 Phyllis Fleming Physics Physics 108
 1 The magnitude of the force of +Q or -Q on qo at r = 3 m is FqQ = kQqo/r2 = (9.0 x 109 N-m2/C2)(2 x 10-6 C)(10-12 C)/(9 m2) = 2 x 10-9 N. The force of +Q on +q0 at P1 is to the right since +Q repels +qo and urges it to the right P2 is to the left since +Q repels +qo and urges it to the left. The force of -Q on + qo at P1 is to the left since -Q attracts +qo and urges it to the left P2 to the right since -Q attracts +qo 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 Fe on a test charge qo divided by the test charge. For all points in Fig. 1 above, the magnitude of the electric field = (Fe)/qo = (kQqo/r2)/qo = kQ/r2 = (9 x 109 N-m2/C2)(2 x 10-6 C)/9 m2 = 2 x 103 N/C. 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 P2 the electric field is to the left because a positive test charge there would be urged to the left. 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 P2 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 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 109 N-m2/C2)Q/r2. For q1 in Fig. 2, E1 = (9 x 109 Nm2/C2)(9 x 10-9 C)/(3 m)2 = 9 N/C to the right because if a positive test charge is placed at P, q1 would urge it to the right.  E2 = (9 x 109 N-m2/C2)(1 x 10-9 C)/ (1 m)2 = 9 N/C to the right because if a positive test charge is placed at P, q2 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 q2 was to the left. If q2 = +1 nC,  the electric field due to q2 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 q1,  E1 = q1/4πεor12 = 9 x 109 N-m2/C2(10-6 C/25.0m2) = 360 N/C.  The direction of E1 is shown in Fig. for #4. E2 = 9 x 109 N-m2/C2(0.0800 x 10-6 C/16.00 m2) = 45 N/C,  and E2 = -45 j N/C, as shown in Fig. for #4. Taking components of E1,  (E1) x = E1 cos 53o = 360 N/C (0.60) = 216 N/C, and (E1)y = E1 sin 53o= 360 N/C (0.80) = 288 N/C.  The component of the resultant E in the y direction Ey = (E1)y + (E2)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 = 48.4o with X-axis.
 5 In the Fig. for #5 above,  E1 represents the electric field due to q1 = 5.00 x 10-6 C,  which is r = 5.00 m from P. E1 = kq1/r2 = 9 x 109 N-m2/C2 (5.00 x 10-6C)                                      25.0 m2 = 1.80 x 103 N/C. E1x = E1 cos Θ= 1.80 x 103 N/C (0.800) = 1.44 x 103 N/C. E1y = E1 sin Θ=1.80 x 103 N/C (0.600) =1.08 x 103 N/C. In the Fig. for #5 above, E2 represents the electric field due to q2 = -3.00 x 10-6 C, which is a distance of r2 = 5.00 m from P. E2 = kq2/r2 = 9 x 109 N-m2/C2 (3.00 x 10-6)/25.0 m2 = 1.08 x 103 N/C in the negative Y direction (Fig. for #5). E2x = 0 and  E2y = -1.08 x 103 N/C. In the Fig. for #5,  E3 represents the electric field due to q3 = 1.60 x 10-6 C, which is a distance of r3 = 4.00 m from P. E3 = kq3/r3 = 9 x 109 N-m2/C2 (1.60 x 10-6)/16.0 m2 = 0.900 x 103 N/C in the positive X direction as shown in Fig. for #5. E3x = 0.900 x 103 N/C  and  E3y = 0. Let the total electric field = E. Ex = E1x + E2x + E3x = (1.44 + 0 + 0.90)103 N/C = 2.34 x 103 N/C. Ey = E1y + E2y + E3y = (1.08 -1.08 + 0)103 N/C = 0.  The resultant electric field is totally in the +X direction and it is equal to 2.34 x 103 N/C. Electric force on q4 = q4E = 2.0 x 10-6 C(2.34 x 103 N/C) = 4.68 x 10-3 N.
 6 λ = charge/length so that charge = λ(length) or dq = λ dx. Treating dq as a point charge, dE = k ( λ dx)/r2. 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 dx1 and dE2 due to the charge at dx2 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. The magnitude of the component of dE in the Y-direction dEy = dE cos Θ= k λdx cos Θ/r2. Only b is constant. The quantities x, r, and Θ are variables. x = b tan Θ  and  dx = b sec2 Θ dΘ. r2 = b2 + x2 = b2 + b2 tan2 Θ = b2(1 + tan2 Θ) = b2 sec2 Θ. dEy = λkdx cos Θ/r2 = λkb sec2 Θ dΘ cos Θ/b2 sec2 Θ= λkdΘ cos Θ/b. When x = - 3b/4,  r = [b2 + (-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. ∫ 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 ds1 is dE1 = dq/4πεoR2 = λo cos ΘR dΘ/4πεoR2. The electric field dE2 at point P due to an element of length ds2  has the same magnitude as that of dE1,  but not the same direction as dE1,  as illustrated in Fig. 6 above.  Since cos (180o - Θ) = - cos Θ the charge is negative in quadrant 2.  dE2 is in towards the arc. Components of the dE's in the Y direction cancel. Components in the X direction are equal and add. (dE1)x = (dE2)x = dE1 cos Θ = λo dΘ (cos2 Θ)/R = ( λo dΘ /4πεoR)(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 m2)+(-0.1m)(-0.04 m2)] = 0.8 N-m2/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 m2. 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-m2/C = charge enclosed/ εo                                      0.8 N-m2/C = Q/(8.85 x 10-12 C2/N-m2). 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, the electric field can not depend on angles Θ and Φ, the only possible direction for E is radially outward (or inward for negative charge), the magnitude of the electric field must be constant everywhere on the Gaussian surface. For E radially outward, E, dA = 0  and E • dA = E dA cos E, dA = E dA cos 0o = E dA. If E is constant, ∫E dA = E ∫ dA = E 4 πr2, where 4 πr2 = 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. charge/volume = ρ = Q/[(4 π/3)(b3 - a3)], 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.   For  r < a,  no charge is enclosed. For  a < r < b,  the Gaussian surface has a volume (4 π/3)(r3 - a3) and encloses an amount of charge =    ρ(4 π/3)(r3 - a3) = [Q/(4 π/3)(b3 - a3)](4 π/3)(r3 - a3) =  Q(r3 - a3)/(b3 - a3). For  r > b,  the Gaussian surface encloses the total charge Q. 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 πr2 = charge enclosed/ εo. For  r < a, E 4 πr2 = 0         E = 0 For  a < r < b,  E 4 πr2 = Q(r3 - a3)/ εo(b3 - a3)         E = Q(r3 - a3)/4πεor2(b3 - a3) For  r > b, E 4 πr2 = Q/ εo         E = Q/4πεor2 For  r = a,  E = Q(a3 - a3)/4πεor2(b3 - a3) = 0 = E For  r = b,  E = Q(r3 - a3)/4πεor2 (b3 - a3) = Q/4πεob2 For  r = b,  E = Q/4πεor2 = Q/4πεob2 Thus, For  r ≤ a, E = 0 For  a ≤ r ≤ b, E = Q(r3 - a3)/4πεor2(b3 - a3) For  r ≥ b, E = Q/4πεor2
 11 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. The volume of the cylinder is the area of the base πr2 times the length L, or πr2L.
 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 = 90o  and  E • dA = 0. For the lateral surface E, dA = 0o  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/ εo   or    E = λ/2πεor.
 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 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. 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/r2. This corresponds to the field outside a charged sphere. 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. The field outside a sphere with charge q is E = kq/r2. q = Er2/k = (144 N/C)(0.25 m2)/(9.0 x 109 N-m2/C2) = 4.0 x 10-9 C. For a cylinder,  E = λ/2πεor  or  λ= 2πεo rE = 2 π(8.85 x 10-12 C2/N-m2)(0.5 m) (144 N/C) = 4.0 x 10-9 C/m. For an infinite sheet of charge,  E = σ/2 εo  or σ = 2 εo E = 2(8.85 x 10-12 C2/N-m2)(144 N/C) = 2.55 x 10-9 C/m2.
 15 The electric field due to an infinite sheet of charge is  σ/2 εo.  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, At point P1 the electric field due to the sheet on the left Eleft is to the left, the electric field due to the center sheet Ecenter is to the right, and the electric field due to the sheet on the right Eright is to the left. Taking to the right to be positive, the electric field is to the left with magnitude E (at P1) equals -σ/2 εo + σ/2 εo - σ/2 εo = - σ/2 εo. At point P2 the electric field due to the sheet on the left Eleft is to the right, the electric field due to the center sheet Ecenter is to the right, and the electric field due to the sheet on the right Eright is to the left. Taking to the right to be positive, the E is to the right with magnitude E (at P2) equals +σ/2 εo + σ/2 εo - σ/2 εo = +σ/2 εo . At point P3 the electric field due to the sheet on the left Eleft is to the right, the electric field due to the center sheet Ecenter is to the left, and the electric field due to the sheet on the right Eright is to the left. Taking to the right to be positive, the electric field is to the left with magnitude E (at P3) equals σ/2 εo - σ/2 εo - σ/2 εo = - σ/2 εo . At point P4 the electric field due to the sheet on the left Eleft is to the right, the electric field due to the center sheet Ecenter is to the left, and the electric field due to the sheet on the right Eright is to the right. The electric field is to the right with magnitude E (at P4) equals σ/2 εo - σ/2 εo + σ/2 εo = +σ/2 εo .
 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 πr2 = charge enclosed/ εo , 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 πr2 = charge enclosed/ εo. (b) For a < r < b, E(4 πr2) = Q/ εo E = Q/4πεor2 = kQ/r2 (c) For  r > b,  charge enclosed = + Q - Q = 0  so E(4 πr2) = 0 or E = 0.
 17 From Gauss' theorem, = charge enclosed/ εo . 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. For r < R, For r < R,  Gauss's theorem gives E2 πrL = (2πρoL/6 εo)(3ar2 - 2cr3),  and        E = (ρo/6 εo)(3ar - 2cr2). For r > R, charge enclosed is found by using limits for the above integral of 0 and R. Charge enclosed = (2πρoL/6) (3aR2 -2cR2),  and now E2 πrL = (2πρoL/6 εo)(3aR2 -2cR3),  and        E = (ρo/6r εo)( (3aR2 -2cR3) The expression of E for  r < R  at  r = R  reduces to  E = (ρo/6 εo)(3aR - 2cR2), 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 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 = 90o. For the cap,  E . dA = E dA. For the lateral surface,  E . dA = 0. By Gauss's theorem, E dA = charge enclosed/ εo = xdA ρ/ εo and E = ρx/ εo. 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/ εo) are constant, so this is an example of simple harmonic motion.         Fnet = ma   or -(ρq/ εo)x = m d2x/dt2   or d2x/dt2 + (ρq/m εo)x = 0. Compare with d2x/dt2 + (k/m)x = 0, for which the frequency  f = 1/2 π (k/m)1/2 and see for this case  f = 1/2 π (ρq/m εo)1/2.

20.

 X Y xo = 0 yo = 0 vox = v voy = 0 ax = 0 ay = Fy/m = eE/m, with ay  up x = L = vt  or  t = L/v                       (Equation 1) y = 1/2 ayt2 = d/2 = 1/2 qE/m t2            (Equation 2)

Substituting Eq. 1 into Eq. 2:  d/2 = 1/2 eE/m (L/v)2 or v = L(eE/dm)1/2.

 21 Electric field E due to the long wire at a distance r = λ/2πεor and the force on the electron with charge e is F = eE = e λ/2πεor radially inward. This force produces a centripetal acceleration such that       Fnet = ma e λ/2πεor = mv2/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 C2/N-m2)]1/2 = 2.8 x 106 m/s.
 22 The electric force on the sphere of charge q is Fe = qE.  The electric field due to the sheet E = σ/2 εo.  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 εo)/mg   or   σ = 2 εomg 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 = kq2/x2 - kq2/(x +d)2 - kq2/(x - d)2 + kq2/x2 Putting the above expression over the common denominator x2(x2 + d2) (x2 - d2) gives: F = {kq2/[ x2(x + d)2 (x - d)2]}{2(x + d)2 (x - d)2 - x2(x - d)2 - x2(x +d)2}   = {kq2/[ x2(x2 - d2)2} {2(x2 – d2)2 - x2(x - d)2 - x2(x +d)2}   = {kq2/[ x2(x2 - d2)2} {2(x4 – 2x2d2 + d4) - x2(x2 - 2dx + d2)          - x2(x2 - 2dx + d2)}   = {kq2/[ x2(x2 - d2)2} {-6x2d2 + 2d4} Since 6x2d2 > 2d4, 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 = = -kq2/x2 + kq2/(x +d)2 + kq2/(x -d)2 - kq2/x2. The magnitude of the force is the same as in (a), but opposite in direction. Notice for x >> d, |F| = |{kq2/[ x2(x2 - d2)2} {-6x2d2 + 2d4}| reduces to 6kq2d2/x4 = 6kp2/x4, where p = the dipole moment.
 24 In general the torque τ= p x E τ = pE sin p, E For p parallel to E (Fig. for 24a above) p, E = 0, sin 0o = 0,  so  τ= 0. For p perpendicular to E (Fig. for 24b above) p, E = 90o, sin 90o = 1,  so τ= pE =[qd](E)          = [(3.2 x 10-19 C)(2 x 10-9 m)](5.0 x 105 N/C)          = 3.2 x 10-22 N-m. The direction of the torque is into the page. For p antiparallel to E (Fig. for 24c above) p, E = 180o, sin 180o = 0,  so τ= 0.
 25 In Fig. for #25(a), we think of Q1 setting up a field E1 at P, where E1 = kQ1/r2 to the right.  When Q2 is placed at P, it experiences a force Fon Q2, where Fon Q2 = Q2E1 = kQ1Q2/r2 to the right. In Fig. for #25(b), we think of Q2 setting up a field E2 at P’, where E2 = kQ2/r2 to the left.  When Q1 is placed at P’, it experiences a force Fon Q1, where Fon Q1 = Q1E2 = kQ1Q2/r2 to the left. This is the same result of the force on Q2 due to Q1 and the force on Q1 due to Q2 found from Coulomb's law.
 26 You can find an electric field due to a distribution of charges by treating dq as a point charge, taking components of the electric field, and then integrating,  or using Gauss's theorem if the charge distributions has spherical, cylindrical, or plane symmetry. Once you know an electric field E, you can find the electric force Fe on a charge q from Fe = qE.
 27 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/x2 = (kQ/L) dx/x2. 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(x2 - L2)](x + L) - (x - L)                                       = [k2LQ/x(x2 - L2)] For  x >> L,  E = [k2LQ/x3. Again we find E inversely proportional to the cube of the distance x. Comparing the above expression with that for field for a dipole, we identify the dipole p = 2LQ.

 Homepage Sitemap
 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming August 8, 2002 April 30, 2003