S12.1 SOLUTIONS TO PROBLEMS 12 (ODD NUMBERS)

Transcription

1 OLUTION TO PROBLEM 2 (ODD NUMBER) 2. The electric field is E = φ = 2xi + 2y j and at (2, ) E = 4i + 2j. Thus E = 2 5 and its direction is 2i + j. At ( 3, 2), φ = 6i + 4 j. Thus the direction of most rapid decrease is 3i + 2j. The directional derivative is dφ ds = φ û, where at (, 2) û = (3i j) and φ = 2i 4j. Thus dφ ds = (2i 4j) (3i j) =. 2.3 (a) Using (2.), gives curl A = y i (3z 3 4xz)j. (b) First form A = 2x 3 yz 3 i x 2 y 2 z 2 j + 3x 3 yz 4 k then use (2.) to gives curl(a) = x 2 z(3xz 3 + 2y 2 )i + 3x 2 yz 2 (2x 3z 2 )j 3xz 2 (y 2 + x 2 z)k. (c) curl curl A = ( A), so using the result of (a) and then again using (2.), gives curl curl A = ( 4x + 9z 2 )i +(4z )k. (d) grad(a curl A) = [A ( A)], so using the result of (a) gives and A ( A) = 2xyz 2 + 3yz 4 = f (x,y,z) [A ( A)] = f x i + f y j + f z k = 2yz 2 i + z 2 (3z 2 2x)j + 4yz(3z 2 x)k. (e) from (2.4a), curl grad =. 2.5 We use the two identities: which is (2.4c), and ( a) = ( a) 2 a () (φa) = φ( a) a ( φ), from Table 2. with φ = a, i.e., a( a) = ( a)( a) a ( a). (2) 2.

2 Using () and (2) in the given expression, gives ( a)( a) a ( a)+ a ( a) a 2 a + a 2 a = ( a)( a). 2.7 ince dz = around the circle, the integral reduces to V dr = (x 2 + y 2 )y dx (x 2 + y 2 )x dy. If we use plane polar co-ordinates in the xy-plane, around the circle we have r = a, so that and x = a cosθ, y = a sin θ, dx = a sin θdθ, dy = a cosθdθ V dr = a 4 (sin 2 θ + cos 2 θ)dθ = a 4. To verify this result, we can Green s theorem in the plane, (.2) in which case we evaluate the double integral over the interior R of the circle of x (x 2 + y 2 )x y (x 2 + y 2 )y = 4(x 2 + y 2 ) = 4r 2. and so the integral is a dr (4r 2 )r dθ= a In the xy-plane, z = and hence dz =, so using dr = dx i +dy j +dz k, gives F dr = (x 2y 2 )dx +(3x + 2y)dy, where we must remember that x and y are not independent variables along the curve specified. If we work in the two dimensional polar co-ordinates, the integration path corresponds to r(θ) = cosθ i + sin θ j, θ π where θ is the angle from the x-axis measured in the counter-clockwise direction. Hence dx = sin θ dθ, dy = cosθ dθ, and substituting in the integral gives 2.2

3 F dr = (sin θ cosθ + 2sin 3 θ + 3cos 2 θ)dθ) π = π sin 2θ + 2sin θ 2sin θ 2 cos2 θ cos2θ dθ 2 2 = cos2θ 2cosθ cos3 θ + 3 θ + 3 sin 2θ 2 4 π = (9π +6) The contour is shown in the figure below. The work done is If we set W = F dr = (xy dx P = xy, P y = x, Q = y 2, y 2 dy). Q x =, then by Green s theorem in the plane (.2), Q W = x P dx dy = x dx dy y area area = dy x dx = 2y dy =. y= 2 y x= 2.3 Taking the curl of F gives curl F = (c +)i +(a 4)j +(b 2)k. For F to be conservative, curl F =, and hence we must choose a = 4, b = 2, c =. Therefore, F = (x + 2y + 4z)i +(2x 3y z)j +(4x y + 2z)k., The scalar potential φ is then found from F = φ, where 2.3

4 so that φ x = x + 2y + 4z, φ = 2 x 2 + 2yx + 4zx + f (y,z), where f (y,z) is a function to be determined. imilarly for the y-component, φ = 2xy 3 2 y 2 zy + g(x,z), and from the z-component, φ = 4xz yz + z 2 + h(x,y). On comparing these three results, we find the general solution φ = 2 x 2 + 2yx + 4zx 3 2 y 2 zy + z 2 +d, where d is an arbitrary constant. 2.5 onsider the face corresponding to x =. The vector ds is outward from this face and normal to it and so can be written dy dz i. Then V ds = (x 2 i + 2 y 2 j + 2 z 2 k) dy dz i = x 2 dy dz, and since x =, the contribution to the integral is, z= y dy dz =. z= y= In a similar way, on the opposite face with x =, the corresponding integral is zero. If this is repeated for the three other pairs of faces, it is straightforward to show that one face of each pair contributes and the other is zero. o, finally, the total 2 surface integral from all three pairs is (a) If we take the axis through the centre to be the z-axis, and use spherical polar coordinates, the distance from a point (r,θ,φ) to the axis is r = r sin θ. Hence from (2.46) and (.44), I M = ρr 2 sin 2 θr 2 sin θ dr dθ dφ = 3M r 4 dr dφ sin 3 θ dθ, 4πa 3 Ω where Ω is the volume of the sphere. The integral a π 2.4

5 π sin 3 θ dθ = ( cos 2 θ)dcosθ = 4 3, so that finally, I M = 2 5 Ma 2. (b) By the parallel axis theorem (2.57), the moment about a tangent to the sphere is I = I M + Ma 2 = 7 5 Ma By the divergence theorem the integral is ( F)dυ = 2.2 From Table 2., we have V V x (4xz)+ y ( y 2 )+ z (yz) dυ = (4z 2y)dυ = (4z 2y) dx dy dz V = dx dy[2z 2 z= 2yz] z= x= y= = dx 2y y 2 y= =. y= x= (ψa) = ψ( a)+ a ψ, which gives the desired identity on substituting a = ψ. Integrating this identity over the volume Ω, and setting 2 ψ =, gives ( ψ) 2 dυ = (ψ ψ)dυ = ψ ψ ds Ω Ω using the divergence theorem. ince ψ = on, this gives ( ψ) 2 dυ = Ω 2.23 (a) and hence ψ = in Ω since ( ψ) 2. Therefore ψ is constant in Ω and must be zero because ψ = on. 2.5

6 (φe) = x (φe x )+ y (φe y )+ z (φe z ) φ = φ E + E x x + E φ y y + E φ z z = φ E + E φ. (b) At very large distances R, the charge distribution can be approximated by a point charge at the origin, so that E = Q 4πε R ˆr, φ = Q 2 4πε R. (c) ρφ dυ = ρφ dυ Ω Ω 2 since ρ = outside Ω. Hence, using Poisson s equation (2.53) ρφ dυ = ε φ div E dυ Ω Ω2 dυ ε E φ = ε div(φe) Ω 2 Ω 2 dυ using the identity derived in (a). Then, using the divergence theorem in the first integral and E = φ in the second integral, gives ρφ dυ = ε E 2 dυ + ε φe ds Ω Ω2, where is the surface of the larger sphere, and ds = R 2 sin θ dθ dφ e r = R 2 sin θ dθ dφˆr in spherical polar co-ordinates by (2.39). ubstituting this, together with the results of part (b), gives ρφdυ = ε E 2 dυ Ω + Q 2 Ω 2 6π 2 ε R 2 R dr π dθ dφ. The integral is 4πR, giving the desired result with c =Q 2 (4πε ) The boundary of is a circle in the xy plane of unit radius and centred at the origin. This can be written as Then x = cosθ, y = sin θ, z =. θ. 2.6

7 A dr = [(2x y)dx yz 2 dy y 2 z dz] = (2cosθ sin θ)( sin θ)dθ = π. Using tokes Theorem, the integral becomes ( A) ˆn ds = (k ˆn) ds, on evaluating the curl using (2.). But k ˆn ds is just the projection of a surface element ds onto the xy plane. Hence the integral becomes (k ˆn) ds = dx R dy, where R is the projection of on the xy-plane, so that we finally obtain x=+ x= dx y=+ ( x 2 ) dy = 4 dx y= ( x 2 ) ( x 2 ) dy = π, o in agreement with the value obtained by direct evaluation of the line integral The work done W is given by W = F dr where is the circuit shown in the figure below. (a) On the circle, Therefore r = a cosθ i +a sin θ j dr = a sin θ i +a cosθ j F = a 2 (sin 2 θ i + cos 2 θ j) F dr = a 3 sin 3 θ dθ +a 3 cos 3 θ dθ =. (b) Using tokes theorem,, W = F dr = curl F ds where ds = dx dy ˆk and curl F = 2(x y)ˆk, so that 2.7

8 since the area is symmetric in x and y. W = 2 (x y)dx dy = (c) ince curl F, the field F is not conservative. For an arbitrary loop, W. 2.8