1 Chapter 28 - Exercise 8 Physics 9 Fall 2009 Homework 3 - s The cube in the figure contains no net charge The electric field is constant over each face of the cube Does the missing electric field vector on the front face point in or out? What is the field strength? Because the electric field is constant, and because the cube contains no net charge, the total flux through the cube has to be zero The flux into the cube is 15 A + 20 A = 35A, where A is the area of the cubes face The flux out of the the cube is 10 A + 15 A + 15 A = 40A This means that more flux is leaving the cube than entering it Because the fluxes in have to balance the fluxes out, we see that the flux on the front face has to point in, and have a magnitude of 5 N/C 1
2 Chapter 28 - Exercise 11 The electric flux through the surface shown in the figure is 25 Nm 2 /C What is the electric field strength? Since the electric field is constant, the net flux through the surface is Φ E = E A = EA cos θ The electric field makes an angle of 60 with the surface of the area, meaning that it makes an angle of θ = 30 with respect to the normal of the surface, which is what we want Now, Φ E = EA cos θ E = Φ E /A cos θ With numbers we find E = Φ E A cos θ = 25 (01) 2 = 2890 N/C cos 30 2
3 Chapter 28 - Exercise 13 A 20 cm 30 cm rectangle lies in the xz plane What is the electric flux through the rectangle if (a) E ( ) = 50î + 100ˆk N/C? (b) E ( ) = 50î + 100ĵ N/C? The rectangle lies in the xz plane, and so its normal points along the y direction So, A = A y ĵ, where A y = 002 003 = 6 10 4 m 2 Thus, A = 6 10 4 ĵ (a) Since the flux is Φ E = E A, and since î ĵ = î ˆk = 0, then the net flux in part (a) is zero! (b) Again, Φ E = E A, and since î ĵ = 0, and ĵ ĵ = 1, we have Φ E = E A = E y A y = 100 6 10 6 = 006 N/m 2 3
4 Chapter 28 - Exercise 29 The figure shows four sides of a 30 cm 30 cm 30 cm cube (a) What are the electric fluxes Φ 1 to Φ 4 through sides 1 to 4? (b) What is the net flux through these four sides? (a) Again, the flux is Φ E = AE cos θ On side 1, the angle between E and the normal is θ 1 = 150 For side 2, it s θ 2 = 60, for side 3 it s θ 3 = 30, and for side 4 it s θ 4 = 120 Thus, Φ 1 = EA cos θ 1 = (500) (03) 2 cos 150 = 039 N/Cm 2 Φ 2 = EA cos θ 2 = (500) (03) 2 cos 60 = 0225 N/Cm 2 Φ 3 = EA cos θ 3 = (500) (03) 2 cos 30 = 039 N/Cm 2 Φ 4 = EA cos θ 4 = (500) (03) 2 cos 120 = 0225 N/Cm 2 (b) The net flux through the four sides is just the sum of the individual fluxes, Φ = Φ 1 + Φ 2 + Φ 3 + Φ 4 = 0, which we know has to be the case, since there is no enclosed net charge 4
5 Chapter 28 - Problem 39 A hollow metal sphere has inner radius a and outer radius b The hollow sphere has charge +2Q A point charge +Q sits at the center of the hollow sphere (a) Determine the electric fields in the three regions r a, a < r < b, and r b (b) How much charge is on the inside surface of the hollow sphere? On the exterior surface? The setup is seen in the figure to the right There is a single point charge, Q at the center of the sphere The inner radius of the sphere is a, and the outer radius is b We need to determine the fields in each of the regions using Gauss s law (a) Gauss s law says that E d A = Q encl Inside the sphere, for r a, the net enclosed charge is entirely from the point charge at the center So, Q encl = +Q Now, the spherical symmetry of the charge says that we should take a sphere of radius r < a as our Gaussian surface Everywhere on that sphere, the electric field is constant Furthermore, the direction of the electric field is perpendicular to the direction of the normal to the Gaussian surface (both point radially) So, E d A = E da = E da = EA, where A is the area of the surface Since the surface area of sphere is 4πr 2, we find that E (r) = 1 Q r 2, as expected Next, let s look outside the sphere, where r b In this case the net enclosed charge is +3Q, from the center point charge, and the charged sphere We again want a spherical Gaussian surface Proceeding as before we find E (r) = 1 3Q r 2 Finally, what happens inside the metal itself? This is a static collection of charges The charges inside the metal sphere feel the positive charge inside the shell They move around due to the electrical force until they have reached a configuration where all the force cancels out in other words, until the electric field is zero What happens? Since there is a positive charge inside, negative charges in the 5
metal shell are attracted to it and move to coat the inner surface with a negative charge, Q, which is just enough to cancel the positive point charge This leaves a net positive charge in the shell which then pushes away from itself, and onto the surface of the sphere So, this charge is now on the surface, in addition to the +2Q that was there before! Thus, the net charge on the surface of the shell is +3Q! So, to recap, 1 Q r a r 2 E (r) = 0 a < r < b 1 3Q r 2 r b (b) As discussed above, there is Q smeared over the inside surface, and +3Q on the outside 6
6 Chapter 28 - Problem 44 A positive point charge q sits at the center of a hollow spherical shell The shell, with radius R and negligible thickness, has net charge 2q Find an expression for the electric field strength (a) inside the sphere, r < R, and (b) outside the sphere, r < R In what direction does the electric field point in each case? (a) Again, let s start by drawing a picture, seen to the right Taking a Gaussian surface inside the shell, r < R, encloses only the positive charge +q So, by Gauss s law, E da = E ( 4πr 2) = q, which says that E = 1 q ˆr, for r 2 r R, and points radially outward (b) Now, outside the shell, a Gaussian surface encloses both charges, and so the net enclosed charge is Q encl = 2q+q = q So, Gauss s theorem says that E da = q Evaluating the integral as before gives E = 1 q ˆr, for r R, and points r 2 radially inward 7
7 Chapter 28 - Problem 45 Find the electric field inside and outside a hollow plastic ball of radius R that has charge Q uniformly distributed on its outer surface Inside the sphere the field is zero there s no charge inside the ball; it s all on the surface So, inside E = 0, for r < R Outside the ball, we do see a net charge of Q Since the ball is spherically symmetric, we choose a spherical Gaussian surface of radius r As before, the electric field is constant everywhere on the surface because of the symmetry So, as we ve seen before, E d A = EA = Q encl = Q, and so we just find the ordinary Coulomb s law, E = 1 like an ordinary point charge, as we ve seen before Q r 2 So, the field looks just 8
8 Chapter 28 - Problem 53 A spherical shell has inner radius R in and outer radius R out The shell contains total charge Q, uniformly distributed The interior of the shell is empty of charge and matter (a) Find the electric field outside of the shell, r R out (b) Find the electric field in the interior of the shell, r R in (c) Find the electric field within the shell, R in r R out (d) Show that your solutions match at both the inner and outer boundaries (e) Draw a graph of E versus r Once again, we will use Gauss s Law to find the electric field in each region (a) Outside of the shell, when r R out, the field looks like that of a point charge, since any spherical Gaussian surface encloses the full charge So, E = 1 Q ˆr r 2 (b) Inside the shell, where r R in, there is no charge and hence nothing to source the electric field So, the electric field is zero, E = 0 (c) Now, this gets a bit tougher Now, we want to use a Gaussian surface of radius R in r R out ; ie, the surface is inside the metal itself Since the total charge on the shell is uniformly distributed, the Gaussian surface encloses a certain amount of charge The charge density is constant, so ρ = Q/V, where V is the volume of the charge distribution The total volume of the sphere is just the difference between two spheres of radii R out and R in, V = 4 3 π (R3 out R 3 in) Thus, ρ = 3Q 4π(R 3 out R3 in) Now the Gaussian surface encloses a volume ( V encl ) = 4π 3 (r3 Rin), 3 which gives an r enclosed charge of Q encl = ρv encl = Q 3 Rin 3 Again, using Gauss s law we Rout 3 R3 in find that E da = EA = E (4πr 2 ) = Q encl, and so E = 1 ( ) Q r 3 Rin 3 ˆr r 2 Rout 3 Rin 3 Again, we find the following field values 0 ( ) r R in 1 Q r E (r) = 3 R 3 r 2 in ˆr R Rout 3 in r R out R3 in 1 Q ˆr r R r 2 out 9
(d) Now, the boundaries are at the values r = R in and R out We want to look at the field inside( the metal, ) when R in r R out When r = R in we get E (R in ) = 1 Q R 3 in R 3 Rin 2 in ˆr = 0 This matches the value inside the sphere On Rout 3 R3 in the other boundary, when r = R out, then E ( ) = 1 R 3 out Rin 3 ˆr = 1 Q ˆr, Rout 2 (e) Q R 2 out R 3 out R3 in which exactly matches our solution for r R out on the surface So, we can trust our solutions! The electric field is plotted in the graph to the right The field starts out at zero at the origin, and remains so until it reaches the inner surface at r = R in Then the field begins to rise, not quite linearly, going as r r 2, until it reaches the outer surface, r = R out This is where it reaches its maximum, after which it begins to fall off with the usual r 2 Coulomb law 10
9 Chapter 28 - Problem 54 An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge Ze Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere (a) Show that the electric field inside this atom is E in = Ze ( 1 r 2 r R 3 (b) What is E at the surface of the atom? Is this the expected value? Explain (c) A uranium atom has Z = 92 and R = 010 nm What is the electric field strength at r = 1 2 R? ) The atom is seen in the figure to the right It contains a positively charged nucleus of charge +Ze, which is embedded inside a uniform sphere of charge Ze We are interested in the field inside the sphere, so we take a spherical gaussian surface of radius r (a) Since the electric field is radial, and everywhere constant on the Gaussian surface, Gauss s law gives the usual result that E da = EA = E (4πr 2 ) = Q encl Now, what s Q encl? This problem is very similar to the last one, except with the addition of +Ze at the center So, the total enclosed charge is Q encl = +Ze negative charge ( The negative charge is just the charge density, times the enclosed volume, 4π 3 r3 = Ze r3 Thus, R 3 ) Ze 4 3 πr3 ) Q encl = Ze (1 r3, R 3 which, using Gauss s law as discussed above, gives E in = Ze ( 1 r r ) 2 R 3 (b) On the surface of the atom, r = R, and so the electric field is zero This is to be expected, since on the surface the enclosed charge is zero (+Ze Ze), and so the electric flux is zero from Gauss s law The atom looks neutral outside the surface 11
(c) When r = R/2, then E ( ) R = Ze ( 4 2 R 1 ) 2 2R 2 = 7 Ze 2 R 2 Plugging in the numbers gives ( ) R E = 7 Ze 2 2 R = 7 92 1602 10 19 9 10 9 2 2 (01 10 9 ) 2 = 464 10 13 N/C So, E = 464 10 13 N/C, which is a very strong electric field! 12
10 Chapter 28 - Problem 58 A sphere of radius R has total charge Q The volume charge density (C/m 3 ) within the sphere is ( ρ = ρ 0 1 r ) R This charge density decreases linearly from ρ 0 at the center to zero at the edge of the sphere (a) Show that ρ 0 = 3Q/πR 2 Hint: You ll need to do a volume integral (b) Show that the electric field inside the sphere points radially outward with magnitude E = Qr (4 3 r ) R 3 R (c) Show that your result of part b has the expected value at r = R (a) Now, the charge density is not constant! So, we can t just say that Q = ρv Instead, we have to integrate, Q = dq = ρdv As we ve seen before, for a sphere dv = 4πr 2 dr, as you can see by imagining a series of thin shells of thickness dr piled on top of each other So, R ( Q = ρdv = 4πρ 0 1 r ) [ ] r r 2 3 dr = 4πρ 0 R 3 r4 R = π 4R 3 ρ 0R 3, 0 and so ρ 0 = 3Q πr 3 (b) Now, we take a Gaussian surface of radius r inside the sphere From Gauss s law, E da = Q encl, the left-hand side is E (4πr 2 ) because the field is radial inside the sphere What s Q encl? Again, we integrate as before, only now out to a radius r, corresponding to our surface r Q encl = ρdv = 4πρ 0 Plugging for ρ 0 gives Thus, from Gauss s law 0 ( 1 r R Q encl = 4πρ 0 3 E (r) = ) r 2 dr = 4πρ 0 [ r 3 ] 3 r4 4R ] [r 3 3r4 = Q [4 r3 3 r ] 4R R 3 R Qr (4 3 r R 3 R ) 0 = 4πρ 0 3 ] [r 3 3r4 4R ( ) (c) At r = R, then E (R) = QR R 4 3 R 3 R = Q So, the electric field at r = R R 2 is that of a point charge, which is exactly we should expect! 13