Exercicio Pag 44 45 e 46 Livro Curso de Fisica Basica 3 Moyses Nussenzveig

Se for o ponto positivo d e y\n\\[ E = \\frac{a}{(a^2 + b^2)} y \\]\n(no ponto P, exato, não foi sã.\nEstatística: $E_d = \\frac{a}{(a^2 + b^2)}$)\n\n3.6\nConsiderando o eixo x, 1.3 m na direção y e 2.4 m\nCampo, reduzido para dq como na 3\nduvida igual ao de mais se rezar o\n\nCampo, partindo pelo lado.\nSe fiz na parte para o lado 2 e lá será\n\nCálculo E = k dq .\\frac{D^2}{(x^2 + d^2)}\nE = k * d.d^2.\\d^2_1 . \\lambda \\ d = dq = \\frac{d^2}{dx}\n\nSubstituindo no comportamento F = k.\\frac{dx}{d}\n\nE = k . dE = k . \\frac{d}{dx}\n\\left(\\frac{dE}{d.D}\\right) dE = KAD\nE = \\frac{(k.d.D + C)}{[D^2 . (A + C)]}\n\nE = \\frac{2kQ.2D}{D^2 + D^2}\n\n\\lambda = \\frac{ \\frac{2D}{\\pi (D)\\cdot 2A^2}}{(D + 2A)}. 3.7\n\\[ \\phi = E . A . \\cos \\theta - fluxo \\]\n\\[ \\phi = \\frac{(Q_{int})}{\\epsilon_0} - fluxo \\]\n\\[ A \\phi = \\frac{q_{in}}{A^2} \\theta \\cdot \\frac{PQ}{S} . dA)\\] ⟹ E = k . \\frac{A^2}{\\pi . A^2} \n\n3.11\n\\[ div(\\mathbf{C} \\times \\mathbf{r}) = \\mathbf{r}\\cdot \\mathbf{c} + \\mathbf{c}\\cdot \\mathbf{C} \\]\n\\[ E = \\sum_{i,j,k} div(p_x.xY + q_y.yZ + r_z.zX) = E \\]\n\\[ + div(\\mathbf{C}) = \\mathbf{E} \\]\\[ . div(\\mathbf{r}) * K \\] 3.12\n(a) \\[ \\phi = E.A, \\quad 0 < r < a \\]\n\\[ A = \\pi r^2 \\quad \\rho_0 = \\frac{3}{4\\pi a^2} \\quad \\rho_p = \\frac{4}{3}\\pi a^3\\] \n(b) \\[ a < r < b \\]\n\\[ E = \\frac{\\rho_0 Z}{3.0} \\quad \\text{tela} \\] \n(c) \\[ b < r < c \\quad \\text{tela do buraco} \\]\n\\[ E = \\frac{\\rho_0 G}{\\epsilon_0 (4\\pi r^2)(a^2-b^2)} \\] \n\\[ A = 4\\pi r^2 \\] 3.13 (0 < r < ∞)\nA) \\( \\Phi_1 = \\int \\rho(r) dV = \\rho(r) = \\rho_0 \\exp(-\\frac{r}{a}) \\; dv = \\int \\sqrt{r^2} dr \\)\n\\( = \\int_0^{\\infty} \\left( \\frac{r^2}{a^3} \\int \\frac{(r^2)^2}{4\\pi r_0^2 \\rho_0} \\left( \\frac{r^4}{r^6} \\right) dr \\right) d\\rho \\)\n\\( = \\int_0^{L} ( -ac^{-1} (\\rho_2 + \\rho_1 \\; r\\); L) \\cdots \\\n\\lim_{r \\to \\infty} \\left( \\frac{L}{L_2} \\right)= \\cdots \\\n\\text{(Isso é um exercício a mais.)}\n\nB) \\( \\Phi_2 = \\frac{\\rho_0}{\\epsilon_0} e^{{-\\frac{\\rho}{\\rho}(r^2)^{2/3}}} (\\rho^1) A \\)\n\\( = \\int ( \\int_0^{L} r^2 r \\left(\\rho^2 + \\alpha + r^2 \\right )) \\cdots \\\n\\cdots \\\n\\text{(E1)} \n\n\\(E = \\frac{-\\rho A^3}{\\epsilon_0 \\sqrt{3}} (r^2 = 1 + \\cdots)\n\\cdots \\\n\\int \\left( \\frac{(r^2)}{a^2} +\\frac{L \\int \\sqrt{r}} {a} \\right) dr\\)\n