Council of Student Organizations De La Salle University Manila

Council of Student Organizations De La Salle University Manila PHYENG2 Quiz 1 Problem Solving: 1. (a) Find the magnitude and direction of the force of +Q on q o at (i) P 1 and (ii) P 2 in Fig 1a below. (b) Find the magnitude and direction of the force of -Q on q o at (i) P 1 and (ii) P 2 in Fig.1b below. Take Q = 2 x 10-6 C and q o = 10-12 C. 2. Now imagine that all the +q o 's are removed from Fig. 1 above. Find the magnitude and the direction of the electric field (a) in Fig. 1a and (b) in Fig. 1b at P 1 and P 2. 3. (a) Find the electric field at point P in Fig. 2 below. (b) Repeat for q 2 = +1 nc. 4. In Fig. 3 below, q 1 = +1.00 µc and q 2 = -0.0800 µc. Find the direction and magnitude of the electric field at point P. 5. Find (a) the electric field, direction and magnitude, at P (4.00 m, 5.00m) in Fig. 4 below due to q 1 = +5.00 x 10-6 C at (0, 2.00 m), q 2 = -3.00 x 10-6 C at (4.00 m, 0), and q 3 = +1.6 x 10-6 C at (0, 5.00 m) and (b) the force on q 4 = +2.0 x 10-6 C placed at P.

6. A straight positively charged line coincides with the X-axis and carries a charge per unit length of λ. While the charge is actually discrete, it is useful to think of the charge spread continuously along the line. Find: (a) the charge dq on a length dx of the line, (b) the magnitude of de 1 or de 2 at point P along the Y-axis for the symmetrically displaced elements of length dx 1 and dx 2 in Fig. 5 below, (c) the direction of de1 + de 2, and (d) the magnitude of the component of de1 or de 2 in the direction of the resultant field. Now we can integrate to find the resultant field. Remember that integration is not a vector addition, so we cannot integrate until we have found the resultant direction of the field. (e) In Fig. 5 we have labeled various lengths and an angle: b, x, r, and Θ. Which of these is a constant? (f) Write x in terms of b and a function of Θ. (g) Find an expression for the resultant dey in terms of k, λ, b, a function of Θ and dθ. (h) If you were integrating with respect to x, the lower limit would be -3b/4 and the upper limit would be +3b/4. What are the lower and upper limits on the sine of Θ? (i) Complete the calculation of the resultant electric field at P in Fig. 5 below.

7. A thin nonconducting wire is bent into a semicircle of radius R (Fig. 6 below). The wire is charged so that the linear density λ = λ o cos Θ, where λ o is a constant. Find the direction and the magnitude of the electric field at the center of the semicircle at point O. 8. A cube with sides of 0.20 m has its center at the origin of a rectangular coordinate system and its faces perpendicular to the coordinate axes. The electric field is E = 100 N/C-m xi. Determine the amount of charge within the cube. 9. In Fig. 7 below, the dashed spherical surface is a Gaussian surface of radius r. The solid sphere is a uniform distribution of charge of radius R. If spherical symmetry is to be preserved, the electric field E at P remains constant when the solid sphere is rotated. If this is true, (a) can the electric field depend on angles Θ or Φ? (b) must the electric field E always be radially outward (or inward if the charge is negative)? and (c) must the electric field be constant in magnitude everywhere on the Gaussian surface of radius r? (d) If the angle between the electric field and the surface, E, da, is zero how can you write E. da? Remember the direction of the surface is always taken as the outward normal. (e) If E is a constant, how may you write for the spherical Gaussian surface? 10. The thick, spherical shell of inner radius a and outer radius b shown in Fig. 8 below carries a uniform volume charge density ρwith total charge Q. Find: (a) ρin terms of Q, a, and b, (b) the charge enclosed by a spherical Gaussian surface of radius (i) r < a, (ii) a < r < b, and (iii) r > b, and (c) the electric field for (i) r < a, (ii) a < r < b, and (iii) r > b. (d) Use the expressions found in (c) to show that these also are the electric r b, and (iii) r b. field for (i) r a, (ii) a

11. What is (a) the lateral surface area and (b) the volume of a right circular cylinder of radius r and length L? 12. Use Gauss's theorem to determine the electric field due to a very long wire carrying a charge per unit length + λ a distance r from the wire. 13. The dashed spherical surface in Fig. 9 is a Gaussian surface, which surrounds a dipole. The solid curves with arrows represent the field lines for the dipole. Can Gauss's theorem be used to find the field of a dipole? Explain your answer. 14. The electric field outside and at a distance of 0.5 m from the center of a charged sphere, the axis of a long charged cylinder or an infinite sheet of charge is 144 N/C. For which of these charge distributions do you find an electric field of (a) 72 N/C, (b) 36 N/C, (c) 144 N/C at a distance of 1.0 m? Explain your answers. Find (d) the charge q on the sphere, (e) the charge per unit length λ on the cylinder and (f) the charge per unit area σ on the infinite sheet. 15. Three infinite sheets of charge are parallel to each other, as is shown in Fig. 10. The sheet on the left has a uniform surface charge density +σ, the one in the middle a uniform surface charge density - σ, and the one on the right a uniform surface charge density of +σ. Find the electric field at (a) P 1, (b) P 2, (c) P 3, and (d) P 4.

16. A conducting sphere of charge +Q and radius a is concentric with a very thin conducting spherical shell of charge - Q and radius b > a. (a) Where are the charges? Find the electric field for (b) r a, (c) a r b, (d) r b. 17. An infinitely long insulating cylinder of radius R has a volume charge density that varies with the radius as ρ = ρ o (a - cr), where ρ o, a, and c are positive constants and r is the distance from the axis of the cylinder. Find the electric field at radial distances for (a) r < R and (b) r > R. 18. Can electric field lines intersect? Why or why not? 19. A solid sheet of nonconducting material of thickness t and infinite length and width has a uniform positive charge density ρ throughout the sheet. (a) Use Gauss's law to find the electric field as a function of the distance x from the center of the sheet for - t/2 < x < t/2. (b) If a small hole is drilled though the sheet and a particle of mass m and charge -q is released from rest on one side of the sheet, what is the frequency of the particle's motion? 20. A uniform electric field E is set up between two metal plates of length L and spacing d, as shown in the figure to the right. An electron enters the region midway between the plates moving horizontally with speed v. Find an expression for the minimum speed the electron must have to get through the region without hitting either plate. Neglect gravitational effects. 21. An electron is moving in a circular path around a long, uniformly charged wire carrying 2.5 nc/m. What is the speed of the electron? 22. A small sphere with mass m carries charge q. It hangs from a silk thread that makes an angle Θ with a large, charged nonconducting sheet, as illustrated in the figure below. Find the sheet's surface charge density σ.

23. Two identical dipoles are placed in a straight line as shown in Fig. 11a below. Find the direction of the electric force on each dipole in Fig. 11a. Repeat for Fig. 11b below. 24. An electric dipole consisting of charges +3.2 x 10-19 C and -3.2 x 10-19 C separated by 2.0 x 10-9 m is in a field of 5.0 x 10 5 N/C. Calculate the torque on the dipole when the dipole moment is (a) parallel and in the same direction as the field (b) perpendicular to the field and (c) parallel and in the opposite direction of the field. 25. Two positive point charges repel each other. Explain this experimental result in terms of the electric field of one charge action on the other charge. 26. There are really two types of problems concerning electric fields: a. Given a distribution of charges, find the electric field due to them. b. Given an electric field E, find the electric force F e on a charge q. Explain two ways of calculating electric fields and how you find the electric force on a charge q in an electric field. 27. One of two identical nonconducting rods of length L carries a total positive charge Q and the other carries a total charge of - Q as shown in Fig. 12 below. (a) Find the electric field at point P in Fig. 12 due to these charge distributions as a function of x. (b) What does this expression in (b) reduce to for x >> L. (c) Identify a dipole moment for this distribution.

Answer: 1. Answers - Electric Fields a. The magnitude of the force of +Q or -Q on q o at r = 3 m is F qq = kqq o /r 2 = (9.0 x 10 9 N- m 2 /C 2 )(2 x 10-6 C)(10-12 C)/(9 m 2 ) = 2 x 10-9 N. The force of +Q on +q 0 at i. P1 is to the right since +Q repels +q o and urges it to the right ii. P2 is to the left since +Q repels +q o and urges it to the left. b. The force of -Q on + q o at i. P1 is to the left since -Q attracts +q o and urges it to the left ii. P2 to the right since -Q attracts +q o and urges it to the right. 2. By definition, the direction of the electric field is the direction in which a positive test charge is urged. The magnitude of the electric field at a point P is equal to the electric force F e on a test charge q o divided by the test charge. For all points in Fig. 1 above, the magnitude of the electric field = (F e )/q o = (kqq o /r 2 )/q o = kq/r 2 = (9 x 10 9 N-m 2 /C 2 )(2 x 10-6 C)/9 m 2 = 2 x 10 3 N/C. a. In Fig. 1a, the electric field at P1 is to the right because a positive test charge there would be urged to the right. At P 2 the electric field is to the left because a positive test charge there would be urged to the left. b. In Fig. 1b, the electric field at P1 is to the left because a positive test charge there would be urged to the left. At P 2 the electric field is to the right because a positive test charge there would be urged to the right. Do not use the sign of a charge setting up a field to determine the direction of a field. Go to the point, imagine placing a positive test charge there, and ask what direction the test charge would be urged by the charge setting up the field. 3. a. In general the electric field E due to a point charge Q a distance r from the point where you wish to find the field is given by

E = (9 x 10 9 N-m 2 /C 2 )Q/r 2. For q1 in Fig. 2, E 1 = (9 x 10 9 Nm 2 /C 2 )(9 x 10-9 C)/(3 m) 2 = 9 N/C to the right because if a positive test charge is placed at P, q 1 would urge it to the right. E 2 = (9 x 10 9 N-m 2 /C 2 )(1 x 10-9 C)/ (1 m) 2 = 9 N/C to the right because if a positive test charge is placed at P, q 2 would urge it to the right. The resultant field at P is E = (9 + 9) N/C = 18 N/C to the right. Again notice the importance of not using the negative sign of the charge in finding the field. If you had used the negative sign, you might have said the field due to q 2 was to the left. b. If q2 = +1 nc, the electric field due to q 2 at P is 9 N/C to the left. Now taking to the right as positive, the resultant field at P is E = (9-9) N/C = 0. 4. In Fig. for #4 above, the hypotenuse of the triangle = (9.00 + 16.00) 1/2 m = 5.00 m. The field due to q 1, E 1 = q 1 /4πε o r 1 2 = 9 x 10 9 N-m 2 /C 2 (10-6 C/25.0m 2 ) = 360 N/C. The direction of E 1 is shown in Fig. for #4. E 2 = 9 x 10 9 N-m 2 /C 2 (0.0800 x 10-6 C/16.00 m 2 ) = 45 N/C, and E 2 = -45 j N/C, as shown in Fig. for #4. Taking components of E 1, (E 1 ) x = E 1 cos 53 o = 360 N/C (0.60) = 216 N/C, and (E 1 ) y = E 1 sin 53 o = 360 N/C (0.80) = 288 N/C. The component of the resultant E in the y direction E y = (E 1 ) y + (E 2 ) y = (288-45) N/C = 243 N/C. E = (216 i + 243 j)n/c. E = {(216) 2 + (243) 2 } 1/2 = 325 N/C. tan Θ = 243/216. Θ = 48.4 o with X-axis. 5.

a. In the Fig. for #5 above, E 1 represents the electric field due to q 1 = 5.00 x 10-6 C, which is r = 5.00 m from P. E 1 = kq 1 /r 2 = 9 x 10 9 N-m 2 /C 2 (5.00 x 10-6 C) 25.0 m 2 = 1.80 x 10 3 N/C. E 1x = E 1 cos Θ= 1.80 x 10 3 N/C (0.800) = 1.44 x 10 3 N/C. E 1y = E 1 sin Θ=1.80 x 10 3 N/C (0.600) =1.08 x 10 3 N/C. In the Fig. for #5 above, E 2 represents the electric field due to q 2 = -3.00 x 10-6 C, which is a distance of r 2 = 5.00 m from P. E 2 = kq 2 /r 2 = 9 x 10 9 N-m 2 /C 2 (3.00 x 10-6 )/25.0 m 2 = 1.08 x 10 3 N/C in the negative Y direction (Fig. for #5). E 2x = 0 and E 2y = -1.08 x 10 3 N/C. In the Fig. for #5, E 3 represents the electric field due to q 3 = 1.60 x 10-6 C, which is a distance of r 3 = 4.00 m from P. E 3 = kq 3 /r 3 = 9 x 10 9 N-m 2 /C 2 (1.60 x 10-6 )/16.0 m 2 = 0.900 x 10 3 N/C in the positive X direction as shown in Fig. for #5. E 3x = 0.900 x 10 3 N/C and E 3y = 0. Let the total electric field = E. E x = E 1x + E 2x + E 3x = (1.44 + 0 + 0.90)10 3 N/C = 2.34 x 10 3 N/C. E y = E 1y + E 2y + E 3y = (1.08-1.08 + 0)10 3 N/C = 0. The resultant electric field is totally in the +X direction and it is equal to 2.34 x 10 3 N/C. b. Electric force on q4 = q 4 E = 2.0 x 10-6 C(2.34 x 10 3 N/C) = 4.68 x 10-3 N.

6. a. λ = charge/length so that charge = λ(length) or dq = λ dx. b. Treating dq as a point charge, de = k ( λ dx)/r 2. c. The direction of an electric field is that in which a positive test charge is urged. If a positive test charge were at P, it would be urged up along the line connecting dq and P. de1 due to the charge at dx 1 and de 2 due to the charge at dx 2 are as shown in Fig. 5 above. As shown in the figure, the X-components of these two elements of electric field cancel. The resultant direction of the electric field due to these two elements of charge or any other two symmetrically displaced elements will be in the positive Y-direction. d. The magnitude of the component of de in the Y-direction dey = de cos Θ= k λdx cos Θ/r 2. e. Only b is constant. The quantities x, r, and Θ are variables. f. x = b tan Θ and dx = b sec 2 Θ dθ. g. r 2 = b 2 + x 2 = b 2 + b 2 tan 2 Θ = b 2 (1 + tan 2 Θ) = b 2 sec 2 Θ. de y = λkdx cos Θ/r 2 = λkb sec 2 Θ dθ cos Θ/b 2 sec 2 Θ= λkdθ cos Θ/b. h. When x = - 3b/4, r = [b 2 + (-3b/4) 2 ] 1/2 = b[16/16 + 9/16] 1/2 = 5b/4, which is also true for x = +3b/4. For lower limit, sin Θ= (-3b/4)/(5b/4) = - 3/5. For the upper limit, sin Θ = (3b/4)/(5b/4) = 3/5. i. λk dθ cos Θ/b = k λ sin Θ/b. Putting in the lower and upper limits, Ey = k λ/b [3/5 - (-3/5)] = 6k λ/5b.

7. The charge on an arc of length ds is dq = λds, where λ is the charge per unit length. If the arc length subtends an angle dθ in a circle of radius R, dq = λ(ds) = λ(r dθ). The electric field at a point P due to an element of length ds 1 is de1 = dq/4πε o R 2 = λ o cos ΘR dθ/4πε o R 2. The electric field de2 at point P due to an element of length ds 2 has the same magnitude as that of de 1, but not the same direction as de 1, as illustrated in Fig. 6 above. Since cos (180 o - Θ) = - cos Θ the charge is negative in quadrant 2. de 2 is in towards the arc. Components of the de's in the Y direction cancel. Components in the X direction are equal and add. (de 1 ) x = (de 2 ) x = de 1 cos Θ = λ o dθ (cos 2 Θ)/R = ( λ o dθ /4πε o R)(1/2 + 1/2 cos 2Θ)dΘ. The total electric field is to the left and 8. We must find the flux = E da leaving the sides of the cube. For all cases E = 100 N/C-m xi. For the right and left sides, da = ±da i and E da = (100 N/C-m)[( xi dai + xi -dai) = (100 N/C-m)[ 0.1m da + (-0.1m)(-dA)]

= (100N/C)[0.1m(0.04 m 2 )+(-0.1m)(-0.04 m 2 )] = 0.8 N-m 2 /C, since i i = 1, x = +0.1m on the right hand side, x = -0.1m on the right hand side, and da = (0.20 m x 0.20 m) = 0.04 m 2. For the top and bottom sides, E da = (100 N/C-m)( xi daj + xi -daj)] = 0, because i j = 0. For the front and back sides, E da = (100 N/C-m)( xi dak + xi -dak) = 0, because i k = 0. For all sides, E. da = 0.8 N-m 2 /C = charge enclosed/ ε o 0.8 N-m 2 /C = Q/(8.85 x 10-12 C 2 /N-m 2 ). Q = 0.8 C(8.85 x 10-12 ) = 7.1 x 10-12 C. 9. For spherical symmetry, the electric field at P must remain constant at point P despite the rotation of the sphere inside the Gaussian surface. If this is to be true, a. the electric field can not depend on angles Θ and Φ, b. the only possible direction for E is radially outward (or inward for negative charge), c. the magnitude of the electric field must be constant everywhere on the Gaussian surface. d. For E radially outward, E, da = 0 and E da = E da cos E, da = E da cos 0 o = E da. e. If E is constant, E da = E da = E 4 πr 2, where 4 πr 2 = the surface area of the Gaussian spherical surface of radius r. 10. In Fig 8 above, the dashed surfaces represent the Gaussian surfaces and the solid lines represent the spherical charged shell of inner radius a and outer radius b. a. charge/volume = ρ = Q/[(4 π/3)(b 3 - a 3 )], where the volume of the spherical shell is the difference between the volumes of a sphere of radius b and a sphere of radius a. b.

i. For r < a, no charge is enclosed. ii. For a < r < b, the Gaussian surface has a volume (4 π/3)(r 3 - a 3 ) and encloses an amount of charge = ρ(4 π/3)(r 3 - a 3 ) = [Q/(4 π/3)(b 3 - a 3 )](4 π/3)(r 3 - a 3 ) = Q(r 3 - a 3 )/(b 3 - a 3 ). iii. For r > b, the Gaussian surface encloses the total charge Q. c. For each of the Gaussian surfaces, E, da = 0 and E da = E da. Also E is constant everywhere on the Gaussian surface, so. For all cases then, E 4 πr 2 = charge enclosed/ ε i. For r < a, E 4 πr 2 = 0 E = 0 ii. For a < r < b, E 4 πr 2 = Q(r 3 - a 3 )/ ε o (b 3 - a 3 ) E = Q(r 3 - a 3 )/4πε o r 2 (b 3 - a 3 ) iii. For r > b, o. E 4 πr 2 = Q/ ε o E = Q/4πε o r d. For r = a, E = Q(a 3 - a 3 )/4πε o r 2 (b 3 - a 3 ) = 0 = E For r = b, E = Q(r 3 - a 3 )/4πε o r 2 (b 3 - a 3 ) = Q/4πε o b 2 For r = b, E = Q/4πε o r 2 = Q/4πε o b 2 Thus, i. For r a, E = 0 ii. For a r b, E = Q(r 3 - a 3 )/4πε o r 2 (b 3 - a 3 ) iii. For r b, 2 11. 2 E = Q/4πε o r a. The circumference of the base equals 2 πr, where r is the radius of the base. If you roll the cylinder out, it becomes a rectangular parallelepiped with width 2 πr and length L. Its area is the

width times the length or 2 πrl. b. The volume of the cylinder is the area of the base πr 2 times the length L, or πr 2 L. 12. In the Fig. for #12 above, the dashed cylindrical surface is the Gaussian surface of length L and radius r. If you are unable to detect the rotation of the long, thin wire, the electric field E must be radially outward for a positively charged wire and the magnitude of the electric field must be constant over the lateral part of the cylindrical Gaussian surface. For the ends of the Gaussian cylinder, E, da = 90 o and E da = 0. For the lateral surface E, da = 0 o and E da = E da. Since E is constant everywhere on the Gaussian surface, The charge enclosed by the Gaussian surface = (charge/length)length = λl. By Gauss's theorem,. From above, E 2 πrl = λl/ ε o or E = λ/2πε o r.. 13. If you rotate the dipole inside the spherical Gaussian surface, the electric field at a fixed point on the Gaussian surface would change. You cannot use a spherical Gaussian surface nor any other type of surface such that the electric field would preserve the symmetry. Thus you cannot use Gauss's theorem to find the electric field of a dipole. 14. a. For r = 0.5 m, E = 144 N/C. For r = 1.0 m, E = 72 N/C. As you double r, E is halved. E must be inversely proportional to r, that is, E 1/r. This corresponds to the field being set up by a long cylinder of charge. b. For r = 0.5 m, E = 144 N/C. For r = 1.0 m, E = 36 N/C. As you double r, E is reduced by a factor of four. E must be inversely proportional to the square of r, that is, E 1/r 2. This corresponds to the field outside a charged sphere. c. For r = 0.5 m and r = 1.0 m, E = 144 N/C. The electric field is constant regardless of the distance from the charge distribution. This corresponds to the field set up by an infinite sheet of charge. d. The field outside a sphere with charge q is E = kq/r 2. q = Er 2 /k = (144 N/C)(0.25 m 2 )/(9.0 x 10 9 N-m 2 /C 2 ) = 4.0 x 10-9 C. e. For a cylinder, E = λ/2πεor or λ= 2πε o re = 2 π(8.85 x 10-12 C 2 /N-m 2 )(0.5 m) (144 N/C) = 4.0 x 10-9 C/m. f. For an infinite sheet of charge, E = σ/2 εo or σ = 2 ε o E = 2(8.85 x 10-12 C 2 /N-m 2 )(144 N/C) = 2.55 x 10-9 C/m 2.

15. The electric field due to an infinite sheet of charge is σ/2 ε o. The direction of an electric field at a point P is that in which a positive test charge placed at that point would be urged. As shown in Fig. 10 above, a. At point P1 the electric field due to the sheet on the left E left is to the left, the electric field due to the center sheet E center is to the right, and the electric field due to the sheet on the right E right is to the left. Taking to the right to be positive, the electric field is to the left with magnitude E (at P 1 ) equals -σ/2 εo + σ/2 ε o - σ/2 ε o = - σ/2 ε o. b. At point P 2 the electric field due to the sheet on the left E left is to the right, the electric field due to the center sheet E center is to the right, and the electric field due to the sheet on the right E right is to the left. Taking to the right to be positive, the E is to the right with magnitude E (at P 2 ) equals +σ/2 εo + σ/2 ε o - σ/2 ε o = +σ/2 ε o. c. At point P 3 the electric field due to the sheet on the left E left is to the right, the electric field due to the center sheet E center is to the left, and the electric field due to the sheet on the right E right is to the left. Taking to the right to be positive, the electric field is to the left with magnitude E (at ) equals P 3 σ/2 εo - σ/2 ε o - σ/2 ε o = - σ/2 ε o. d. At point P4 the electric field due to the sheet on the left E left is to the right, the electric field due to the center sheet E center is to the left, and the electric field due to the sheet on the right E right is to the right. The electric field is to the right with magnitude E (at P 4 ) equals σ/2 εo - σ/2 ε o + σ/2 ε o = +σ/2 ε o.

16. In the figure above, the Gaussian surfaces for r < a, a < r < b, and r > b are shown by dashed surfaces. Using the same arguments we did in Problem 10, we establish that the electric field is perpendicular to the Gaussian surfaces and constant everywhere on them. The definition of a conductor is that charges are perfectly free to move in them. They move until there is no electric field inside the conductor that would produce a motion of charges. Since there is no electric field inside the sphere of radius a, there can be, in agreement with Gauss' theorem, no charges enclosed by a surface inside the sphere. That is, if = (0)4 πr 2 = charge enclosed/ ε o, the charge enclosed must equal zero. You can use this argument until r goes to a and decide the +Q is on the outside of the sphere of radius a. In addition, since there can be no charges inside of the spherical shell, the - Q must be on the inner surface of the spherical shell, as shown in the above figure. (a) From all of this we conclude that for r < a, E = 0. The Gaussian surface with radius r such that a < r < b, encloses +Q. By Gauss' theorem, E 4 πr 2 = charge enclosed/ ε o. (b) For a < r < b, E(4 πr 2 ) = Q/ ε o E = Q/4πε o r 2 2 = kq/r (c) For r > b, charge enclosed = + Q - Q = 0 so E(4 πr 2 ) = 0 or E = 0.

17. From Gauss' theorem, = charge enclosed/ ε o. By symmetry conditions, E, da = 0 and E is constant everywhere along the Gaussian surface. So for both r > R and r < R,. The charge enclosed by the Gaussian surface will differ for r < R and r > R. For r > R, the Gaussian surface will enclose the total charge in a cylinder of length L. Since the density of charge ρis a function of r, we need to find an expression for the charge enclosed. Because the density of charge ρ = charge/volume, for a volume dv, the charge enclosed dq = ρ dv. We must find the volume dv of a cylindrical shell of length L radius r and thickness dr, as shown in the lower drawing in Fig.11 above. If you slit this shell and rolled it out, you would have a plate of width 2 πr, length L, and thickness dr. Thus dv = 2 πrl dr. a. For r < R, For r < R, Gauss's theorem gives E2 πrl = (2πρ o L/6 ε o )(3ar 2-2cr 3 ), and E = (ρ o /6 ε o )(3ar - 2cr 2 ). b. For r > R, charge enclosed is found by using limits for the above integral of 0 and R. Charge enclosed = (2πρoL/6) (3aR 2-2cR 2 ), and now E2 πrl = (2πρoL/6 ε o )(3aR 2-2cR 3 ), and E = (ρ o /6r ε o )( (3aR 2-2cR 3 ) The expression of E for r < R at r = R reduces to E = (ρo/6 ε o )(3aR - 2cR 2 ), as does the expression of E for r > R. 18. The tangent to an electric field line at a point gives the direction of the electric field at that point. If two field lines intersect at that point, then the electric field would have two different directions at that point. Since this is impossible, electric field lines cannot intersect.

19. a. Because there is equal amount of positive charge on the sheet from x = - t/2 to x = t/2, the electric field at x = 0 is 0. By symmetry and the definition of the direction of an electric field, the electric field is to the right for x > 0 and to the left for x < 0. For the cap of the "beer can" Gaussian surface, E, da = 0. For the lateral surface E, da = 90 o. For the cap, E. da = E da. For the lateral surface, E. da = 0. By Gauss's theorem, E da = charge enclosed/ ε o = xda ρ/ ε o and E = ρx/ ε o. b. The force on the negative charge q = Eq = -(ρq/ εo)x, where the minus sign means the force is opposite to the displacement x. The quantities in the parenthesis (ρq/ ε o ) are constant, so this is an example of simple harmonic motion. Fnet = ma or -(ρq/ ε o )x = m d 2 x/dt 2 or d 2 x/dt 2 + (ρq/m ε o )x = 0. Compare with d 2 x/dt 2 + (k/m)x = 0, for which the frequency f = 1/2 π (k/m) and see for this case f = 1/2 π (ρq/m ε o ) 1/2. 1/2 20. X Y x o = 0 yo = 0 vox = v voy = 0 ax = 0 ay = F y /m = ee/m, with a y up x = L = vt or t = L/v (Equation 1) y = 1/2 a y t 2 = d/2 = 1/2 qe/m t 2 (Equation 2) 2 Substituting Eq. 1 into Eq. 2: d/2 = 1/2 ee/m (L/v) or v = L(eE/dm) 1/2.

21. Electric field E due to the long wire at a distance r = λ/2πε o r and the force on the electron with charge e is F = ee = e λ/2πε o r radially inward. This force produces a centripetal acceleration such that Fnet = ma e λ/2πε o r = mv 2 /r or v = (e λ/2 πm εor) 1/2 = [1.6 x 10-19 C(2.5 x 10-9 C/m)/2 π(9.1 x 10-31 kg)(8.85 x 10-12 C 2 /N-m 2 )] 1/2 = 2.8 x 10 6 m/s. 22. The electric force on the sphere of charge q is F e = qe. The electric field due to the sheet E = σ/2 ε o. For the sphere to remain at rest, the net force on it must equal zero, that is T + mg + qe = 0, where T is the tension in the string. From the geometry of the figure above, we see that tan Θ = qe/mg = (qσ/2 ε o )/mg or σ = 2 ε o mg tan Θ/q. 23. Taking to the right to be positive for the force of the charges in the dipole at the left on the charges in the dipole at the right in Fig. 11a above, F = kq 2 /x 2 - kq 2 /(x +d) 2 - kq 2 /(x - d) 2 + kq 2 /x 2 Putting the above expression over the common denominator x 2 (x 2 + d 2 ) (x 2 - d 2 ) gives: F = {kq 2 /[ x 2 (x + d) 2 (x - d) 2 ]}{2(x + d) 2 (x - d) 2 - x 2 (x - d) 2 - x 2 (x +d) 2 } = {kq 2 /[ x 2 (x 2 - d 2 ) 2 } {2(x 2 d 2 ) 2 - x 2 (x - d) 2 - x 2 (x +d) 2 } = {kq 2 /[ x 2 (x 2 - d 2 ) 2 } {2(x 4 2x 2 d 2 + d 4 ) - x 2 (x 2-2dx + d 2 ) - x 2 (x 2-2dx + d 2 )} = {kq 2 /[ x 2 (x 2 - d 2 ) 2 } {-6x 2 d 2 + 2d 4 } Since 6x 2 d 2 > 2d 4, the net force on the dipole to the right will be to the left. By Newton's third law, the net force on the dipole to the left will be equal in magnitude, but to the right. For Fig. 11b, again taking to the right to be positive for the forces of the charges in the dipole at the left on the charges of the dipole at the right, F = = -kq 2 /x 2 + kq 2 /(x +d) 2 + kq 2 /(x -d) 2 - kq 2 /x 2.

The magnitude of the force is the same as in (a), but opposite in direction. Notice for x >> d, F = {kq 2 /[ x 2 (x 2 - d 2 ) 2 } {-6x 2 d 2 + 2d 4 } reduces to 6kq 2 d 2 /x 4 = 6kp 2 /x 4, where p = the dipole moment. 24. In general the torque τ= p x E τ = pe sin p, E a. For p parallel to E (Fig. for 24a above) p, E = 0, sin 0 o = 0, so τ= 0. b. For p perpendicular to E (Fig. for 24b above) p, E = 90 o, sin 90 o = 1, so τ= pe =[qd](e) = [(3.2 x 10-19 C)(2 x 10-9 m)](5.0 x 10 5 N/C) = 3.2 x 10-22 N-m. The direction of the torque is into the page. c. For p antiparallel to E (Fig. for 24c above) p, E = 180 o, sin 180 o = 0, so τ= 0. 25. In Fig. for #25(a), we think of Q 1 setting up a field E 1 at P, where E 1 = kq 1 /r 2 to the right. When Q 2 is placed at P, it experiences a force F on Q2, where F on Q2 = Q 2 E 1 = kq 1 Q 2 /r 2 to the right.

In Fig. for #25(b), we think of Q 2 setting up a field E 2 at P, where E 2 = kq 2 /r 2 to the left. When Q 1 is placed at P, it experiences a force F on Q1, where F on Q1 = Q 1 E 2 = kq 1 Q 2 /r 2 to the left. This is the same result of the force on Q 2 due to Q 1 and the force on Q 1 due to Q 2 found from Coulomb's law. 26. a. You can find an electric field due to a distribution of charges by i. treating dq as a point charge, taking components of the electric field, and then integrating, or ii. using Gauss's theorem if the charge distributions has spherical, cylindrical, or plane symmetry. b. Once you know an electric field E, you can find the electric force F e on a charge q from F e = qe. 27. a. The charge per unit length on either rod λ = Q/L. The charge dq on a length dx = λdx = (Q/L)dx. For an element of charge dq, de = kdq/x 2 = (kq/l) dx/x 2. The field due to the positively charged rod is to the right, with magnitude: The field due to the negatively charged rod is to the left, with magnitude: The total electric field E = kq/x[1/(x - L) - 1/(x + L) = [kq/x(x 2 - L 2 )](x + L) - (x - L) = [k2lq/x(x 2 - L 2 )] 3 b. For x >> L, E = [k2lq/x. Again we find E inversely proportional to the cube of the distance x. c. Comparing the above expression with that for field for a dipole, we identify the dipole p = 2LQ.