Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2010 Problem Set 2 Solutions
Problem 1 (10 points): Concept Questions. Explain your reasoning. Concept Question 1: A pyramid has a square base of side a, and four faces which are equilateral triangles. A charge Q is placed on the center of the base of the pyramid. What is the net flux of electric field emerging from one of the triangular faces of the pyramid? Answer: Consider the eight faced closed surface consisting of two pyramids with the charge at the center. The total flux by Gauss’s law is just Q / ! 0 . Since each face is identical, the flux through each face is one eight the total flux or Q / 8! 0 .
!a ˆ Concept Question 2: A charge distribution generates a radial electric field E = 2 e ! r / br r where a and b are constants. What is the total charge giving rise to this electric field? Answer: In order to find the total charge, I choose a Gaussian surface that extends over all space. Because the electric field is radially symmetric, I choose my Gaussian surface to be a sphere of radius r and I will take the limit as r ! " . The flux is given by !! a a a ˆ ˆ lim " E # da = lim " 2 e% r / br # dar = lim " 2 e% r / b da = lim 2 e% r / b 4& r 2 = 4& a lim e% r / b = 0 $$ $$ r $$ r r !" r !" r !" r !" r r !" r r r When I take the limit as r ! " , the exponential term goes to zero, and so the flux goes to zero. Therefore the charge enclosed is zero. Problem 2 (10 points): Nonuniformly charged sphere A sphere of radius R has a charge density ! = !0 (r / R) where !0 is a constant and r is the distance from the center of the sphere. a) The total charge inside the sphere is the integral r =R r =R !0 4" r = R 3 !0 4" R 4 2 2 3 Q = $ ! 4" r dr = $ !0 (r / R)4" r dr = $0 r dr = R 4 = !0" R R r= r #=0 r #=0 b) There are two regions of space: region I: r < R , and region II: r > R so we apply Gauss’ Law to each region to find the electric field. For region I: r < R , we choose a sphere of radius r as our Gaussian surface. Then, the electric flux through this closed surface is ! ! 2 " EI ! dA = EI ! 4# r . "" Since the charge distribution is nonuniform, we will need to integrate the charge density to find the charge enclosed in our Gaussian surface. In the integral below we use the integration variable r ! in order to distinguish it from the radius r of the Gaussian sphere. r$=r Qenc 1 r $ = r "0 4# r $ = r 3 " 4# r 4 "0# r 4 1 2 2 . = " 4# r $ dr $ = "0 ( r $ / R)4# r $ dr $ = = r $ dr $ = 0 !0 ! 0 r $% 0 ! 0 r $% 0 R! 0 r $% 0 4 R! 0 R! 0 = = = Notice that the integration is primed and the radius of the Gaussian sphere appears as a limit of the integral. PS021 Recall that Gauss’s Law equates electric flux to charge enclosed: "E "" ! I !Q ! dA = enc . #0
2 #0" r 4 So we substitute the two calculations above into Gauss’s Law to arrive at: E I ! 4" r = . R$ 0
! ! r2 ˆ ˆ We can solve this equation for the electric field EI = E I r = 0 r , 0 < r < R . 4 R" 0 ! !0 r 2 The electric field points radially outward and has magnitude EI = , 0 < r < R. 4" 0 For region II: r > R : we choose the same spherical Gaussian surface of radius ! ! r > R , and the electric flux has the same form " EII ! dA = E II ! 4# r 2 . "" All the charge is now enclosed, Qenc = Q = !0" R 3 , so the right hand side of
Qenc Q "0# R3 Gauss’s Law becomes . Then Gauss’s Law becomes = = !0 !0 !0 E II ! 4" r 2 = #0" R3 . We can solve this equation for the electric field $0 ! !0 R 3 ˆ ˆ EII = E II r = r , r > R . In this region of space, the electric field points radially outward and has 4" 0 r 2 ! ! R3 magnitude EII = 0 2 , r > R , so it falls off as 1/ r 2 as we expect since outside the charge 4" 0 r distribution, the sphere acts as if it all the charge were concentrated at the origin.
Problem 3 (10 points): NP Junction The charge density in this problem is:
# + !0 0 < z< a % !( x, y, z ) = !( z ) = $" !0 " a< z< 0 % z >a &0 so the electric field is a superposition of two slabs of opposite charge density. ! Outside both slabs, the field of a positive slab E P (due to the Ptype semiconductor) is constant and points ! away and the field of a negative slab E N (due to the Ntype semiconductor) is also constant and points towards the slab, so when we add both contributions we find that ! the electric field is zero outside the slabs. The fields E P are shown on the figure to the right. The superposition PS022 ! of these fields ET is shown on the top line in the figure. The electric field can be described by ! !0 z < "a #! "a < z< 0 ! #E 2 . ET ( z ) = $ ! 0< z < a #E1 #! x >d %0 For region ! a < z < 0 : The Gaussian surface is shown on the left hand side of the figure on the previous page. Notice that the field is zero outside. So for our choice of Gaussian surface, on the cap ˆ ˆ inside the slab the unit normal for the area vector points in the positive zdirection, thus n = +k . So ! !! ˆˆ ˆ the dot product becomes E2 ! nda = E2, z k ! kda = E2, z da . Therefore the flux is " E ! da = E2, z Acap . ""
closed surface The charge enclosed is Qenclosed !0 = " #0 Acap ( a + z ) !0 where the length of the Gaussian cylinder is a + z since z < 0 . Substituting these two results into Gauss’s Law yields E2, z Acap = the electric field in the Ntype is given by E2, x = ! "0 Acap ( a + z ) #0 . Hence ! "0 ( a + z ) . The negative sign means that the electric #0 ! ! "0 ( a + z ) ˆ field point in the –z direction so the electric field vector is E2 = k . Note when z = !a then #0 ! ! ! # !0 a ˆ E2 = 0 and when z = 0 , E2 = k . We make a similar calculation for the electric field in the P"0 type noting that the charge density has changed sign and the expression for the length of the Gaussian cylinder is a ! z since z > 0 . Also the unit normal now points in the –zdirection. So the dot product ! ˆˆ ˆ becomes E ! nda = E ( " k ) ! kda = " E da . Thus Gauss’s
1 1, z 1, z Law becomes ! E1, z Acap = field is E1, z = ! + "0 Acap ( a ! z ) #0 . So the electric "0 ( a ! z ) . The vector description is then #0 ! !! #! a ˆ Note when z = a then E1 = 0 and when z = 0 , E1 = 0 k . "0 The graph of the electric field is shown in the figure.
Problem 4 (10 points): Coaxial Cylinders (a) The electric field is zero inside the inner conducting cylinder. (b) We use a Gaussian cylinder of length l and radius a < r < b . Then, the flux PS023 ! ! "0 ( a ! z ) ˆ E1 = k #0 is !! E ! dA = E 2# rl . The charge enclosed is given by Qenc = ! l = ( q / L)l . "" So Gauss’ Law becomes "E "" ! I !Q ! ql q1 ˆ ! dA = enc $ E 2% rl = $ E= r;a<r <b #0 L# 0 L2%# 0 r Problem 5 (10 points): Solid Sphere with a Cavity At first glance this charge distribution does not seem to have any of the symmetries that enable us to use Gauss’s law. However we can consider this charge distribution as the sum of two uniform spherical distributions of charge. The first is a sphere of radius 2 R centered at the origin with a uniform volume charge density ! . The second is a sphere of radius R centered at the point along the yaxis a distance R from the origin (the center of the spherical cavity) with a uniform volume charge density " ! . When we add together these two distributions of charge we obtain the uniform charged sphere with a spherical cavity of radius R as described in the problem because the positive and negative charges cancel where they overlap. We can then add together the electric fields from these two distributions at any point in the cavity to obtain the electric field of the original distribution at that point inside the cavity (superposition principle). Each of these two distributions are spherically symmetric and therefore we can use Gauss’s Law to find the electric field associated with each of them.. We do need to be careful when we add together the electric fields. As you will see that process is somewhat subtle and a good vector diagram will help considerably. So let’s begin by choosing a point P inside the cavity. We will now apply Gauss’s Law to our first distribution, the sphere of radius 2 R centered at the origin with a uniform volume charge density ! . The point P is a distance r < 2R from the origin. We choose a sphere of radius r as our Gaussian surface with r < 2R . Then, the electric flux through this closed ! ! surface is " E ! " dA = E! " 4$ r 2 , where E! denotes the outward normal ## component of the electric field at the point P associated to the spherical distribution with uniform volume charge density ! . Because the charge distribution is uniform, the charge enclosed in the Gaussisan surface is
Qenc ! (4" r 3 / 3) . So we substitute the two calculations = #0 #0 above into Gauss’ law to arrive at: E! $ 4" r 2 = ! !r ˆ ˆ ˆ field E ! ( P ) = E! r = r . where r is a unit vector at the point P pointing radially away from the 3" 0 origin.
We now apply Gauss’s Law to our second distribution, a sphere of radius R centered at the point along the yaxis a distance R from the origin with a uniform volume charge density " ! . The point P is a distance r ! < R from the center of the cavity. We choose a sphere of radius r ! as our Gaussian surface with r ! < R . Then, the electric flux through this PS024 ! (4" r 3 / 3) . We can solve this equation for the electric #0 closed surface is electric field at the point P associated to the spherical distribution with uniform volume charge density " ! . Because the charge distribution is uniform, the charge enclosed in the Gaussisan surface is Qenc ! (4" r $3 / 3) . Therefore applying Gauss’s Law yields =% #0 #0 ! ! E! " # dA = E! " # 4% r & 2 , where E" ! denotes the outward normal component of the " $$ ! (4" r %3 / 3) . We can solve this equation for the electric #0 ! ! r$ ˆ ˆ ˆ field E# ! ( P) = E# ! r$ = # r$ where r! is a unit vector at the point 3" 0 P pointing radially away from the center of the cavity.
E$ ! & 4" r 2 = $ The electric field associated with our original distribution is then ! ! ! ! !r ! r$ ! !!! ˆ ˆ ˆ ˆ ˆ ˆ (rr # r $r$) = (r # r$) where r is a E ( P ) = E ! ( P ) + E # ! ( P ) = E ! r + E# ! r $ = r# r$ = 3" 0 3" 0 3" 0 3" 0 ! vector from the origin to the point P and r! is a vector from the center !!! of the cavity to the point P . From our diagram we see that a = r " r! . ! !! Therefore the electric field at the point P is given by E( P) = a. 3" 0 This is a remarkable result. The electric field inside the cavity is uniform. The direction of the electric field points from the center of entire sphere to the center of the cavity. This direction is uniquely specified and is an example of ‘broken symmetry’. Problem 6 (10 points): Expt. 1: Equipotential Lines and Electric Fields PreLab Questions A. Equipotentials Curves – Reading Topographic Maps. (a) You can tell how steep something is by looking at how quickly it passes through constant height contours (~ equipotentials). The steepest section is along Octavia between Pacific and Washington. The most level street is Jackson between Buchanan and Octavia, which runs parallel to the 275 foot contour and hence is very flat. (b) Looking at Octavia, it passes through 5 contours (125 feet) in two blocks (about 0.12 miles) so it has a slope of ~1000 ft/mi. (c) Work is change in potential energy (and hence height). The change in height walking 3 blocks S on Laguna is almost nothing (you go up but come back down again). West on Clay from Franklin you rise 50 feet in the block, so that is more work. B: Equipotentials, Electric Fields and Charge One group did this lab and measured the equipotentials (the magenta circles) at V = 0.25 V, 0.5V and then from V = 1 V to V = 10 V in 1 V increments. They followed the convention that red was their positive electrode (V = +10 V) and blue was ground (V = 0 V). (a) Sketch eight electric field lines on it (equally spaced around the inner conductor). Solution: See black arrows (b) What, approximately, is the magnitude of the electric field at r = 1 cm, 2 PS025 cm, and 3 cm, where r is measured from the center of the inner conductor? Express the field in V/cm. Solution: At r = 1 cm, V ~ 4 V and we move 1 V in about 1/5 cm. E ~ 5 V/cm At r = 2 cm, V ~ 1.5 V and we move about 1/2 V in 1/2 cm. E ~ 1 V/cm At r = 3 cm, V ~ 0.7 V and we move about 0.2 V in 1/2 cm. E ~ 0.4 V/cm (c) What is the relationship between the density of the equipotential lines, the density of the electric field lines, and the strength of the electric field? Solution: The denser the equipotential lines and hence electric field lines, the stronger the field
Electric Field (V/cm)
4 3 (d) Plot the field strength vs. 1/r2 for the three points from part (a). If the field were created by a single point charge what shape should this sketch be? Is it? Solution: It should be (and is!) a straight line (e) Approximately how much charge was on the inner conductor when the group made their measurements? Solution: q E = ke 2 , so slope is ke q = 5 V cm. q " 5 # 10!12 C r C: Finding the Electric Field from the Electric Potential The graph shows the variation of an electric potential V with distance x. (a) What is Ex in the region x > !1 m ? (Be careful to indicate the sign of Ex .) Solution: In the region x > !1 m , V ( x) = 5 V ! (5 V " m 1 ) x . So d Ex = ! V ( x) = 5 V " m 1 dx (b) What is Ex in the region x < !1 m ? (Be careful to indicate the sign of Ex .) 5 2 Linear Fit: Slope ~ 5 V cm 1 0 0.0 0.2 0.4 0.6 0.8 1.0 1/r2 (cm2) Solution: In the region x < !1 m , V ( x) = 20 V+(10 V ! m 1 ) x . So d Ex = ! V ( x) = !10 V " m 1 . dx (c) A negatively charged dust particle with mass mq = 1"10!13 kg and charge q = !1" 10!12 C is released from rest at x = +2 m . Will it move to the left or to the right? Solution: For x > !1 m , the electric field is pointing in the positive xdirection, so a negatively charged particle will experience a force pointing in the negative xdirection, hence it will move to the left. PS026 ...
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This note was uploaded on 10/05/2010 for the course MIT 802FALL taught by Professor Stephens during the Fall '10 term at MIT.
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