Tarefa 2 Física 3 Pdf

Maria Lucia Pinto 126562, Turno D \nEg = Ez = 0, Ex = 4 N/C * x \nU = \\int E F dx \nU = \\int 4x dx = 4/2 x² |^{4}_{0}, U = 32 \nYA - VB = 32 √5 \nVB - VA = -32 √15 \n\n dv = xdq = K(d)dx \nV = \\int_{0}^{t} \\frac{K}{\\sqrt{x²+y²}} \ dx \nV = KC \\frac{\\sqrt{x²+y²}}{r} = \\sqrt{\\frac{V_{x}^2+y^2}} V = (10^{-7}m)^{2}(0.356m)^{2} + (0.356m)^{2} = 0.356m \nV = 8.99 * 10^{9} 499i^{2} → V = 3.16 * 10^{-2} \n\ney - dy_p = -C d ( \\sqrt{(x^2+y^2)} ) = C (1-y) \nye = 0.298 N/C $\\Phi_{32} = 6 - 10i dA - 2xi dA + 0 \n\\Phi_{33} = 6 - 1 dA - 6zidA \n\\Phi_{32} = -1 dA + 2xidA \n\\Phi_{y32} = 10 dA - .8A \n\\Phi_{y32} = -10(1-0)(3-1) - 2(1-0)(3-1) \n\\Phi_{y32} = -10(2-2) -2 = 20+4 \n\\Phi_{y32} = 24 N.m²/C \n\\Phi_{y3} = \\Phi_{y31} + \\Phi_{y32} = 36 - 24, \\Phi_{y3} = 12N*m²/C Flujo em x y\n-dA\n\ndxy1 = h f . dA = Sf [ (1 + 2) - 2 ]\n\ndxy2 = -6 b dA\n\ndxy1 = 3b dA = 36 dA ( 4 - 1 )\ndyy1 = 96\n\ndyy1 = 36\n\ndxy2 = -3b\n\ndxy2 = g b 3 k ^ k [ dA ] = b f - 1 dA\n\ndxy2 = - 6 [ (4 - 1)( 1 - 0 )]\ndyy1 = - 3b , dxy1 = 6b\n\ndxy4 + dxy2\n\n\ndyy4 = 96 - 36 , dxy = 6b\n\nd_t = dxy3 + 5 x2 + dxy\n\n\nd_t = 12 + 0 | 6 b 12 + 6b\n\nd_t = Guna , tanno.\n(d_t1) E = 9 b\n=> 12 + 6b) E = 24 E + 6b = 24 - 12\n=> b = 2N / (c . m)