are two parallel capacitors, so the total capacitance is the the sum C = 2π0 ln b a [L− x + rx] 2