Department of Physics IIT Kanpur, Semester II,

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1 Department of Phsics IIT Kanpur, Semester II, 7-8 PHYA: Phsics II Solutions # Instructors: AKJ & SC Solution.: (a) At the top of the hill, the gradient of the height function should be ero, that is, h(, ). This gives Or, h(, )ˆ + h(, ) h(, )ŷ ( 6 8)ˆ + ( 8 + 8)ŷ, and Solving these two equations we get and. Thus the top of the hill is located at and. (b) The height of a hill is defined as the height of its top, which for the given hill is located at and. Therefore the height of the hill is h(, ) e /.5. (c) The steepest slope at an point is in the direction of the gradient h(, ). Now, the gradient vector at (, ) is h(, ) h(, )(/6)( ˆ + ŷ). The direction of this gradient vector is given b tan θ, or θ 5., Therefore the slope is steepest in the direction θ 5. (d) The slope of h(, ) at (,) in the direction n is h(, ) n h(, )(/6)( ˆ + ŷ) (ˆ + ŷ). Solution.: We the magnitude of the separation vector given as Therefore the gradient is ( ) + ( ) + ( ). ˆ + ŷ + ẑ ( ) + ( ) + ( ) ( )ˆ + ( )ŷ + ( )ẑ ( ) + ( ) + ( ). Thus we see that the gradient is just a unit vector parallel to. Solution.: We φ ( + )ˆ + ŷ + ẑ. Using the definition of gradient in the cartesian coordinates sstem and equating the two sides of the equation, we get φ(,, ) +, φ(,, ), φ(,, ).

2 Integrating the above equations eilds φ(,, ) + + f(, ) φ(,, ) + g(, ) φ(,, ) + h(, ), where f(, ), g(, ) and h(, ) are arbitrar functions. Choose, f(, ), g(, ) and h(, ). The scaler function is φ(,, ) + + constant. We note that the scalar function in this case can onl be obtained up to a constant. Also, b taking the gradient of this scalar function, we can verif our result. Solution.: ln r ( ˆ + ŷ + ) ẑ ln + + ˆ + ŷ + ẑ + + r r. r ( ˆ + ŷ + ) ẑ + + ˆ ŷ ẑ ( + + ) r / r. () Solution.5: (a) ( ˆr r n ) ( ˆ + ŷ + ẑ ( + + ) n+ (n + )( + + ) ( + + ) n+ ) ( + + ) / ( + + ) n/ + + ( + + ) n+ ( + + )(n + )( + + ) ( + + ) n+ n + n ( + + ) n+ r n+. + (n + )( + + ) ( + + ) n+ ( + + ) n+ + (n + )( + + ) ( + + ) n+ (b) For n, (ˆr/r n ) decreases with r and tends to ero as r approaches infinit. However, for n, (ˆr/r ) everwhere ecept at r. As we see in the above equation, the divergence at r is of tpe / and is therefore not defined. So, we cannot use the above result for calculating divergence at r. The divergence (ˆr/r ) at r can be calculated b use of Gauss s theorem and in fact can be shown to be infinit. Mathematicall, the divergence in this case can be represented as the Dirac-delta function: (ˆr/r ) πδ (r). (c) One eample of the vector-fields of tpe E ˆr/r is the electric field due to a point charge. Therefore, this tpe of vector fields are ver relevant in electrodnamics.

3 Solution.6: We E B. Taking the curl on both sides, we get t or, or, ( E) ( B) t ( E) E E t E E t since E. Similarl, one can show that B B t since B. The above two equations are known as the wave-equations and the describe the propagation of electromagnetic fields in vacuum. Solution.7: We v ()ˆ + ()ŷ + ()ẑ. Stokes theorem sas ( v) da v dl. For this problem the area integral needs to be calculated over the shaded region and the line integral needs to be calculated along the closed path consisting of three line segments, as indicated in Fig.. (iii) (ii) (i) FIG. : Let us first calculate the left hand side of the above equation. We ( v v v ) ( v ˆ + v ) ( v ŷ + v ) ẑ ( )ˆ + ( )ŷ + ( )ẑ Therefore, ( v) da ( ˆ ŷ ẑ) ddˆ dd ( )d 8 ()

4 We now calculate the line integral along the three line segments. (i) ; v dl ; v dl (ii), ; v dl d; v dl (iii) ; v dl ; v dl d v dl ( )d 8 Therefore, we v dl () Comparing equations () and (), we verif the Stoke s theorem. Solution.8: We need to verif the divergence theorem, which sas that ( v)dτ v da. FIG. : For the given problem, the volume integral needs to be performed over the octant of the sphere and the area integral needs to be calculated over the four surfaces that completel surround the octant. Let us first calculate the volume integral. We v r cos θˆr + r cos φˆθ r cos θ sin φˆφ. We use the formula for divergence in the spherical coordinate sstem: v r r (r v r ) + θ (sin θv θ) + r r (r r cos θ) + r (r cos θ) + r cos θ φ (v φ) θ (sin θr cos φ) + (cos θr cos φ) + ( r cos θ cos φ) Substituting the above epression for divergence in the volume integral, we get ( v)dτ (r cos θ)r sin θdrdθdφ φ ( r cos θ sin φ)

5 We need to integrate the volume integral over the octant of the sphere as shown in Fig. We therefore get, ( v)dτ r dr π/ π π cos θ sin θdθ The area surrounding the octant consists of four surfaces and thus we need to calculate the area integrals separatel over these four surfaces: (i) The surface in the plane: da drdφˆθ; v da (r cos φ)drdφ. But θ π/. Therefore, we π/ v da r cos φdrdφ r dr cos φdφ (ii) The surface in the plane: da rdrdθˆφ; v da ( r cos θ sin φ)rdrdθ. But φ π/. Therefore, we v da r cos θdrdθ r dr π/ π/ dφ cos θdθ (iii) The surface in the plane: da rdrdθˆφ; v da ( r cos θ sin φ)( rdrdθ). But φ. Therefore, we v da da () (iv) The spherical curved surface: da r sin θdθdφˆr; v da (r cos θ)(r sin θdθdφ). But r. Therefore, we π/ π/ v da ( cos θ)( sin θdθdφ) cos θ sin θdθ dφ π π. Therefore, we v da + + π π (5) Comparing equations () and (5), we verif the divergence theorem. 5