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HW2%20solutions

# HW2%20solutions - Physics 204 morning summer 2008 Homework...

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Unformatted text preview: Physics 204, morning summer 2008 Homework 2 solutions Chapter 18 Problems 33, 62, 29, 63, 65 33. SSM REASONING Since the charged droplet (charge = q) is suspended motionless in the electric field E, the net force on the droplet must be zero. There are two forces that act on the droplet, the force of gravity W = mg , and the electric force F = qE due to the electric field. Since the net force on the droplet is zero, we conclude that mg = q E . We can use this reasoning to determine the sign and the magnitude of the charge on the droplet. SOLUTION a. Since the net force on the droplet is zero, and the weight of magnitude W points F downward, the electric force of magnitude F = q E must point upward. Since the electric field points upward, the excess charge on the droplet must be positive in order for the force F to point upward. b. q = mg Using the expression mg = q E , we find that the magnitude of the excess charge on the droplet is mg (3.50 ×10 –9 kg)(9.80 m/s 2 ) = = 4.04 ×10 –12 C E 8480 N/C The charge on a proton is 1.60 × 10–19 C, so the excess number of protons is 1 proton ( 4.04 ×10–12 C ) ⎛ 1.60 ×10–19 C ⎞ = ⎜ ⎟ ⎝ ⎠ 2.53 ×107 protons 62. REASONING AND SOLUTION The electric field is defined by Equation 18.2: E = F/q0. Thus, the magnitude of the force exerted on a charge q in an electric field of magnitude E is given by F = q E (1) The magnitude of the electric field can be determined from its x and y components by using the Pythagorean theorem: 2 2 E = Ex + E y = a. ( 6.00 ×103 N/C ) + (8.00 ×103 N/C ) 2 2 = 1.00 × 104 N/C From Equation (1) above, the magnitude of the force on the charge is F = (7.50 × 10–6 C)(1.00 × 104 N/C) = 7.5 × 10 –2 N b. From the defining equation for the electric field, it follows that the direction of the force on a charge is the same as the direction of the field, provided that the charge is positive. Thus, the angle that the force makes with the x axis is given by 3 ⎛ Ey ⎞ −1 ⎛ 8.00 × 10 N/C ⎞ ⎟ = tan ⎜ ⎜ 6.00 × 103 N/C ⎟ = 53.1° ⎟ ⎟ ⎝ ⎠ ⎝ Ex ⎠ θ = tan −1 ⎜ ⎜ 29. REASONING AND SOLUTION a. In order for the field to be zero, the point cannot be between the two charges. Instead, it must be located on the line between the two charges on the side of the positive charge and away from the negative charge. If x is the distance from the positive charge to the point in question, then the negative charge is at a distance (3.0 m + x) meters from this point. For the field to be zero here we have k q− ( 3.0 m + x ) = q+ x2 or 16.0 μ C ( 3.0 m + x ) = 4.0 μ C x2 q− ( 3.0 m + x ) 2 = q+ x2 2 or 4.0 = ( 3.0 m + x )2 x2 Suppressing the units for convenience and rearranging this result gives 4.0x 2 = ( 3.0 + x ) 2 k q+ Solving for the ratio of the charge magnitudes gives q− = 2 4.0x 2 = 9.0 + 6.0 x + x 2 or or 3x 2 − 6.0 x − 9.0 = 0 Solving this quadratic equation for x with the aid of the quadratic formula (see Appendix C.4) shows that x = 3.0 m or x = −1.0 m We choose the positive value for x, since the negative value would locate the zero‐field spot between the two charges, where it can not be (see above). Thus, we have x = 3.0 m . b. Since the field is zero at this point, the force acting on a charge at that point is 0 N . 63. REASONING The two charges lying on the x axis produce no net electric field at the coordinate origin. This is because they have identical charges, are located the same distance from the origin, and produce electric fields that point in opposite directions. The electric field produced by q3 at the origin points away from the charge, or along the −y direction. The electric field produced by q4 at the origin points toward the charge, or along the +y direction. The net electric field is, then, E = –E3 + E4, where E3 and E4 can be determined by using Equation 18.3. SOLUTION The net electric field at the origin is E = − E3 + E4 = = ( −k q3 r32 + k q4 r42 )( 2 (5.0 × 10−2 m ) − 8.99 × 109 N ⋅ m 2 /C2 3.0 × 10−6 C ) + (8.99 × 109 N ⋅ m2 /C2 )(8.0 × 10−6 C ) 2 ( 7.0 × 10−2 m ) = +3.9 × 106 N/C The plus sign indicates that the net electric field points along the +y direction . 65. REASONING AND SOLUTION The average force F on the proton can be determined from the impulse‐ momentum theorem (Equation 7.4): F t = (mv) Therefore, the magnitude of the force is F= mv − mv0 Δt (5.0 ×10−23 kg ⋅ m/s ) − (1.5 ×10−23 kg ⋅ m/s ) = 5.6 ×10−18 N = 6.3 × 10−6 s From the definition of electric field, E = F , we find that the magnitude of the field is q0 E= F 5.6 ×10−18 N = = 35 N/C q0 1.60 × 10−19 C ...
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