Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Magnetic monopole wikipedia , lookup
Speed of gravity wikipedia , lookup
Geomorphology wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Lorentz force wikipedia , lookup
Field (physics) wikipedia , lookup
Maxwell's equations wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Homework 1 Solutions, Electromagnetic Theory I Dr. Christopher S. Baird, Fall 2011 University of Massachusetts Lowell Problem 1 Considering only the fields in a two-dimensional plane, draw the electrostatic field lines as accurately as possible for the following systems: (a) A triangle-shaped object with a total charge +Q fixed uniformly over its surface. (b) A point charge -2Q placed a distance d away from the triangle already described. (c) The point charge and triangle already described placed inside a thin, hollow conducting circular shell with no net charge. SOLUTION: When constructing field lines diagrams, we should keep several points in mind. Very near point charges, the field lines extend out radially. Also, field lines are perpendicular to the surface of conductors at the surface. We should remember that far away from an object, the object looks like a point charge, so the field line pattern far away should become radial. When we draw field lines, they should start on positive charges and end on negative charges or lead off the page. They should never cross or have unnatural kinks. Lastly, conducting shells shield the interior field pattern, and therefore all internal information from the outside world, except for the total net internal charge. This is because the charges on the shell's inner surface move around until they cancel the fields. (a) +Q (b) -2Q +Q (c) -2Q +Q Problem 2 A thin wire ring of radius R is centered on the origin, lies in the x-y plane, and contains a total charge +Q fixed permanently and uniformly along its extent. An external field is applied, Ex = (E0/L2)(x2+y2)2z/L, Ey = 3 E0(z/L+1)3, Ez = E0 sin(z/L). Here, L is a length. What is the total force that the ring experiences? SOLUTION: First we note that the the uniform charge is permanently fixed, so that the introduction of the external field does not alter the charge distribution. We must first find a mathematical expression for the ring charge in terms of Dirac deltas. Because the ring is round and centered on the origin, the most natural coordinate system is spherical coordinates (r, θ, ϕ). The ring is infinitely thin in the radial direction and the polar angle direction, so that we must use Dirac deltas. The ring is constant in shape and in charge density in the azimuthal direction, so our expression for the charge density should not include the azimuthal angle in any way. The general expression is: δ( x−x ')=δ (u−u ')δ( v−v ')δ (w−w ')U V W For spherical coordinates, u = r, v = θ, w = ϕ and U = 1, V = 1/r, W = 1/(r sin θ) so that we have: δ( x−x ')=δ (r −r ') δ(θ−θ) δ( ϕ−ϕ ') r r sin θ Apply this to our case and multiply by some constant A so that the total charge works out: ρ(x)= A δ(r −R) δ(θ−π/2) r Now we determine A by forcing the total charge to be equal to +Q. +Q=∫ ρ dV 2π π ∞ +Q=∫ ∫∫ ρ(r ,θ , ϕ)r 2 sin θ dr d θ d ϕ 0 0 0 2π π ∞ +Q=∫ ∫∫ [ A δ( r− R) 0 0 0 δ(θ−π/2) 2 ]r sin θ dr d θ d ϕ r +Q=2 π R A A= Q 2π R We have our final form for the charge density. Note that once the Dirac deltas are applied, their arguments will be set to zero, so we can go a head and do that in advance, setting r = R: ρ(x)= Q δ( r− R)δ(θ−π /2) 2 π R2 The total force is the integral of all the force components that each part of the ring feels: F=∫ ρ E dV Expand into components: F=̂i ∫ ρ E x dV + ̂j∫ ρ E y dV +k̂ ∫ ρ E z dV Expand out integrals into Cartesian coordinates: ∞ ∞ ∞ −∞ ∞ −∞ ∞ −∞ ∞ F=̂i ∫ dx ∫ dy ∫ dz ρ( x , y , z ) E x ( x , y , z ) +̂j ∫ dx ∫ dy ∫ dz ρ( x , y , z ) E y ( x , y , z ) −∞ ∞ −∞ ∞ −∞ ∞ −∞ −∞ −∞ +k̂ ∫ dx ∫ dy ∫ dz ρ( x , y , z ) E z (x , y , z ) Plug in the electric field: ∞ ∞ ∞ F=̂i ∫ dx ∫ dy ∫ dz ρ( x , y , z )[( E 0 / L2 )( x 2+ y 2 )2 z / L ] −∞ −∞ ∞ ∞ −∞ ∞ +̂j ∫ dx ∫ dy ∫ dz ρ(x , y , z )[3 E 0 ( z / L+1)3 ] −∞ ∞ −∞ ∞ −∞ ∞ +k̂ ∫ dx ∫ dy ∫ dz ρ(x , y , z)[ E 0 sin( z / L)] −∞ −∞ −∞ Our charge density expression is in spherical coordinates, so we need to transform this force equation into spherical coordinates: 2π π ∞ Q F=̂i ∫ d ϕ∫ d θ∫ dr r 2 sin θ [ δ( r −R)δ(θ−π/2)][(E 0 / L 2)(r 2 sin2 θ)2(r cos θ / L)] 2 2 π R 0 0 0 2π π ∞ Q ̂ + j ∫ d ϕ∫ d θ ∫ dr r 2 sin θ [ δ( r−R)δ(θ−π / 2)][3 E 0 (r cos θ / L+1)3 ] 2 2π R 0 0 0 2π π ∞ Q +k̂ ∫ d ϕ∫ d θ∫ dr r 2 sin θ [ δ(r− R) δ(θ−π /2)][ E 0 sin (r cos θ / L)] 2 π R2 0 0 0 Apply the Dirac deltas: [( ) ] 2 R F=E 0 Q ̂i +3 ̂j L Problem 3 Jackson 1.1 q Use Gauss's theorem ∮S E⋅n da= and ∮ E⋅d l=0 to prove the following: 0 a) Any excess charge placed on a conductor must lie entirely on its surface. (A conductor by definition contains charges capable of moving freely under the action of applied electric fields.) b) A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from fields due to charges placed inside it. c) The electric field at the surface of the conductor is normal to the surface and has a magnitude σ /ϵ0 , where σ is the charge density per unit area on the surface. SOLUTION: a) First, the problem contains the unstated assumption that what is wanted is the location of the charges in static equilibrium. This assumption is valid because this problem is found in the chapter on electrostatics. Static equilibrium (the lack of movement) can only exist when free charges are present if there are no electric fields. Therefore the electric field inside a conductor is zero. One can draw any arbitrary, closed surface completely inside the conductor and it will always have zero electric field at every point on the surface. The integral of the electric field over this surface is zero and by Gauss's law, the total charge is therefore zero. Because the surface is arbitrary, it can be chosen to be infinitesimally small or be chosen to go through any point in the conductor's interior. Therefore every point inside a conductor has zero charge, and all the charge must then reside on the surface. b) Assume for the moment that the electric field external to the conductor is perpendicular to the conductor's surface at the surface (the proof is left to part c). Because the conductor is hollow, the interior region of the conductor has by definition no free charges. If we choose a closed mathematical surface just inside and parallel to the inner surface of the hollow conductor, it therefore contains a total charge of zero. Gauss's Law then becomes: ∮S E⋅n da=0 Because the mathematical surface of the integral is parallel to the surface of the conductor, and the electric field is perpendicular to the surface of the conductor, the electric field must be parallel to the normal: n E This simplifies Gauss's law to: ∮S E da=0 The magnitude of a vector, in this case E, is always positive. There is no way to get the integral of a permanently positive function to equal zero except if the function itself is zero at every point. Thus the electric field is zero at every point on the mathematical surface. The surface can be chosen to show that all points in the hollow region have zero electric fields. Another way of proving this is to consider a hollow conductor with no charges external to it. According to part a, there are no electric fields in the hollow region. Now bring charges in from infinity and place them just external to the conductor. Their electric fields can never penetrate the interior of the conductor according to part a, and thus can never reach the hollow region beyond the conducting shell. The fields in the hollow region remain zero. Note, this is true only for perfect conductors. In practice, if a conductor is thin enough and non-perfect enough, the fields actually do penetrate through a conducting shell. On the other hand, a charge inside the hollow region of a closed conductor induces a charge on the conductor that creates an external field. If the charge free region is now the infinite region external to the conductor, we can never draw a closed surface around infinity, and can thus never use Gauss's law to prove there is zero field. Another way of doing this is to draw a Gaussian integration surface outside and around the entire conductor. The surface now encloses the internal charges and by Gauss's law there must therefore be non-zero fields outside the conductor. c) Inside the conductor there are no electric fields. Outside the conductor, there are no free charges, and therefore there can be electric fields. What happens at the surface of the conductor? If there is a component of the electric field tangential to the conductor's surface, it would accelerate charges along the surface, and there would be no static equilibrium. There is therefore no tangential component, and electric fields are always normal to the conductor's surface. Draw a pillbox surface half-in and half-out of the surface of the conductor and let us integrate the electric field over the surface. There is no electric field tangential to the conductor's surface, thus the sides of the pillbox contributes nothing to the integral. Also, the electric field is zero inside the conductor, thus the bottom of the pillbox contributes nothing to the integral. All that is left is the top of the pillbox. If the pillbox is small enough, the surface normal of its top and the electric field are parallel so that Gauss's law becomes: q ∫top E da= 0 where q is the charge contained inside the pillbox. Now shrink the pillbox until it is infinitesimally small. The electric field is constant over an infinitesimally small surface and can be taken out of the integral, so that the integral is evaluated to just be the total area of the pillbox top: E Atop = q 0 Rearranging: E= q 1 A top 0 Define as the charge per unit area q / Atop and the equation becomes: E= ϵσ 0 Problem 4: Jackson 1.3 Using Dirac delta functions in the appropriate coordinates, express the following charge distributions as three-dimensional charge densities ρ(x). (a) In spherical coordinates, a charge Q uniformly distributed over a spherical shell of radius R. (b) In cylindrical coordinates, a charge λ per unit length uniformly distributed over a cylindrical surface of radius b. (c) In cylindrical coordinates, a charge Q spread uniformly over a flat circular disc of negligible thickness and radius R. (d) The same as part (c), but using spherical coordinates. SOLUTION: The easiest method to use is to set a Dirac delta for every dimension that has an infinitely thin appearance. Multiply this by some arbitrary parameter, integrate over the whole object, set this equal to the total charge, then solve for the arbitrary parameter. (a) The charge distribution is only thin in the radial direction. ρ(r , θ , ϕ)= Aδ(r −R) Now integrate over all space and set it equal to the total charge Q. 2π π ∞ Q=∫ ∫ ∫ ρ(r , θ , ϕ)r 2 sin θ dr d θ d ϕ 0 0 0 ∞ Q=4 π A∫ δ(r −R)r 2 dr 0 Q=4 π R2 A A= Q 4 π R2 r , ,= Q r −R 2 4 R This answer should be obvious now. It is just the total charge divided by the area of a sphere times the delta. (b) ρ(r , ϕ , z )= Aδ(r −b) 2π ∞ λ=∫ ∫ ρ( r , ϕ , z )r dr d θ 0 0 2π ∞ λ= A∫ d θ ∫ δ(r −b) r dr 0 0 λ= A 2 π b A= λ 2πb r , , z = r−b 2b Again, this should be obvious that this is the surface charge density time the delta, where the surface charge density is the linear charge density divided by the circumference of the cylinder. (c) We must use the step function H in the radial direction. r , , z = A z H R−r ∞ 2 ∞ Q= ∫ ∫ ∫ r , , z r dr d dz −∞ 0 0 ∞ 2 −∞ 0 ∞ Q= A ∫ z dz ∫ d ∫ H R−r r dr 0 R Q= A 2 ∫ r dr 0 A= Q R2 r , , z = Q z H R−r R2 Again, it should be obvious that this is the deltas times the surface charge density, which is the total charge divided by the area of the disc. (d) Try: r , ,= A − /2 H R−r r 2 ∞ Q=∫ ∫ ∫ r , , r 2 sin dr d d 0 0 0 2 ∞ Q= A ∫ d ∫ − /2sin d ∫ H R−r r dr 0 A= 0 0 Q R2 ρ(r , θ , ϕ)= Q δ( θ−π/2) H ( R−r ) r π R2 Problem 5 Jackson 1.11 Use Gauss's theorem to prove that at the surface of a curved charged conductor, the normal derivative of the electric field is given by ( 1 ∂E 1 1 =− + E ∂n R1 R2 ) where R1 and R2 are the principal radii of curvature of the surface. SOLUTION: We will find the normal derivative of the electric field by taking the limit of the finite difference: E ( x+Δ r n)−E ̂ ( x) ∂E = lim ∂ n Δr →0 Δr where Δr is a small length increment in the normal direction. This is the definition of a derivative according to fundamental theorem of calculus. We set a square Gaussian pillbox just above the surface of a point on the curved charges conductor (not straddling), and use Gauss's theorem to integrate over the pillbox. Make the upper and lower surfaces curved so that their curvature matches the conductor's surface. Set the location of the center of the lower surface at x and the center of the upper surface at x+Δ r n̂ . Set the sides of the pillbox normal to the conductor's surface so they do not contribute. The pillbox is above the surface and therefore contains no charge. n ∮S E⋅n da=0 Δr ∫top E⋅n da+∫bottom E⋅n da=0 Shrink the pillbox down in the usual way so that the electric field becomes constant across its surface and comes out of the integral. Be careful and remember that the normal to the bottom Gaussian surface is in the opposite direction as the conductor's normal, so we need a negative sign to account for this. E top ∫top da− E bottom ∫bottom da=0 E (x+Δ r n̂ )( R1+Δ r )(R2 +Δ r )∫∫ d θ1 d θ 2=E (x) R1 R2∫∫ d θ 1 d θ 2 E (x+Δ r n̂ )= E (x) R1 R2 ( R1+Δ r )( R 2+Δ r ) Plug this into the normal derivative definition: R1 R2 R1 R2 −1 ( R1+Δ r )(R2 +Δ r ) ∂E = lim E ∂ n Δr →0 Δr −R1−R 2−Δ r ∂E = lim E ∂ n Δ r → 0 ( R1+Δ r )( R2+Δ r ) ( 1 ∂E 1 1 =− + E ∂n R1 R2 ) For very small curvatures of radius, such as the tip of a pointed conductor or the edge of a conducting cube, the derivative becomes very large. This means that the field is changing very quickly, so that the field lines are diverging away from the point or edge. For very large curvatures of radius, such as approaching a flat surface, this equation tells us the derivative approaches zero. This means that the electric field is approximately constant outside near-flat conducting surfaces.