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Engenharia Eletrônica ·
Eletromagnetismo
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Karina Xavier da Silva\nEngenharia Eletrônica\nFísica 3 - Lista 1\nCap 21\nF_AB = F_A - L_1 - L_2 - L_3\nL_0 = L_12\nqa/q2 = ?\nF = 0\nF2 = F1 + F2\n0 = F1 - F3\nF2 = - F3\nkqaqb² = -kqbqax²\n(x12)² - (x2)²\nq1² = q2²\n(q1 + q2)² x = 0\nq1 + q2 = 0\nq1 = -q2\nLogo q1 e q2 tem sinais opostos\ne |q1| = |4q1|. 21.8\nA = 40\nB = -60\nC = 0\nA e B Fixas\nA 49 -60\nExp. 1 A d B d - 69\nC 0\n-69 + 29 = -40\n2\nExp. 1 -\nExp 2 Rim\nFAB12 = k(8)/(39) - 3 kQ²/d²\nExp 1. Rim\nFAB1 = k(20)(-20) - 4 kQ²/d²\nFAB2 = -3/8\n= 0,375 21.9\nFAB = 0.193 N\nA - 50m B \nFixas \nForce\nF=0 O\nFabra = 0.036N m²km (sinus) C\nE = - 0.193 N d - dm (sinus.app)\nF = k * d² - b 0.193.(.0.7)² =\,\n-8.79.10⁻²\nF' = 0.036N m²km (sinus)\n0.036(0.5)²\, (qA + qB)² = 9.10⁻¹² 2 = k(qa + qB)\n= 8.79 10³\ng1 = 2.15 10⁶ - g0 = 0\nqB² - 2.10-9 = 3.15 10⁻¹₂ = 0\nD = 410 + 12.10⁻¹² = 16.10⁻¹² 21.20\nd = \nB A\n Fyb\n Fxa\n\ng/qb = v \nF = Aec Fg - BEA\n\nSistema 1\nθ = 150° Fx0 - Bθ0\n\nFtot = Fg - Fc = Fg (1 - Fc/Fg) = Fg (1 - x/4)\n\nFg = k * Qc/ d² = xQc Fg\n\nk = 9 * 10⁹\nF0 kb0 \nF = x 4/5 * 9.81\n\nComo Fxy = 2F0 temos: Usando Fg como referência de F0\n\n2F0 = F0 (1 - x/4) = 2 = 1 - x/4 ⇒ x = -4\n\nb\n\nqc/qb x\n\ncomo 2 ⇒ θ = 0° Fxy = 1,25F0\n\nFtot = Fg + Fc = F0 (1 + Fc/Fg)\n\n1,25F0 = F0(1 + x) \n\nx = +1\n 21.21 Cusa orgânica nos condutores // Dist. carga nos uníque.\n\nm₁ = 4 cm = 4 * 10⁻² m\nm₂ = 6 cm = 6 * 10⁻² m\nb = 8 * 10⁶ C/m²\n\nρ = b/m\n\nq = ?\n\nDizem-se nos suprimentos ρ = q/v\n\nD² = ρ bn² dV\nV = (4/3) π m³\n\nDv = (1/8) 𝑙⁴ m ² h = (1/4π) m² d\n\ndq = 4π b n dn\n\nPara calcular a carga total integramos:\n\n∫dq = ∫(4π b n² dn)\n\nq = (4π mb²/2) 6 * 10⁹ / 2 * 4 * 10⁵ =\n\nq = 4π 3 * 10⁶ [10] =\n\nq = 3,8 * 10⁸ C\n 22.22 θ=30°\nAd = 5 cm = 0.215 m\n\nb₂ = 5 * 10⁻⁹\n\nb₁ = − 1 * 10⁻⁹\n\nD = ?\nF = 0\n\nEI = 0\n\nFxy = F3 cos θ + F1 cos θ - F2 = 0\n\nFxy = k1 q1 cos θ + k2 q2 cos θ - kq₃ q₁\n\n1/(d²) + 1/(d²) = 1/(d + D)²\n\nk1 [ d2 cos θ + q1 cos θ - q2 = 0]\n\n(2/4) * 4 = 0\n\ncos²θ[ q₁ + q₂ ]= q2 \n\n1/(d²) + 1/(2d) + d² = 0\n\ncos²(2d)(16 * 10⁻⁹ + 16 * 10⁻⁹) = 8 * 10⁻⁹\n\n4/10⁴ + 4/(10⁵) + D² 0.65 [ 3.2 10^9 ] = 8 10^10\n\n4.10^4\n\n\n\n\n\n\n0.532 10^5 = 5.5 10^9\n\n4.10^4 4.10^4 + d^2\n\n4.10^4 4.10^4 + d^2 = 15.39 10^\n\nD^2 4.10^4 - 11.39 10^6 = 0\n\nD = 1610^4 9556 10^\n\nD = -4.10^2 + 7.85 10^2\n\nD^2 = 1.92 10^{-2} m\nD^ = \\Theta n cos(theta)\n\nD = 1.92\n 17/8 e\nF_total = 1.5 10^5 N, x - \\infty\n\nq_0 = + 8e\ny_s = 0.3 m\n\nq_2 = ? en unidades de e incluyendo o sincl\n\ny\ny\n\ny^' y^'' x\n\nF_z = F_{F_1} - F_{F_2}\n\n0 = k q_2 q_3 + k q_1 q_2\n\n(d_1^2 d_2^2)\n\n0 = d_3^2 + g \n\n0 = q_2 + q_1\n\nPara gran tenemos que\n\n8e = |q_1| = |q_3| log(0)\\\n|q_1| = -8 e\n\nF_z = k a_1 a_2 + k a_1 a_2\n\nF_2 = k a_1 a_2 -> q_2 = F_d^2 = 1.5 10^5 a_2\nd_{x} = q_1\n\nq_2 = -13 e\n|q_1| = 13 e\n q_1 = + 4e\nd_1 = 2 mm - 0.10^5 m\nq_2 = + 6e\nd_2 = 6 mm - 0.10^5 m\n\n---\n\nF_{1x} = ?\n\nF - -\n\nF_z - u_z\nd_d = u - 2 =\n\nd_1^2 = d_2^2 + d_2^2 -> d = \\sqrt{d_1^2 + d_2^2}\n\nF_{1z} = 5.52 10^{-2} = 1.38 10^2 N\n6.10^6\n\nF = F_z\nd = ...\n\nF_x = 1.98 10^{-2} 6.10^5 = 8.28 10^5\n\nF_x = 1.31 10^{-2} N 21.92\n\n l = L sin θ\n\ng\n\nT.cos θ = mg\nT.sin θ = F\n\nF = 1 / (4πϵ₀x²) → T.sin θ = 1 / (4πϵ₀x²)\n\nT = √(mg) ↔ tan θ = q² / (4πϵ₀x²)\n\nT.sin θ = 1 / (4πϵ₀x²)\n\nT.cos θ / mg\n\nx = q²\n\n2L / (4πϵ₀x²mg) → x³ = 2L.q² / (4πϵ₀mg)\n\nlog x = (L.q² / (2πϵ₀mg))^(1/3)
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Karina Xavier da Silva\nEngenharia Eletrônica\nFísica 3 - Lista 1\nCap 21\nF_AB = F_A - L_1 - L_2 - L_3\nL_0 = L_12\nqa/q2 = ?\nF = 0\nF2 = F1 + F2\n0 = F1 - F3\nF2 = - F3\nkqaqb² = -kqbqax²\n(x12)² - (x2)²\nq1² = q2²\n(q1 + q2)² x = 0\nq1 + q2 = 0\nq1 = -q2\nLogo q1 e q2 tem sinais opostos\ne |q1| = |4q1|. 21.8\nA = 40\nB = -60\nC = 0\nA e B Fixas\nA 49 -60\nExp. 1 A d B d - 69\nC 0\n-69 + 29 = -40\n2\nExp. 1 -\nExp 2 Rim\nFAB12 = k(8)/(39) - 3 kQ²/d²\nExp 1. Rim\nFAB1 = k(20)(-20) - 4 kQ²/d²\nFAB2 = -3/8\n= 0,375 21.9\nFAB = 0.193 N\nA - 50m B \nFixas \nForce\nF=0 O\nFabra = 0.036N m²km (sinus) C\nE = - 0.193 N d - dm (sinus.app)\nF = k * d² - b 0.193.(.0.7)² =\,\n-8.79.10⁻²\nF' = 0.036N m²km (sinus)\n0.036(0.5)²\, (qA + qB)² = 9.10⁻¹² 2 = k(qa + qB)\n= 8.79 10³\ng1 = 2.15 10⁶ - g0 = 0\nqB² - 2.10-9 = 3.15 10⁻¹₂ = 0\nD = 410 + 12.10⁻¹² = 16.10⁻¹² 21.20\nd = \nB A\n Fyb\n Fxa\n\ng/qb = v \nF = Aec Fg - BEA\n\nSistema 1\nθ = 150° Fx0 - Bθ0\n\nFtot = Fg - Fc = Fg (1 - Fc/Fg) = Fg (1 - x/4)\n\nFg = k * Qc/ d² = xQc Fg\n\nk = 9 * 10⁹\nF0 kb0 \nF = x 4/5 * 9.81\n\nComo Fxy = 2F0 temos: Usando Fg como referência de F0\n\n2F0 = F0 (1 - x/4) = 2 = 1 - x/4 ⇒ x = -4\n\nb\n\nqc/qb x\n\ncomo 2 ⇒ θ = 0° Fxy = 1,25F0\n\nFtot = Fg + Fc = F0 (1 + Fc/Fg)\n\n1,25F0 = F0(1 + x) \n\nx = +1\n 21.21 Cusa orgânica nos condutores // Dist. carga nos uníque.\n\nm₁ = 4 cm = 4 * 10⁻² m\nm₂ = 6 cm = 6 * 10⁻² m\nb = 8 * 10⁶ C/m²\n\nρ = b/m\n\nq = ?\n\nDizem-se nos suprimentos ρ = q/v\n\nD² = ρ bn² dV\nV = (4/3) π m³\n\nDv = (1/8) 𝑙⁴ m ² h = (1/4π) m² d\n\ndq = 4π b n dn\n\nPara calcular a carga total integramos:\n\n∫dq = ∫(4π b n² dn)\n\nq = (4π mb²/2) 6 * 10⁹ / 2 * 4 * 10⁵ =\n\nq = 4π 3 * 10⁶ [10] =\n\nq = 3,8 * 10⁸ C\n 22.22 θ=30°\nAd = 5 cm = 0.215 m\n\nb₂ = 5 * 10⁻⁹\n\nb₁ = − 1 * 10⁻⁹\n\nD = ?\nF = 0\n\nEI = 0\n\nFxy = F3 cos θ + F1 cos θ - F2 = 0\n\nFxy = k1 q1 cos θ + k2 q2 cos θ - kq₃ q₁\n\n1/(d²) + 1/(d²) = 1/(d + D)²\n\nk1 [ d2 cos θ + q1 cos θ - q2 = 0]\n\n(2/4) * 4 = 0\n\ncos²θ[ q₁ + q₂ ]= q2 \n\n1/(d²) + 1/(2d) + d² = 0\n\ncos²(2d)(16 * 10⁻⁹ + 16 * 10⁻⁹) = 8 * 10⁻⁹\n\n4/10⁴ + 4/(10⁵) + D² 0.65 [ 3.2 10^9 ] = 8 10^10\n\n4.10^4\n\n\n\n\n\n\n0.532 10^5 = 5.5 10^9\n\n4.10^4 4.10^4 + d^2\n\n4.10^4 4.10^4 + d^2 = 15.39 10^\n\nD^2 4.10^4 - 11.39 10^6 = 0\n\nD = 1610^4 9556 10^\n\nD = -4.10^2 + 7.85 10^2\n\nD^2 = 1.92 10^{-2} m\nD^ = \\Theta n cos(theta)\n\nD = 1.92\n 17/8 e\nF_total = 1.5 10^5 N, x - \\infty\n\nq_0 = + 8e\ny_s = 0.3 m\n\nq_2 = ? en unidades de e incluyendo o sincl\n\ny\ny\n\ny^' y^'' x\n\nF_z = F_{F_1} - F_{F_2}\n\n0 = k q_2 q_3 + k q_1 q_2\n\n(d_1^2 d_2^2)\n\n0 = d_3^2 + g \n\n0 = q_2 + q_1\n\nPara gran tenemos que\n\n8e = |q_1| = |q_3| log(0)\\\n|q_1| = -8 e\n\nF_z = k a_1 a_2 + k a_1 a_2\n\nF_2 = k a_1 a_2 -> q_2 = F_d^2 = 1.5 10^5 a_2\nd_{x} = q_1\n\nq_2 = -13 e\n|q_1| = 13 e\n q_1 = + 4e\nd_1 = 2 mm - 0.10^5 m\nq_2 = + 6e\nd_2 = 6 mm - 0.10^5 m\n\n---\n\nF_{1x} = ?\n\nF - -\n\nF_z - u_z\nd_d = u - 2 =\n\nd_1^2 = d_2^2 + d_2^2 -> d = \\sqrt{d_1^2 + d_2^2}\n\nF_{1z} = 5.52 10^{-2} = 1.38 10^2 N\n6.10^6\n\nF = F_z\nd = ...\n\nF_x = 1.98 10^{-2} 6.10^5 = 8.28 10^5\n\nF_x = 1.31 10^{-2} N 21.92\n\n l = L sin θ\n\ng\n\nT.cos θ = mg\nT.sin θ = F\n\nF = 1 / (4πϵ₀x²) → T.sin θ = 1 / (4πϵ₀x²)\n\nT = √(mg) ↔ tan θ = q² / (4πϵ₀x²)\n\nT.sin θ = 1 / (4πϵ₀x²)\n\nT.cos θ / mg\n\nx = q²\n\n2L / (4πϵ₀x²mg) → x³ = 2L.q² / (4πϵ₀mg)\n\nlog x = (L.q² / (2πϵ₀mg))^(1/3)