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AL 05 PHY MC

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請幫幫忙 05 Q11 why the angle i should be less than a certain value?05 Q12P,Q and R are small identical metal spheres. P and Q are fixed at a certain separation in vacuum and they carry charges of the same magnitude. the attractive force between them is F. sphere R is initially uncharged. it frist touches P and... 顯示更多 請幫幫忙 05 Q11 why the angle i should be less than a certain value? 05 Q12 P,Q and R are small identical metal spheres. P and Q are fixed at a certain separation in vacuum and they carry charges of the same magnitude. the attractive force between them is F. sphere R is initially uncharged. it frist touches P and then it touches Q. what is the electrostatic force between P and Q after R is taken away? 05 Q15 why ans is 28*10^-6C 更新: Q(15) When K is closed . why 1mA? and 6V & 0V?

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(11) First of all, for total internal reflection to occur, it is a must that n1 > n2. Then, let r be the angle of refraction at the n1-air interface (Suppose that the light incidence is from air). Now we have: r = sin-1(sin i/n1) Then, at the n1-n2 interface, the angle of incidence is 90° - r and, for total internal reflection to occur, 90° - r must be greater than certain value, say θ (critical angle). Now, we have: 90° - r > θ 90° - sin-1(sin i/n1) > θ sin-1(sin i/n1) < 90° - θ sin i/n1 < sin (90° - θ) sin i < n1 sin (90° - θ) i < sin-1[n1 sin (90° - θ)] So i should be smaller than a certain value. (12) Suppose that P and Q initially carried charges of +q and -q respectively. Then we have: q2/4πε0r2 = F where r is the fixed separation between P and Q. Now, when R touches P, it brought +q/2 away from P and then when it touches Q, charge amount of -q/2 is shared among R and Q (since +q/2 is neutralized by part of the negative charges), thus R carries -q/4 away and now: P carries +q/2 and Q carries -q/4 So new force magnitude is: F' = (q/2)(q/4)/4πε0r2 = F/8 (15) When K is closed and at steady state, there's a steady current of 1 mA in the circuit and for the capacitors, one of them has its potential difference = 6V and the other being zero. So total charge in the capacitors is 2μ × 6 = 12μC And when K is opened, at steady state, NO current flows in the circuit and hence both capacitors have their potential difference = 10V So total charge now = 2 × 2μ × 10 = 40μC So additional charge = 28 μC 2008-02-13 16:32:59 補充： Q15) When K is closed:Complete circuit with capacitors being open circuit, so 4k and 6k are in series.Then current = 10/10k = 1 mAPotential drop across the first C = 4V, so p.d. across it = 6VSecond C is shorted by K, so p.d. = 0

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