The Electric Potential Part 1, Solution Manual

The electric potential Finding the potential The line integral along the first path is (we’ll suppress the z component of the argument) ∫ (x1 ,y1 ) (0,0) E · ds = ∫ x1 Ex (x, 0) dx + 0 = 0+ ∫ y1 0 ∫ y1 Ey (x1 , y) dy 0 (3x12 − 3y 2 ) dy = 3x12y1 − y13. (99) The line integral along the second path is ∫ (x1 ,y1 ) (0,0) E · ds = = ∫ ∫ y1 Ey (0, y) dy + 0 y1 0 (0 − 3y 2 ) dy + ∫ x1 Ex (x, y1 ) dx 0 ∫ x1 0 6xy1 dx = −y 31 + 3x21 y1 . (100) These two results are equal, as desired. The electric potential ϕ, if taken∫to be zero at (0, 0), is just the negative of our result, because we define ϕ by ϕ = − E · ds, or equivalently E = −∇ϕ. Hence ϕ(x, y) = y 3 − 3x2 y. The negative gradient of this is ) ( ∂ϕ ∂ϕ ∂ϕ , , = (6xy, 3x2 − 3y 2 , 0), (101) −∇ϕ = − ∂x ∂y ∂z which does indeed equal the given E. An alternative method of finding ϕ is to integrate the components of E to find the general form that ϕ must take. Since −∂ϕ/∂x equals Ex = 6xy, we see that −ϕ must take the form of 3x2 y + f (y, z), where f (y, z) is an arbitrary function of y and z. Likewise, since −∂ϕ/∂y equals Ey = 3x2 − 3y 2 , we see that −ϕ must take the form of 3x2 y − y 3 + g(x, z). Finally, since −∂ϕ/∂z equals Ez = 0, we see that −ϕ must take the form of 0 + h(x, y), that is, ϕ is a function of only x and y. The only function consistent with all three of these forms is −ϕ = 3x2 y − y 3 (plus a constant), in agreement with the above result. B l C -3q l 2q A Line integral the easy way The charges are shown in Fig. 32. The line integral of E equals the negative of the change in potential. The potentials at C and D are ) ( q 2q 3q 1 = +√ (−1.586), ϕC = − ℓ 4πϵ0 ℓ 4πϵ0 2ℓ ) ( q 3q 1 2q = −√ (−0.121). (102) ϕD = ℓ 4πϵ0 ℓ 4πϵ0 2ℓ D Figure 32 So the line integral from C to D equals ∫ D E · ds = −(ϕD − ϕC ) = ϕC − ϕD = C q (−1.464). 4πϵ0 ℓ With the given values of q and ℓ, this becomes ( −9 ) 1 10 C (−1.464) = −264 V. 4πϵ0 0.05 m (103) (104) The negative result makes sense, because the field between C and D points at least partly upward, while the ds in the line integral from C to D points downward. If you actually want to calculate the line integral, the y component of the field, as a function of y, is (taking the origin to be at the lower left corner, and ignoring the 4πϵ0 ’s): 2q 3q y ℓ−y Ey = 2 (105) + . ·√ · √ y + ℓ2 (ℓ − y )2 + ℓ2 y 2 + ℓ2 (ℓ − y)2 + ℓ2 ∫0 ∫D The integral C E · ds = ℓ Ey dy is readily calculated, and you can show that the result is ϕC − ϕD, where these ϕ’s are given in Eq. (102). Plot the potential The potential as a function of z is ) ( 1 6 6C 12 C 12 ϕ(z) = − − , =⇒ 4πϵ0 ϕ = |z| |z − 3| |z| |z − (3 m)| 4πϵ0 4πε0φ 15 10 5 -5 5 -5 -10 Figure 33 10 15 z (106) where we have ignored the units (which are C/m) in the second expression. The plot of 4πϵ0 ϕ is shown in Fig. 33. ϕ goes to +∞ at z = 0, and −∞ at z = 3. You can quickly verify that ϕ = 0 at z = 2 and z = 6. √ You can also show that ϕ achieves a local maximum of ϕ ≈ (0.34)/4πϵ0 at z = 6 + 3 2 ≈ 10.24. Since Ez = −∂ϕ/∂z, the field is zero at z = 10.24, so a charge placed there will be in equilibrium (unstable with regard to motion in the z direction if the charge is positive, stable if it is negative). Extremum of ϕ By symmetry, the E field at points on the y axis has no x or z component. And we know that Ey equals −∂ϕ/∂y. So if ϕ has a local maximum or minimum at some point on the y axis, then ∂ϕ/∂y, and hence Ey , equals zero. The full vector E therefore also equals zero. At the point (0, y, 0) with y > 1, the Ey component equals (ignoring the factor of 1/4πϵ0 , along with the units of the various quantities) Ey = 2 1 y−1 , −2· ·√ y2 (y − 1)2 + 12 (y − 1)2 + 12 (107) where the last factor gives the y component of the field from the two negative charges. Setting Ey = 0, moving one of the terms to the other side of the equation, and squaring, we find (y 2 − 2y + 2)3 y4 = . (108) (y − 1)2 Another way of obtaining this relation is to (as you can check) write down the potential (again ignoring the factor of 1/4πϵ0 and units), ϕ(0, y, 0) = 2 1 , −2· √ y (y − 1)2 + 12 φ (109) and then set ∂ϕ/∂y = 0. The result is Eq. (108), of course, because Ey = −∂ϕ/∂y. We can solve Eq. (108) numerically; Mathematica gives the numerical result of y = 1.621. Plots of ϕ(y) and Ey (y) (times 4πϵ0 ) for points on the y axis are shown in Fig. 34. For large y, we know that both ϕ and Ey must be negative, because for large y we have a charge 2 C at a distance y, and two −1 C charges at a distance essentially equal to y − 1. So the negative charges win out. You can show that Ey reaches a maximum negative value at y = 2.153; you will again need to solve an equation numerically. The existence of such a point between y = 1.621 and y = ∞ follows from a continuity argument similar to the one involving ϕ: Since Ey = 0 at these two points, Ey must have a local maximum or minimum somewhere between. Center vs. corner of a square Dimensional analysis tells us that for a given charge density σ, the potential at the center of a square of side s must be proportional to Q/s, where Q is the total charge, σs2 . (This is true because the potential has the units of q/4πϵ0 r, and Q is the only charge in the setup, and s is the only length scale.) Hence ϕ is proportional to σs2 /s = σs. So for fixed σ it is proportional to s. Equivalently, if we imagine∫ increasing the size of the square by a factor f in each direction, the integral ϕ ∝ (σ da)/r picks up a factor of f 2 in the da and a factor of f in the r, yielding a net factor of f in the numerator. Said in yet another (equivalent) way, if we imagine increasing the size of the square by a factor f in each direction, and if we look at corresponding pieces of the small and large versions, then the piece in the large version has f 2 times as much charge, but it is f times as far away from a given point, so its contribution to the potential is f 2 /f = f times the contribution in the small version. If we assemble 4 squares of side b, we make a square of side 2b. The potential at the center is 4ϕ1 (the sum of 4 corner potentials of the side b square). But from the above reasoning that ϕ ∝ s, this potential of 4ϕ1 must also be 2 times the center potential of the side-b square. So we have 4ϕ1 = 2ϕ0 , or ϕ0 = 2ϕ1 . Hence the desired ratio is 2. It therefore takes twice as much work to bring a charge in from infinity to the center, as it does to a corner. From Eqs. (2.25) and (2.30), the analogous statement for a disk of charge is that it takes π/2 as much work to bring a charge in from infinity to the center, as it does to the edge. But that problem can’t be solved via the above scaling/superposition argument. The above result for the square actually holds more generally for any rectangle with uniform charge density. All of the steps in the above logic are still valid, so the potential at the center is twice the potential at a corner. 2.5 2.0 1.5 1.0 0.5 - 0.5 1 2 3 4 y Ey 5 4 3 2 1 0 1 2 Figure 34 3 4 y Escaping a cube, toward an edge √ Let the cube have side length 2ℓ. Then the center is a distance 3ℓ√from each corner. So the potential at the center (ignoring the factor of e/4πϵ0 ℓ) is 8/ 3 = 4.6188. Let’s calculate the potential as a function of x, where √ x is the distance from the center to the midpoint of an edge (at which point x = 2ℓ). There are three different distances involved. You can obtain these by considering the plane that√ contains four corners and √ also the displacement x, which yields two distances each of 1 + ( 2 ± x)2 ; and also the plane that contains four √ corners and is perpendicular to the displacement x, which yields four distances of 3 + x2 . The potential is therefore (ignoring the e/4πϵ0 ℓ) φ ϕ(x) = √ 4.60 4.55 4.50 0.2 0.4 0.6 0.8 1.0 1.2 1.4 Figure 35 x 2 2 4 . +√ +√ √ √ 3 + x2 1 + ( 2 − x)2 1 + ( 2 + x)2 (110) √ √ As a double check, this does equal 8/ 3 when x = 0. Plugging in x = 2 gives the potential at the midpoint of an edge as 4.4555. Although this is smaller than the 4.6188 potential at the center, the plot in Fig. 35 shows that ϕ achieves a maximum value about halfway out to the edge. The maximum happens to be located at x = 0.761, and the value is ϕmax = 4.6242. Since this is larger than the value of ϕ at the center, the proton will not escape if it heads directly toward the midpoint of an edge. Field on the earth The radius of the earth is r = 6.4 · 106 m, so we have E = ϕ = 1 4πϵ0 1 4πϵ0 Q ( = 9 · 109 r2 ( Q = 9 · 109 r kg m3 ) 1C V , = 2.2 · 10−4 m s2 C2 (6.4 · 106 m)2 3) kg m 1C = 1400 V. s2 C2 (6.4 · 106 m) (111) This value of 1400 V is larger than you might think it should be, given the large radius of the earth. The point is that a coulomb is a large quantity of charge, on an everyday scale. Or equivalently ϵ0 has a small value in the system of units we use. Interstellar dust The potential is Q = −0.15 V =⇒ Q 4πϵ0 R ) ( s2 C 2 (−0.15 V)4π 8.85 · 10−12 (3 · 10−7 m) kg m3 (112) = −5 · 10−18 C. = The charge of an electron is −1.6 · 10−19 C, so Q corresponds to n = (5 · 10−18 C)/(1.6 · 10−19 C) = 31 electrons. The field at the surface is Q/4πϵ0 R2 . We could plug in the value of Q we just found, or we could just realize that E= ϕ V −0.15 V (Q/4πϵ0 R) = = −5 · 105 , = m R R 3 · 10−7 m (113) which is a rather large field. Basically, the small R matters more in E than in ϕ, because it is squared in E . Interestingly, note that the relation ϕ = ER says that it takes the same amount of work to drag a test charge out to infinity from the surface of a sphere, as it takes to drag the charge a distance R at full field strength (the value at the surface). Closest approach The size of the electron cloud around the nucleus is much larger than the nucleus. This implies that the potential associated with the cloud is much smaller than the potential associated with the nucleus (due to the 1/r factor). We may therefore ignore the potential due to the electron cloud in the following calculation. Closest approach is reached in a head-on approach, because in that case the proton ends up instantaneously at rest, which means that all of the initial kinetic energy has been converted into potential energy. The initial kinetic energy is of the proton is eV0 , where V0 = 5 · 106 V. The potential energy at closest approach is e(47e)/4πϵ0 r. These are equal when 47(1.6 · 10−19 C) 47e = 1.35 · 10−14 m. = s2 C2 )(5 · 106 V) 4πϵ0 V0 (1.11 · 10−10 kg m3 (114) This is larger than the radius of the silver nucleus, which is about 5 · 10−15 m, so it was reasonable to consider coulomb repulsion only. The strength of the electric field at this position is V0 = 47e 4πϵ0 r E= =⇒ r = 5 · 106 V 47e (47e/4πϵ0 r) V0 = = 3.7 · 1020 V/m, = = r 1.35 · 10−14 m r 4πϵ0 r2 (115) which is huge. The acceleration of the proton at this position is a= F eE (1.6 · 10−19 C)(3.7 · 1020 V/m) = = 3.5 · 1028 m/s2 , = m 1.67 · 10−27 kg m (116) which is gigantic. Gold potential Since volume is proportional to r3 , the amount of charge inside radius r is Q(r3 /a3 ). The field at radius r is effectively due to the charge inside r, so for r ≤ a the field is E= 1 Qr 1 Qr3 /a3 = . 4πϵ0 r2 4πϵ0 a3 (117) Outside the sphere the field is simply Q/4πϵ0 r2 . The potential at the surface of the sphere relative to infinity is ∫ a ∫ a Q dr 1 Q E dr = − ϕ(a) − ϕ(∞) = − = , (118) 2 4πϵ0 a ∞ ∞ 4πϵ0 r and the potential at the center of the sphere relative to the surface is ∫ 0 ∫ 0 1 Q 1 Qr dr ϕ(0) − ϕ(a) = − E dr = − = . 4πϵ0 2a a3 a a 4πϵ0 (119) Adding the two preceding equations gives ϕ(0) − ϕ(∞) = (1/4πϵ0 )(3Q/2a). For the problem at hand, this yields (assuming ϕ(∞) = 0 as usual) ϕ(0) = kg m3 ) 3(79 · 1.6 · 10−19 C) 1 3Q ( = 2.84 · 107 V, = 9 · 109 2 2 2(6 · 10−15 m) s C 4πϵ0 2a or 28.4 megavolts. (120) A sphere between planes The electric field due to the sheets is nonzero only between them, so the field to the right of the system is due only to the sphere. The potential at the point where the sphere touches the right sheet is therefore Qsphere /4πϵ0 R = 4πR2 σ/4πϵ0 R = σR/ϵ0 . The sphere produces no internal field, so the field in its interior is due only to the sheets. It therefore takes on the constant value of σ/ϵ0 , pointing to the left. The potential difference between the surface of the sphere and its center is then −(σ/ϵ0 )R, with the center at a lower potential. The total potential at the center, relative to x = +∞, is therefore σR/ϵ0 − σR/ϵ0 = 0. The potential at the point where the sphere touches the left sheet, relative to the center of the sphere, is −(σ/ϵ0 )R. And the potential at x = −∞, relative to the contact point on the left sheet, is −4πR2 σ/4πϵ0 R = −σR/ϵ0 , due to the sphere’s field. The potential at x = −∞, relative to the center (which we found has the same potential as x = +∞) is therefore −2σR/ϵ0 . As far as the potential at x = −∞ goes, there was actually no need to make any mention of the sphere. We could have chosen a path from x = +∞ to x = −∞ that goes nowhere near the sphere, in which case the potential simply decreases by −(σ/ϵ0 )(2R) due to the field between the sheets. In any event, for any path, the potential due to the sphere has equal values at x = ±∞, so it provides no net difference in the potential at these points. But the sheets do provide a net difference, because there is no way to get from x = +∞ to x = −∞ without passing through the (infinite) sheets. For compact charge distributions (that is, ones that don’t extend to infinity), it is possible to set ϕ = 0 at all points at infinity. But if a charge distribution extends to infinity (as in the present setup), then all points at infinity are not equivalent; we cannot consistently assign them all the value of ϕ = 0. E and ϕ for a cylinder (a) Consider a coaxial cylinder with length ℓ and radius r < a. The charge contained inside is πr2 ℓρ. The area of the cylindrical part of the surface is 2πrℓ, and since E is perpendicular to the surface by symmetry, the flux is 2πrℓE. So Gauss’s law gives the internal electric field as ∫ πr 2 ℓρ ρr q =⇒ 2πrℓE = =⇒ E = (for r < a). (121) E · da = ϵ0 ϵ0 2ϵ0 We’ll also need the external field for part (b). For this field, consider a cylinder of radius r > a. This contains a fixed amount of charge πa2 ℓρ, so Gauss’s law gives ∫ πa2 ℓρ ρa2 q =⇒ 2πrℓE = =⇒ E = (for r > a). (122) E · da = 2ϵ0 r ϵ0 ϵ0 This is the same as the field from a line of charge (namely λ/2πϵ0 r) with linear density λ = πa2 ρ. Note that the internal and external fields agree at r = a. (b) If ϕ = 0 at r = 0, then we have ∫ r ∫ r ρr2 ρr dr For r < a : ϕ(r) = − E dr = − =− , 4ϵ0 0 2ϵ0 0 ∫ r ∫ a For r > a : ϕ(r) = − E dr − E dr a 0 ∫ r 2 2 ρa dr ρa ρa2 ρa2 = − =− ln(r/a). (123) − − 2ϵ0 r 2ϵ 4ϵ0 4ϵ 0 0 a z This goes to −∞ as r → ∞. It also goes to −∞ for any given value of r if a → 0 while the charge per unit length (πa2 ρ) is held constant. P1 2d d Potential from a rod At point P1 in Fig. 36 we have ϕ1 = 1 4πϵ0 ∫ d −d (124) At point P2 with a general x value, we have (using the integral table in Appendix K) (√ ) ∫ d (√ ) d λ x2 + d 2 + d λ dz λ 1 2 + z 2 + z  √ = √ . ln ln x = ϕ2 = 4πϵ0 4πϵ0 −d 4πϵ0 −d x2 + z 2 x2 + d 2 − d (125) These two potentials are equal when √ √ √ x 2 + d2 + d √ = 3 =⇒ 4d = 2 x2 + d2 =⇒ x = 3 d. (126) x 2 + d2 − d Ellipse potentials Let z ′ specify the location of a point on the rod. Then the potential at the general point (x, 0, z) is (you should verify this integral by differentiating it) ϕ= 1 4πϵ0 ∫ d −d √ x2 λ dz′ + (z ′ − z)2 P2 λ d λ d λ λ λ dz  ln(2d − z) = − ln ln 3. = =− −d 4πϵ0 4πϵ0 4πϵ0 3d 2d − z = = (√ ) d λ ln x2 + (z ′ − z )2 + (z ′ − z )  4πϵ0 −d (√ ) λ x2 + (d − z)2 + d − z . (127) ln √ 4πϵ0 x2 + (d + z)2 − d − z You can verify that if x = 3d/2 and z = d, this equals (λ/4πϵ0 ) ln 3, which is the same as the value of ϕ in Exercise 2.43. And the sum of the distances from (3d/2, 0, d) to the ends of the rod at (0, 0, d) and (0, 0, −d) equals 3d/2 + 5d/2 = 4d, which is the same as the sum of the distances for each of the points in Exercise 2.43. So this is all consistent with the equipotential curve in the x-z plane being an ellipse. If x and z lie on the ellipse described by x2 /(a2 − d2 ) + z 2 /a2 = 1, then solving for x2 gives x2 = (a2 − z 2 )(a2 − d2 )/a2 . Using this, you can show that the quantity under the square root in Eq. (127) can be written as x2 + (d ± z)2 = ( a2 ± zd a )2 . (128) We therefore have ) ) ) ( ( ( 2 λ λ λ (a − z)(a + d) a+d a − zd + a(d − z) = = . ϕ= ln ln ln (a − z)(a − d) a−d a2 + zd − a(d + z) 4πϵ0 4πϵ0 4πϵ0 (129) As desired, this result is independent of z and x, so ϕ is constant on the ellipse described by x2 /(a2 − d2 ) + z 2 /a2 = 1. The ellipse containing the point (3d/2, 0, d), along with the two points from Exercise 2.43, has its a value (which is the length of the major axis, which lies along the z axis) equal to 2d. -d Figure 36 x A stick and a point charge (a) A little piece dx of the stick at position x (where x is negative) is a distance a − x from the point x = a. Adding up the contributions to the electric field from all the pieces of the stick gives the stick’s electric field at x = a as (with λ = Q/ℓ being the linear charge density) ) ( ∫ 0 λ dx 1 0 1 λ λ 1 = Estick = = −  2 4πϵ0 a a + ℓ 4πϵ0 a − x −ℓ −ℓ 4πϵ0 (a − x) Q λℓ = . (130) = 4πϵ0 a(a + ℓ) 4πϵ0 a(a + ℓ) Between the two objects, the field from the stick points rightward, and the field from the point charge points leftward. We want these two fields to cancel. The distance from the point x = a to the point charge is ℓ − a, so we want Q Q = 4πϵ0 a(a + ℓ) 4πϵ0 (ℓ − a)2 =⇒ (ℓ − a)2 = a(a + ℓ) =⇒ ℓ2 − 2aℓ + a2 = a2 + aℓ =⇒ a = ℓ/3. (131) (b) There are no other points. Ignoring the inside of the stick, the field can’t be zero anywhere else on the x axis because to the right of the point charge, both objects produce rightward-pointing fields, so they can’t cancel. And to the left of the stick, both objects produce leftward-pointing fields, so again they can’t cancel. For points not on the x axis, both objects produce a field that has a nonzero component pointing away from the x axis, so again the sum can’t be zero. The existence of an E = 0 point inside the stick, mentioned in the statement of the exercise, follows from a continuity argument: The field points rightward (and in fact diverges) just to the right of the right end of the stick. And it points leftward just to the left of the left end of the stick. So somewhere in between it must be zero. Figure 37 (c) Some equipotential curves and field lines are shown in Fig. 37 (but see Footnote 1). The field lines are everywhere perpendicular to the equipotential curves (because E = −∇ϕ, and the gradient of a function is perpendicular to the level surfaces). The equipotential curves make the transition from two surfaces (one around each object) to one surface (essentially a sphere at infinity) by intersecting at the E = 0 point at x = ℓ/3 that we found in part (a). The field is indeed zero wherever equipotential curves intersect, because the field must be perpendicular to both lines at the crossing, and the only vector that is perpendicular to two different directions is the zero vector. Zoomed-in views of the equipotentials and field lines near the E = 0 point are shown in Fig. 38 and Fig. 39, respectively. The field lines that start off heading nearly along the x axis toward the point x = ℓ/3 end up taking nearly a right turn and then head off to infinity. Their direction is vertical near the x axis due to the symmetric nature of the crossing. And it is vertical at infinity because the two objects have equal charge; you can use Gauss’s law to show this. But it isn’t vertical in between. y x Figure 38 Right triangle ϕ The potential at P equals the area integral, ϕP = The dy integral equals ∫ σ da σ = 4πϵ0 4πϵ0 r ∫ y b dx 0 ∫ ax/b 0 √ dy . x2 + y 2 (132) x ) (√ ) (√  (√ )ax/b a a2 x2 + (ax/b)2 + (ax/b) x2 + y 2 + y  ln . √ = ln 1+ 2 + = ln b b x 2 + 02 + 0 0 (133) Note that this is independent of x. That is, all vertical strips with thickness dx give the same contribution to the potential at P . To see intuitively why this is the case, consider two strips, and look at two infinitesimal bits of these strips that subtend the same angle. If the right bit is, say, twice as far from P as the left bit, then it has twice the charge (because the thickness dx is fixed, but the height of the right bit is twice as large). These two factors of 2 cancel in the expression ϕ = (1/4πϵ0 )(q/r), which means that the two bits of the strips give the same contribution to the potential. It then follows that the two complete strips give the same contribution. Figure 39 Due to the independence of x, the dx integral in Eq. (132) simply brings in a factor of b, so we arrive at ) (√ σb a 2 + b2 + a . (134) ϕP = ln b 4πϵ0 √ Since a2 + b2 /b = 1/ cos θ, and a/b = tan θ = sin θ/ cos θ, we can write ϕP as ( ) σb 1 + sin θ ϕP = ln , (135) 4πϵ0 cos θ as desired. Alternatively, you can derive this result directly, by performing an integral over x and θ, ∫ instead of x and y (by substituting y = x tan θ). You will need to use the integral dθ/ cos θ = ln[(1 + sin θ)/ cos θ]. a P b Remark: It is interesting to compare the potentials at the two vertices of a given very thin right triangle, or equivalently at the left vertices of the two triangles shown in Fig. 40. In the first case we can take the a ≪ b limit of Eq. (134), which you can show (with the help of ln(1 + ϵ) ≈ ϵ) leads to ϕ ≈ σa/4πϵ0 . (You can verify that this is consistent with the result you would obtain for a thin pie piece. You should also think about why it is a P b Figure 40 independent of b.) In the second case we have the reverse situation with b ≪ a, which leads to ϕ ≈ (σb/4πϵ0 ) ln(2a/b). So if we take a = ℓ and b = 100ℓ in the first case, we obtain ϕ ≈ σℓ/4πϵ0 . And if we take b = ℓ and a = 100ℓ in the second case, we obtain ϕ ≈ (ln 200)σℓ/4πϵ0 . It makes sense that the second case has the larger ϕ, because the charge is generally closer to that vertex P in that case. But the log behavior isn’t obvious. A square and a disk The result from Exercise 2.46 is ϕ = (σb/4πϵ0 ) ln[(1+sin θ)/ cos θ], where θ is measured with respect to the side with length b. We can divide the given square into 8 triangles, all of which have θ = 45◦ and b = s/2. The potential at the center is therefore ) ( (3.525)σs σ(s/2) 1 + sin 45◦ ϕsquare = 8 . (136) = ln ◦ cos 45 4πϵ0 4πϵ0 For the disk, we can slice it into concentric rings, which gives the potential at the center as ∫ d/2 ∫ 1 πσd dq 1 2πrσ dr ϕdisk = . (137) = = 4πϵ0 0 4πϵ0 r r 4πϵ0 Setting ϕdisk = ϕsquare yields d/s = (3.525)/π = 1.122. As was to be expected, the disk is larger than the inscribed circle, for which d = s; but smaller than the circumscribed √ circle, for which d = 2 s = (1.414)s. Field from a hemisphere l R θ Figure 41 r P As in Problem 2.7, our strategy will be to find the potential at radius r, and then take the derivative to find the field. The calculation is the same as in Problem 2.7, except that the limits of integration are modified. If we define θ in the same way is in Fig. 12.28, it now runs from π/2 to π. Following the steps in the solution to Problem2.7, the potential at point P in Fig. 41 is (we’ll keep things in terms of the density σ ) ∫ π 2πR2 σ sin θ dθ ϕ(r) = √ 2 2 π/2 4πϵ0 R + r − 2rR cos θ π  σR √ 2 R + r2 − 2rR cos θ  = 2ϵ0 r π/2 ) √ σR ( 2 2 = (R + r) − R + r . (138) 2ϵ0 r We are concerned with √ we want to know the field at the center. For √small r, because small r we can write R2 + r2 = R 1 + r2 /R2 ≈ R(1 + r2 /2R2 ). So the potential near the center is ) ) σR ( σR ( ϕ(r) = 1 − r/2R (139) r − r2 /2R = 2ϵ0 2ϵ0 r The field at the center is then E(r) = − dϕ σ . = 4ϵ0 dr (140) You can check that you arrive at the same result if you take P to be on the left side of the center. You will need to be careful about the limits of integration and various signs.