Texto de pré-visualização
1ª Lista, Física 3 Prof. Mário fe. 2017-1 ① a) \vec{r_1} = x \hat{x} \, \vec{r_2} = x \hat{x} \vec{r_1} - \vec{r_2} = (x-d/2) \hat{x} \vec{r_{01}} = -d/2 \hat{x} \ + \frac{1}{| \vec{r_1} - \vec{r_2}|} = x-d/2 \vec{F_0} = F_{0(z)} + F_{0(e)} \vec{F_0} = \frac{1}{4 \pi \varepsilon_0} \frac{q_0}{(x-d/2)^2 \hat{x} + \frac{1}{4 \pi \varepsilon_0} \frac{q (-q)}{(x+d/2)^3} \vec{F_0} = \frac{q q_2}{4 \pi \varepsilon_0} \frac{x-d/2}{|x-d/2|} - \frac{(x+d/2) \hat{x}}{|x+d/2|^3} \text{b) } |x| >> d/2 \, x < 0 |x-d/2| \approx |x| , |x+d/2| \approx |x| \vec{F_0} = \frac{q_0 q}{4 \pi \varepsilon_0} \frac{-d/2 \hat{x} - d/2}{|x|^3} \vec{F_0} = -\frac{qqd}{4 \pi \varepsilon_0} \frac{\hat{x}}{|x|^3}, \ p = q d \hat{x} \text{ momento dipolar } \vec{F_0} = \frac{q_0 \hat{p}}{4 \pi \varepsilon_0 |x|^3} ② \frac{q (0,d/2)}{q (0,-d/2)} \, = \frac{-q (\vec{x}_2 + d/2)}{\vec{x}_2 - d/2)} \vec{F_2(11)} = \frac{q}{4\pi \varepsilon_0} \left( \frac{q \hat{x} - \vec{x} + d \hat{x}}{|x|^3} - \frac{q \hat{x} \hat{x}}{|x|^3} \right) \vec{E_{21a}} = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q }{|x|^3} - \frac{q \, \hat{x} \hat{x} - d \hat{x}}{|\vec{x}^2+d^2|^3} + q g\frac{\hat{x}\hat{x}}{|x^2+d^2|^3} \right) \vec{F_2(11)} = \frac{q }{4 \pi \varepsilon_0} \left( \frac{x \hat{x} - c \hat{x}}{|x|^4} - 1 \right) \ast \frac{1}{x^3} - \frac{1}{(x^2+d^2)^{3/2}} = \frac{1 - (-d/2 + 1/x^3)}{|x-d/2} = \frac{1 + d/2}{5} x^{-3} ③ \tilde{ \approx -\frac{3}{2 x^5}} \, \frac{-3}{x^5} \vec{F_2(1)} = \frac{\bar{q}}{4 \pi \varepsilon_0} \left[ \left(\frac{-3}{x^5} \bigg/ \left[ \ \right) \right] = \frac{q d q \hat{x}}{|4 \pi \varepsilon_0 |x^4 | \text{ } \, \frac{\bar{q}}{4 \pi \varepsilon_0} \text{p} = d q \text{p}' = d q' \begin{center} \end{center} \vec{F} = 0 \hat{0} + \vec{F_\Delta} \hat{y} \text{força resultante } \text{todas em y} \sum \text{F_y } = 2 |\sum | \text{ \ } \text{em cada par}, \end{array} cosθ = x/a sinθ = y/a a/y = sinφ x = acosθ r = 0 d = (a − x) \hat{x} F = (q)da(−\hat{x}) / (4πε₀|r|²)³/² da = adθ −x x^ = acosθ\hat{x} + asinθ\hat{y} |x| = a x = ∫₀^π a dθ a = ∫₀^π λ a dθ F = (−q)∫₀^π λ a (a(sinθ\hat{x} − acosθ\hat{y})) / (4πε₀ a³) F = (qλ a²) / 4πε₀ a² [(cosθ + sinθ \hat{x})]₀^{π/₁₀} F = qλ / (4πε₀)∫₀^π 0 − dθ − (−1 − 1) \hat{y} F = qλ / (4πε₀ a) = qλ / 2πa (\hat{y}) F = q_a Q) / (9πε₀ a²) Integrals: ∬ (1 - cos2φ) cosθ dθ dφ = (2 - sin2φ), (π/2 | 0), π/2 - π/4 ... 0 F = -q0 * λo / (4πεo a) * πŝ, = -q0 * λo / (2εo a) ŝ, (6) 0 ≤ θ ≤ π, 0 ≤ φ ≤ π/2, ṝ = a * cosθ x̂ + a * sinθ ŷ, Ϭ = a * cosφ ẑ - aŷ, |r̂| = a, F = q0 * da (-ŷ), (4πεo |r̂|^3) como d << r |rc - rd| ≈ rc |r - rc| ≈ r E̲ = q/4πε₀ [(d/r ẑ + x̂) / r³ - (d/r ẑ + x̂)] / r³ E̲ = qd/4πε₀ (x̂/d ẑ/r ̂/r,ẑ/r) E̲ = (-q/d ẑ/r³) – p/4πε₀r³ 1) Cuando y = 0 ? 2) Cuando x + y = 0 ? r² = x² + y², x³ẑ E̲ = [q/4πε₀ (x xₐ / x x̂ + (3₋ l(x ӝ))/2) – q/4πε₀ ] x/x xₐ/x̉ₖ 1) E̲ = q/4πε₀ [-l ẑ/x² + y²]⁽³/₂⁾ 2) E̲ = q/4πε₀ [(-3₋l₁/x̂²) – q/4πε₀ (3/l x/²)] E̲ = q/l ẑ/4πε₀r³
Texto de pré-visualização
1ª Lista, Física 3 Prof. Mário fe. 2017-1 ① a) \vec{r_1} = x \hat{x} \, \vec{r_2} = x \hat{x} \vec{r_1} - \vec{r_2} = (x-d/2) \hat{x} \vec{r_{01}} = -d/2 \hat{x} \ + \frac{1}{| \vec{r_1} - \vec{r_2}|} = x-d/2 \vec{F_0} = F_{0(z)} + F_{0(e)} \vec{F_0} = \frac{1}{4 \pi \varepsilon_0} \frac{q_0}{(x-d/2)^2 \hat{x} + \frac{1}{4 \pi \varepsilon_0} \frac{q (-q)}{(x+d/2)^3} \vec{F_0} = \frac{q q_2}{4 \pi \varepsilon_0} \frac{x-d/2}{|x-d/2|} - \frac{(x+d/2) \hat{x}}{|x+d/2|^3} \text{b) } |x| >> d/2 \, x < 0 |x-d/2| \approx |x| , |x+d/2| \approx |x| \vec{F_0} = \frac{q_0 q}{4 \pi \varepsilon_0} \frac{-d/2 \hat{x} - d/2}{|x|^3} \vec{F_0} = -\frac{qqd}{4 \pi \varepsilon_0} \frac{\hat{x}}{|x|^3}, \ p = q d \hat{x} \text{ momento dipolar } \vec{F_0} = \frac{q_0 \hat{p}}{4 \pi \varepsilon_0 |x|^3} ② \frac{q (0,d/2)}{q (0,-d/2)} \, = \frac{-q (\vec{x}_2 + d/2)}{\vec{x}_2 - d/2)} \vec{F_2(11)} = \frac{q}{4\pi \varepsilon_0} \left( \frac{q \hat{x} - \vec{x} + d \hat{x}}{|x|^3} - \frac{q \hat{x} \hat{x}}{|x|^3} \right) \vec{E_{21a}} = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q }{|x|^3} - \frac{q \, \hat{x} \hat{x} - d \hat{x}}{|\vec{x}^2+d^2|^3} + q g\frac{\hat{x}\hat{x}}{|x^2+d^2|^3} \right) \vec{F_2(11)} = \frac{q }{4 \pi \varepsilon_0} \left( \frac{x \hat{x} - c \hat{x}}{|x|^4} - 1 \right) \ast \frac{1}{x^3} - \frac{1}{(x^2+d^2)^{3/2}} = \frac{1 - (-d/2 + 1/x^3)}{|x-d/2} = \frac{1 + d/2}{5} x^{-3} ③ \tilde{ \approx -\frac{3}{2 x^5}} \, \frac{-3}{x^5} \vec{F_2(1)} = \frac{\bar{q}}{4 \pi \varepsilon_0} \left[ \left(\frac{-3}{x^5} \bigg/ \left[ \ \right) \right] = \frac{q d q \hat{x}}{|4 \pi \varepsilon_0 |x^4 | \text{ } \, \frac{\bar{q}}{4 \pi \varepsilon_0} \text{p} = d q \text{p}' = d q' \begin{center} \end{center} \vec{F} = 0 \hat{0} + \vec{F_\Delta} \hat{y} \text{força resultante } \text{todas em y} \sum \text{F_y } = 2 |\sum | \text{ \ } \text{em cada par}, \end{array} cosθ = x/a sinθ = y/a a/y = sinφ x = acosθ r = 0 d = (a − x) \hat{x} F = (q)da(−\hat{x}) / (4πε₀|r|²)³/² da = adθ −x x^ = acosθ\hat{x} + asinθ\hat{y} |x| = a x = ∫₀^π a dθ a = ∫₀^π λ a dθ F = (−q)∫₀^π λ a (a(sinθ\hat{x} − acosθ\hat{y})) / (4πε₀ a³) F = (qλ a²) / 4πε₀ a² [(cosθ + sinθ \hat{x})]₀^{π/₁₀} F = qλ / (4πε₀)∫₀^π 0 − dθ − (−1 − 1) \hat{y} F = qλ / (4πε₀ a) = qλ / 2πa (\hat{y}) F = q_a Q) / (9πε₀ a²) Integrals: ∬ (1 - cos2φ) cosθ dθ dφ = (2 - sin2φ), (π/2 | 0), π/2 - π/4 ... 0 F = -q0 * λo / (4πεo a) * πŝ, = -q0 * λo / (2εo a) ŝ, (6) 0 ≤ θ ≤ π, 0 ≤ φ ≤ π/2, ṝ = a * cosθ x̂ + a * sinθ ŷ, Ϭ = a * cosφ ẑ - aŷ, |r̂| = a, F = q0 * da (-ŷ), (4πεo |r̂|^3) como d << r |rc - rd| ≈ rc |r - rc| ≈ r E̲ = q/4πε₀ [(d/r ẑ + x̂) / r³ - (d/r ẑ + x̂)] / r³ E̲ = qd/4πε₀ (x̂/d ẑ/r ̂/r,ẑ/r) E̲ = (-q/d ẑ/r³) – p/4πε₀r³ 1) Cuando y = 0 ? 2) Cuando x + y = 0 ? r² = x² + y², x³ẑ E̲ = [q/4πε₀ (x xₐ / x x̂ + (3₋ l(x ӝ))/2) – q/4πε₀ ] x/x xₐ/x̉ₖ 1) E̲ = q/4πε₀ [-l ẑ/x² + y²]⁽³/₂⁾ 2) E̲ = q/4πε₀ [(-3₋l₁/x̂²) – q/4πε₀ (3/l x/²)] E̲ = q/l ẑ/4πε₀r³