Homework 3: HRW Chapter 23, Problem # 11, 29, 39, 49 11. None of the constant terms will result in a nonzero contribution to the flux (see Eq. 23-4 and Eq. 23-7), so we focus on the x dependent term only: Enonconstant = (−4.00y2 ) i^ (in SI units) . The face of the cube located at y = 4.00 has area A = 4.00 m2 (and it “faces” the +j^ direction) and has a “contribution” to the flux equal to Enonconstant A = (−4)(42)(4) = –256 N·m/C2. The face of the cube located at y = 2.00 m has the same area A (however, this one “faces” the –j^ direction) and a contribution to the flux: −Enonconstant A = − (−4)(22)(4) = 64 N·m/C2. Thus, the net flux is Φ = (−256 + 64) N·m/C2 = −192 N·m/C2. According to Gauss’s law, we therefore have 12 2 2 2 9enc 0(8.85 10 C /N m )( 192 N m C) 1.70 10 C.qε− −= Φ = × ⋅ − ⋅ = − × 29. The charge densities of both the conducting cylinder and the shell are uniform, and we neglect fringing effect. Symmetry can be used to show that the electric field is radial, both between the cylinder and the shell and outside the shell. It is zero, of course, inside the cylinder and inside the shell. We take the Gaussian surface to be a cylinder of length L, coaxial with the given cylinders and of radius r. The flux through this surface is 2 ,rLEΦ = π where E is the magnitude of the field at the Gaussian surface. We may ignore any flux through the ends. Gauss’ law yields enc 0 02 ,q r LEε π ε= Φ = where qenc is the charge enclosed by the Gaussian surface. (a) In this case, we take the radius of our Gaussian cylinder to be 3 22 12.00 20.0 (20.0)(1.3 10 m) 2.6 10 m.r R R− −= = = × = × The charge enclosed is qenc = Q1+Q2 = –Q1 = –3.40×10−12 C. Consequently, Gauss’ law yields 12enc12 2 2 203.40 10 C0.214 N/C,2 2 (8.85 10 C / N m ) (11.0 m)(2.60 10 m)qELrε π−− −− ×= = = −π × ⋅ × or | | 0.214 N/C.E= (b) The negative sign in E indicates that the field points inward.(c) Next, for r = 5.00 R1, the charge enclosed by the Gaussian surface is qenc = Q1 = 3.40×10−12 C. Consequently, Gauss’ law yields 0 enc2 ,r LE qπ ε= or 12enc12 2 2 303.40 10 C0.855 N/C.2 2 (8.85 10 C / N m ) (11.0 m)(5.00 1.30 10 m)qELrπε π−− −×= = =× ⋅ × × (d) The positive sign indicates that the field points outward. (e) We consider a cylindrical Gaussian surface whose radius places it within the shell itself. The electric field is zero at all points on the surface since any field within a conducting material would lead to current flow (and thus to a situation other than the electrostatic ones being considered here), so the total electric flux through the Gaussian surface is zero and the net charge within it is zero (by Gauss’ law). Since the central rod has charge Q1, the inner surface of the shell must have charge Qin = –Q1 = –3.40×10−12 C. (f) Since the shell is known to have total charge Q2 = –2.00Q1, it must have charge Qout = Q2 – Qin = –Q1 = –3.40×10−12 C on its outer surface. Cylindrical symmetry of the system allows us to apply Gauss’ law to the problem. Since electric field is zero inside the conducting shell, by Gauss’ law, any net charge must be distributed on the surfaces of the shells. 39. Since the non-conducting charged ball is in equilibrium with the non-conducting charged sheet (see Fig. 23-49), both the vertical and horizontal components of the net force on the ball must be zero. The forces acting on the ball are shown in the diagram to the right. The gravitational force has magnitude mg, where m is the mass of the ball; the electrical force has magnitude qE, where q is the charge on the ball and E is the magnitude of the electric field at the position of the ball; and the tension in the thread is denoted by T. The electric field produced by the plate is normal to the plate and points to the right. Since the ball is positively charged, the electric force on it also points to the right. The tension in the thread makes the angle θ (= 30°) with the vertical. Since the ball is in equilibrium the net force on it vanishes. The sum of the horizontal components yields qE – T sin θ = 0 and the sum of the vertical components yields cos 0T mgθ− =. We solve for the electric field E and deduce σ, the charge density of the sheet, from E = σ/2ε0 (see Eq. 23-13). The expression T = qE/sin θ, from the first equation, is substituted into the second to obtain qE = mg tan θ. The electric field produced by a large uniform sheet of charge is given by E = σ/2ε0, so0tan2qmgσθε= and we have ()()()12 2 2 6 2089 22 8.85 10 C / N m 1.0 10 kg 9.8 m/s tan 302 tan2.0 10 C5.0 10 C/m .mgqε θσ− −−−× ⋅ × °= =×= × Since both the sheet and the ball are positively charged, the force between them is repulsive. This is balanced by the horizontal component of the tension in the thread. The angle the thread makes with the vertical direction increases with the charge density of the sheet. 49. At all points where there is an electric field, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric field is to be found. The field is uniform on the surface, so 24E dA r E⋅ = π∫, where r is the radius of the Gaussian surface. For r < a, the charge enclosed by the Gaussian surface is q1(r/a)3. Gauss’ law yields 321 130 04 .4q q rrr E Ea aπε πε  =⇒=     (a) For r = 0, the above equation implies E = 0. (b) For r = a/2, we have 9 2 2 15213 2 20( / 2) (8.99 10 N m /C )(5.00 10 C)5.62 10 N/C.4 2(2.00 10 m)q aEaπε−−−× ⋅ ×= = = ×× (c) For r = a, we have 9 2 2 1512 2 20(8.99 10 N m /C )(5.00 10 C)0.112 N/C.4 (2.00 10 m)qEaπε−−× ⋅ ×= = =× In the case where a < r < b, the charge enclosed by the Gaussian surface is q1, so Gauss’ law leads to 21 120 04 .4q qr E Erπε πε= ⇒ = (d) For r = 1.50a, we have 9 2 2 1512 2 20(8.99 10 N m /C )(5.00 10 C)0.0499 N/C.4 (1.50 2.00 10 m)qErπε−−× ⋅ ×= = =× × (e) …