and 2 2 C 13 2 sin 2 1.81 10 N/C 0.858 3.11 10 N/C E E ______________________________________________________________________________ 42. REASONING AND SOLUTION The magnitude of the force on q 1 due to q 2 is given by Coulomb's law: 1 2 12 2 12 k q q F r (1) The magnitude of the force on q 1 due to the electric field of the capacitor is given by 1C 1 C 1 0 F q E q (2) Equating the right hand sides of Equations (1) and (2) above gives 1 2 1 2 0 12 k q q q r
160 ELECTRIC FORCES AND ELECTRIC FIELDS Solving for r 12 gives 0 2 12 12 2 2 9 2 2 6 –2 2 = [8.85×10 C /(N m )](8.99×10 N m /C )(5.00×10 C) = = 5.53×10 m (1.30×10 C/m ) k q r × × ______________________________________________________________________________ 43. REASONING The electric field is given by Equation 18.2 as the force F that acts on a test charge q 0 , divided by q 0 . Although the force is not known, the acceleration and mass of the charged object are given. Therefore, we can use Newton’s second law to determine the force as the mass times the acceleration and then determine the magnitude of the field directly from Equation 18.2. The force has the same direction as the acceleration. The direction of the field, however, is in the direction opposite to that of the acceleration and force. This is because the object carries a negative charge, while the field has the same direction as the force acting on a positive test charge. SOLUTION According to Equation 18.2, the magnitude of the electric field is 0 F E q According to Newton’s second law, the net force acting on an object of mass m and acceleration a is F = ma . Here, the net force is the electrostatic force F , since that force alone acts on the object. Thus, the magnitude of the electric field is 3 3 2 5 6 0 0 3.0 10 kg 2.5 10 m/s 2.2 10 N/C 34 10 C F ma E q q The direction of this field is opposite to the direction of the acceleration. Thus, the field points along the x axis . 44. REASONING The external electric field E exerts a force F E = q E (Equation 18.2) on the sphere, where q = +6.6 C is the net charge of the sphere. The external electric field E is directed upward, so the force F E it exerts on the positively charged sphere is also directed upward (see the free-body diagram). Balancing this upward force are two downward forces: the weight m g of the sphere (where m is the mass of the sphere and g is the acceleration due to gravity) and the force F s exerted on the sphere by the spring q = +6.6 C m g F s F E Free-body diagram
Chapter 18 Problems 161 (see the free-body diagram). We know that the spring exerts a downward force on the sphere
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