8.022pset3solution - Massachusetts Institute of Technology...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 3:Electric potential TA: Yi Mao, [email protected] 1. Vector calculus identities. n S n S1 C l S2 n (a) (b) Figure 1: Proof of several vector calculus identities. (a) f = 0. Proof: Method 1 - by brute force. You can work it out explicitly in Catesian coordinates. Method 2: by Stoke's theorem. Consider any surface S whose boundary is line l. S ( f ) ndA = ^ l f dl = l df = 0, (1) where n is the normal unit vector on the surface. Shrinking the surface S and its ^ boundary l to within an infinitesimal neighbor of a point, eq.(1) is still valid. Since S is arbitrary, so is n. Hence we have f = 0 at any point. ^ (b) ( F ) = 0. Method 1 - by brute force. You can work it out explicitly in Catesian coordinates. Method 2 - by Gauss's theorem and Stoke's theorem. Consider any (simply connected) volume V whose boundary is the closed surface S. A closed curve C on the surface cuts S in two as in figure 1. V ( F )dV = = = S ( ( F ) ndA ^ F ) ndA + ^ C S2 S1 C ( F ) ndA ^ (2) (3) F dl + F (-dl) = 0. The minus sign in eq.(2) is because the direction of C and that of n of S2 don't agree ^ with our "normal" system: in order for the sense of circulation around C and the outward pointing of n to make sense, we would have to use a "left-hand rule". This is fine, ^ but requires switching sign. (Alternatively, we could have made n on S2 point in rather ^ than out; we would find the same result.) Again, shrinking the volume to within an infinitesimal neighbor of a point, eq.(3) is still valid. Hence ( F ) = 0 at any point. (c) Associativity of divergence. (f F ) = = i=1 (f Fi ) i=1 xi 3 3 f 3 Fi f + Fi xi i=1 xi = f (d) Associativity of curl. (f F ) x F + f F. (4) (f Fz ) - (f Fy ) y z f Fz Fy f - + Fz - Fy = f y z y z = = f F x + f F x . (5) You can work out the equality for other two components. Hence compactly write (f F ) = f F + f F . 2. Purcell 3.1: Charges in a spherical conductor. + + + + + + + + + + + + + + + + (qb qc) + - -- - - - qb - - -- - - - - -- - -q b + + r >> radius of A + A - - - - - qc - - -- - - - -qc + + Figure 2: A spherical conductor with two spherical cavities. The charge distribution is the following. A charge qb lies in the center of one cavity; hence it induces a surface charge of totally -qb on the spherical boundary of the cavity. This distribution is strictly uniform due to the symmetry. Similar thing takes place for the charge qc and the induced surface charge of totally -qc . Since the spherical conductor A is totally electrically neutral, a surface charge of +(qb + qc ) distributes on the outer spherical shell of A. Since qd is far away from A, this surface charge distribution is approximately uniform. qb only "feels" forces acted by the surface charge -qb ; similar for qc ; qd only "feels" the force by the outer shell of A; A feels the reaction from qb , qc , qd . Therefore the forces on qc , qd , A and qd are Fb = F c = 0 strictly; qd (qb + qc ) r2 approximately and depending on r being large compared to radius of A. The forces between A and qd are of the opposite direction. Fd = F A + - - - - - + + + qd 3. Purcell 3.3: Nasty field line problem. z Q E Q E E E r= h E E r Figure 3: The top part of Gaussian surface (dashed lines) is chosen to be infinitely close to the field lines (solid lines) except around the point charge Q. The boxed figure in the upper-right corner shows the Gaussian surface around Q: as r 0, it is a hemisphere. The bottom part of Gaussian surface is on the plane. Method 1: Establish a surface shown in figure 3. As the top surface parallelly and infinitely closely approaches the field lines that starts horizontally from Q, electric field is tangent to the Gaussian surface at each point, so the electric flux is zero for the top surface except around Q. As points are close to Q, the field is like a single point charge field since contribution from the image charge is small compared to that from Q; hence we should use a spherical surface to calculate the flux. Also, since we consider horizontal field lines out of Q, the solid angle of this spherical surface should be 4/2 = 2. So 1 = - lim E1 2 0 2 = - lim 0 Q 2 2 2 = -2Q. (6) The minus sign is because the normal unit vector of the surface is toward Q (i.e. outward from inside Gaussian surface.) The electric field on the plane is along -z direction, and its magnitude is given by equation (7) of Purcell Ch.3. 2 = + r 0 |E2 |2r dr 2Qh 2r dr 0 (r + h2 )3/2 4Qh . = 4Q - 2 (r + h2 )1/2 = r 2 (7) By Gauss's law, = 1 + 2 = 0. So we solve (8) and get r= 3h. (8) Method 2: You may instead consider a Gaussian surface that approaches the field lines from the outside and "pops up" around Q instead of "popping down" in Method 1. Then the Gauss's law reads = 4Q, but be careful of the sign of 1 in this case. The magnitudes of 1 and 2 do not change. So you can make it by yourself as an exercise. 4. Purcell 3.5: Work pulling a charge away from a conducting plane. Method 1: by definition of work, we conclude that the second student is absolutely correct. Before calculating it explicitly, let's try to understand the difference between these two methods. The work required to seperate to infinite distance two charges Q and -Q is correctly Q2 /2h, the amount of absolute value of electric potential energy. However, recall how we made it: we always assume one of the two charges is fixed, since moving it would simultaneously cost energy so that the work done on the other charge = the potential energy. Our problem is exactly the unwanted case. Remember that image charge is always located symmetrically on the other side of the conducting plane. When we move real charge, the image charge is simultaneously moving. Equivalently in a two-charge system, a mysterious hand is moving the image charge in the same rate as we do the real charge. Consequentely we done the work in the amount of half of the absolute value of"total" potential energy, and the mysterious hand "offers" the other 2 half, due to the symmetry. So explicit calculation should get the result W = 1 Q . 2 2h Let's verify this. Q2 Q2 W = dz = . 4h h (2z)2 Aside: You may ask who kindly offers the mysterious hand that "does" the other "half" of "work"? Your intuition is correct nobody. Is this a paradox? Not really. The answer is, we don't need a free lunch. More precisely, the fields in two different configurations a charge above a conducting plane and a charge-image-charge pair agree only in the region z > 0, not everywhere. So in the real (charge-plane) configuration, the actual total potential energy is not -Q2 /2h, but half of it only, because E = 0, as z < 0. This means the work done to move Q away from the conducting plane is exactly the potential energy of the system; no more work is need in real configuration. The extra "half" of work done by mysterious hand is just because we included an extra imaginary amount of potential energy in the charge-image-charge configuration. Method 2: field perspective. The second student's approach is by definition correct. The work Q2 /2h of the first student's is actually the absolute value of electric field energy of a (+Q)(-Q) pair (up to a cancellation of infinities). The real charge-plane configuration share only half of it, Q2 /4h, as its field energy. The work done to move Q away from the plane in real configuration is equal to the real field energy, i.e. W = Q2 /4h. 5. Purcell 3.9: More charges near a conducting plane. y +q -q y=0 (X-Z) 0 a x -q a +q +q -q 45 x=0 (Y-Z) +q o -q (a) (b) Figure 4: (a) four charges at the corner of a square. (b) a charge inside a right-angle corner and its three image charges. (i) For four charges located in the corners of a square, shown in figure 4.a, the two equipotential surfaces are Y-Z plane (x = 0) and X-Z plane (y = 0). Since any point P (0, y, z) on Y-Z plane, U = + = 0. (ii) A metal sheet is bent through a right angle, and a point charge +q is located in the bisector of the corner, shown in Figure 4.b. According to (i), we need three image charges (two -q, one +q) for the boundary conducting planes to be equipotential. The field in the region (x > 0, y > 0) is then that of the four-charge system. The sketch of field lines in the concerned region is drawn in figure 4.b. (iii) A metal sheet is bent through an angle of 120 , and a point charge +q is located in the bisector of the corner, shown in Figure 5. Note that the angle 120 is so exact q a2 + (a - y)2 + z 2 q + + a2 + (a - y)2 + z 2 -q -q a2 + (a + y)2 + z 2 a2 + (a + y)2 + z 2 Figure 6: a capacitor of parallel square plates with a conducting square plate inserted. 6. Energy of a conductor in a capacitor. 1 s/4 2 s Wide s/2 s/4 Narrow x that the image charge with repect to one plane exactly lies on the extention of the other plane; hence we have no way to include a complete set of image charges to make the conducting planes equipotential. (a) All plates are conducting, so they're individually equipotential. Figure 5: a charge inside a 120 corner and its two image charges. -q L = |1 - 2 | = 4w s = 4n s/4 2 Q = n xL + w (L - x)L all 60 o -q +q (9) (10) (11) We have n = w (b) En = 4n = Ew (c) C= (d) U= (e) F =- 2 2Q L(x + L) Q = . L(x + L) (12) (13) 8Q L(x + L) 4Q = 4w = L(x + L) (14) (15) Q L(x + L) = . 4s 2sQ2 Q2 = . 2C L(x + L) dU 2Q2 s x= ^ x. ^ dx L(x + L)2 (16) (17) (18) 2Q s So a force F = -F = - L(x+L)2 x is required to keep the inserted plate from moving. ^ ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online