Solutions to PS 2 Physics 201

Transcription

1 Solutions to PS Physics 1 1. ke dq E = i (1) r = i = i k eλ = i k eλ = i k eλ k e λ xdx () (x x) (x x )dx (x x ) + x dx () (x x ) x ln + x x + x x (4) x + x ln + x (5) x + x To find the field for x, we first want to rewrite this result in terms of the small parameter x. Doing so yields E = i k eλ ln 1 x + 1+ x x 1 x (6) Next, we perform a Taylor expansion in terms of x about the point x =. For the logarithm term, we find ln 1 x 1+ x x x (7) and for the second term x x = 1 x (8) 1 x = 1+ (9) x = x + x x (1)

2 Putting this all together, we arrive at a final approximation for E given by E = i k eλ + + (11) x x x x = i k eλ 4 (1) x = i λ (1) π x as desired. Comparing this to the expression for a dipole field aligned with the axis of a dipole, we find p i 4π x = i λ π x p = λ (14) (15). To find the work done, we use τ = p E (16) = pe sin θk (17) = k(1 9 )(.5) sin π 6 N m (18) = k.5 1 N m (19) W done = W e = U () π = U(π) U (1) 6 = pe cos π + pe cos π () 6 = pe 1+ () = N m (4) Finally, for the frequency of small oscillations, we use Newton s second law τ = d dt (5) pe sin θ = I θ (6) (7)

3 expanding sin θ for small θ, we find pe I θ = θ (8) ω = pe I (9) () Plugging in the numbers, we have d d I = m + m (1) = md () = kg m () and thus pe ω = I = 5 1 rad s =1 9 rad s (4) (5) (6). By spherical symmetry, we know automatically that the electric field everywhere will be purely in the radial direction. Using this fact, we can apply Gauss law by finding the flux through a sphere of radius r centered about the origin, E da =4πr E r = Q enclosed (7) For r<a, we have that 4 Q enclosed = πr Q 4 (8) πa = Q r a (9) and therefore E(r <a)= e r k e Qr a (4)

4 4 For a r<b, we have that Q enclosed = Q, and so E(a r<b)= e r k e Q r (41) For b r<c, the field must be zero since this describes the interior of a conductor, meaning that a charge of +Q must reside on the interior surface of the conducting shell. Therefore E(b r<c)= (4) Finally, for r c, it must be that Q enclosed =+Q, and therefore E(r c) =e r k e Q r (4) Below is a sketch showing where the charges reside, and some field lines. +Q +Q -Q 4. Exploiting the cylindrical symmetry of the problem tells us that the field directed radially (i.e. in the e r direction) away from the axis of the cylinders, and that πre r = Q enclosed (44)

5 5 where is the length of our cylindrical Gaussian surface. Thus, since the cylinders are hollow, we know that there is no charge enclosed for r<a, and thus E(r <a)= (45) For a r<b, we have that Q enclosed = λ (46) and therefore E(a r<b)=e r k e λ r astly, for r b, we have that Q enclosed =,so (47) E(r b) = (48) To find the surface charge density σ on the inner cylinder, we note that we can express the total charge on a length of the cylinder as either Q =πaσ (49) or as Q = λ (5) Equating these two expressions, we find that σ = λ πa (51) We can substitute this result into our expression for the field between the cylinder to find 4πk e aσ E(a <r<b)=e r r aσ = e r r For b a a, we have that between the cylinders r a a, and thus (5) (5) aσ E(a <r<b)=e r r (54) aσ = e r (a + r a) (55) σ e r (56)

6 6 This is equal in magnitude to the field of a parallel plate capacitor of the same charge density. Furthermore, on a very small scale e r does not vary significantly with the polar angle, and thus may be approximated as a cartesian unit vector. Thus, this setup locally approximates a parallel plate capacitor 5. By Gauss law, Φ e = 1C (57) Thus, by the symmetry of the cube, we must have that the flux through one of the faces is given by Φ e = 1C 1 N m = C (58) 6. For r<rwe have that Q enclosed = r π π =4πA r ρ(r)r sin θdφdθdρ (59) r 4 dr (6) = 4πA 5 r5 (61) Evaluating this same expression at r = R gives that the total charge Q is Q = 4πA 5 R5 (6) and thus for r<r Thus, Gauss law tells us that Q enclosed = Q r5 R 5 (6) For r>r, we have that Q enclosed = Q, and therefore k e Qr E(r <R)=e r (64) R 5 E(r R) =e r k e Q r (65)

7 7 7. By the Pythagorean theorem, the radius of each disc as a function of z is given by and thus the area A of each disc is r(z) = R z (66) A(z) =π R z (67) The volume of the sphere is then given by integrating over all discs contained in the sphere, i.e. from z = R to z = R. This yields R V = π R z dz (68) R = π R R (69) as desired = 4 πr (7) 8. Knowing that Gauss law follows from Coulomb s law, we can define an analogue of Gauss s law for the gravitational field G. Examining the form of both Newton s and Coulomb s law, we have G = GM r e r (71) and E = q 4π r e r (7) where the fields E and G are the forces on a unit charge (unit mass) due to charge q (mass M). By comparison, we see that M plays the same roll as q, and likewise G plays the same roll as 1 4π. From this we arrive at Gauss s law for gravitation, G da = 4πGM enclosed (7) 9. Using Gauss law, we have that for r =.5m, Q enclosed =1µC, and thus Next, for r =m, Q enclosed = 1µC, and thus E =ˆr k e(1µc) (.5m) = N C e r (74) E = ˆr k e(1µc) (m) =. 1 N C e r (75)

8 8 1. We can examine this situation as a solid sphere of uniform charge density ρ and radius R superimposed with a solid sphere of uniform charge density ρ with radius R.et r 1 denote the vector from the center of the larger sphere to a point within the smaller sphere, and let r denote the vector from the center of the smaller sphere to that same point. First, we use Gauss law to find the field E + due to the larger sphere. At a distance r 1, we have that the charge enclosed is given by Q enclosed = 4 πr 1ρ (76) and thus the field E + is given by E + = ρr 1 e r1 = ρ r 1 (77) Similarly, for the field E due to the smaller, negatively charged sphere, we find E = ρr e r = ρ r (78) Summing together these two contributions to find the total field in the cavity, we get E = E + + E = But from the figure, we can see that ρ (r 1 r ) (79) r 1 r = R ˆx (8) Thus, E = ρr ˆx (81) 6 which describes a uniform field in the ˆx direction