General Physics: Solutions to problem set 3

Transcription

1 PHYS 101 General Physics: Solutions to problem set 3 Problem 1: The reflected beam is parallel to the incident beam (simple geometry) Figure 1: Problem 2: The reflected beam isat r = 60 andthe refracted beamis at t, where 1.33sin60 = 1.52sin t. Thus t = All beams are in the same plane. Problem 3: Total internal reflection The angle of refraction satisfies sin t = n glass sin i /n air = 1.52sin i. For i = 40, we have sin t = hence t = 78. For i = 41.1, we have sin t = , hence t = 88 and for i = 42 we have sin t = 1.017, which is impossible. In the last case refraction does not occur and we have total internal reflection of the beam. The critical angle for this phenomenon is when sin t = 1, which means 1.52sin crit = 1, hence crit = Problem 4 The third fringes are situated at: y = L mλ d, y = L mλ d Substituting (m = 3) we find y = 7.74 cm and y = 9.18 cm. Thus the separation is y = y y = 1.40 cm. Overlapping fringes occur when y = y, hence L mλ d = λ Lm d m m = λ λ = =

2 where both 51 and 43 have greatest common divisor 1. Thus the first occurrence of overlapping bright fringes is at m = 51 and m = 43. Substituting we find the position on the screen λ = 1.32 m. Note however that for this position we have y/l 1, which means the approximation tan = sin = that we used in the proof for the bright fringes in not valid. Thus we must write instead the positions for the bright fringes: dsin = mλ where sin = tan = y/l is not valid. Substituting for m = 51 we find sin = i.e. = 61.3 and tan = 1.82 = y/l. Hence y = 2.75 m. This is the position on the screen for the first overlap between the bright fringes of both lights, which is the 51 st fringe of 430 light and 43 th of the 510 light. Problem 5: The electric field at the centre between the two identical charges is just zero (by symmetry). At position 4 cm, the electric field is the sum of two components: E 1 = k q, E d 2 2 = k q 1 d 2 2 where d 1 = 0.02 m and d 2 = 0.06 m and both fields are in the same direction (positive x-axis). Thus E = N/C. The electric force on the negative charge placed on the y-axis a distance 2 cm from the origin has only a y-component (i.e. F x = 0). The y-component is F y = 2 k q2 d 2 sin45 = N F cm F2 2 cm F + 2 cm + Figure 2: Problem 6 2

3 d+ǫ d ǫ ǫ Figure 3: Putting a third positively charged particle in the centre and displacing it a distance ǫ on the positive direction, we have the total force on it is: F = = kq 2 (d+ǫ) kq2 2 (d ǫ) 2 kq 2 d 2 (1+ǫ/d) kq 2 2 d 2 (1 ǫ/d) 2 = kq2 d 2 ( (1+ǫ/d) 2 (1 ǫ/d) 2) = kq2 d 2 ( (1 2ǫ/d) (1+2ǫ/d)+O((ǫ/d) 2 ) ) In the last equality we ignored O(ǫ 2 /d 2 ) terms as they are much smaller than O(ǫ/d) terms. Thus F = kq2 d 2 (1 2ǫ/d 1 2ǫ/d) = kq2 d 2 ( 4ǫ/d) = Cǫ where the constant C = 4kq 2 /d 3. This equation looks like Hook s law. Hence the particle will exhibit a SHM with angular frequency ω = C/m =, where m is the mass of the charge in the middle. The period of oscillation is T = 2π/ω. Problem 7 We divide the wire into small elements of length dx of charge dq = λdx. By symmetry we can deduce that the electric field has only y component (see figure 4) since the component de x cancels against the x component of the field from the charge at x. Hence we only compute the y-component, which is given by: de y = decos = kdq x 2 +r 2 r x2 +r = krλ dx 2 (x 2 +r 2 ) 3/2 The total electric field can be found by integrating the above expression. Since the wire is very long (i.e. L r, with L its length) the limits of integration are. We can however 3

4 d E de y de x r dq = λdx dx x Figure 4: calculate the field for a finite wire of length L by integrating from L L. The integral is mathematics (integration by parts): We have (check by differentiation): dx (x 2 +r 2 ) 3/2 = x r 2 r 2 +x 2 Applying the limits we find: Problem 8 E y = krλ r 2 [ ] + x r2 +x 2 = 2kλ r = λ 1 2πǫ 0 r We divide the disk into infinitesimal elements of area da = rdrd and charge dq = σda. The electric field at the point on the z-axis (axis of disk) has only a z-component since all other components cancel by symmetry. Thus we only need to compute E z, which is given by: Integrating over we get 2π. Thus dq rdrd d de z = k cosφ = σk d 2 +r2 d 2 +r 2 r2 +d 2 R 1 E z = 2πσkd 0 (r 2 +d 2 ) 3/2rdr The integral is straight forward and the result is: ( ) 1 E z = 2πσkd d 1 = σ d2 +R 2 2ǫ 0 4 ( 1 ) 1 1+R2 /d 2

5 d E z d E φ d d dq rd dr Figure 5: When d R, we have 1/ 1+R 2 /d 2 0, and so the result simplifies to E z = σ 2ǫ 0 This result is very important when studying parallel plates capacitors. Problem 9 For fig 6 the equivalent resistance is Figure 6: R = ( )Ω = 17.12Ω

6 d I 1 I 3 I 1 I 3 I 2 c I 2 +I 3 Figure 7: For the network in fig 7 we use Kirchhoff s rule for junctions and loops. For junctions the currents are clearly indicated in the diagram so the rule is satisfied. For loops we consider the loops adca, dbcd and acba (around the power supply) and write V = 0 (and use Ohm s law): I 1 +I 3 3I 2 = 0 I 1 I 3 5(I 2 +I 3 ) I 3 = 0 3I 2 +5(I 2 +I 3 ) V = 0 where V is the voltage across the network. Simplifying the second and third equations we get: This set of equations can be solved and we get: I 1 3I 2 +I 3 = 0 I 1 5I 2 7I 3 = 0 8I 2 +5I 3 = V I 1 = 13 I 2 = 4 I 3 = 1 Thus the total current passing through the network satisfies: I 1 +I 2 = 17 hence V = I i.e. the equivalent resistance is 27/17Ω = 0.63Ω. 6