QUIZ 3 SOLUTIONSA.C. NAJMI1. Problem 1This problem is worth 10 pts total, with four points for part a, three points forparts b and c.1.1. To find the magnitude of the charge on the drop, we simply need to add upthe forces on the drop and set them equal to zero.XFi= FE− Fg= qE − mg = 0,(1)→ qE = mg,(2)→ q =mgE= 3.27 × 10−19C,(3)1.2. b. This was commonly answered incorrectly. For full points, you had toexplain that the direction of the electric field gives the force on a positive testcharge. Since the drop is suspended, we know the electric force on the drop isupward, but the field is downward, thus we know that the charge on the oil dropmust be negative since a negative charge will have an electric force acting on it inthe opposite direction of the electric field lines.1.3. c. If the oil drop is slowly evaporating, then the mass is decreasing in time.Since the electric force on the particle has no mass dependence, it remains constant.However, the gravitational force depends linearly on the mass. Thereforce, if themass is decreasing, then the gravitational force is decreasing in magnitude so theoil drop will float upwards.2. Problem 2This problem was worth 10 points. 4 points for drawing a free body diagram andwriting down Newton’s laws in the usual cartesian directions. 3 points for solvingthese in a meaningful way , and 3 points for numerical calculations, units, and thefinal answer.We start by writing down Newton’s laws in the x and y directions:Tx= T cos(θ) = Fg= mg,(4)Ty= T sin(θ) = FE= qE,(5)since we aren’t interested in the tension, we can take the quotient of these equationsto obtain:tan(θ) =qEmg,(6)→ E =mgqtan(θ) =1300p(3)9≈ 250NC.(7)12 A.C. NAJMI3. Problem 3This problem is worth 10 points. 1 point each for each part of part a, 3 pointsfor part b, 2 points for part c, and 2 points for part d. For part b, 1 point forwriting down the electric potential equation, 1 point for identifying that each chargecontributed, 1 point for the correct answer. For part c, full credit given if youranswer is consistent with the answer for the third part of part a, that is to sayin the opposite direction. For part d, 1 point for writing down the work from apotential change, 1 point for using the potential you found in b to find the answer.3.1. a. We apply the formula:|E| =kq1r21= 9 × 103NC,(8)|E| =kq2r22= 9 × 103NC,(9)using r1= 4m, r2= 3m, q1= −16µC, q1= 9µC, we find that the two electric fieldmagnitudes are equal, although they are orthogonal since each acts only along thex or y axis respectively. Since the negative charge is along the x axis, the electricfield from q1must point in the positive x direction, and similarly the electric fieldfrom q2must point along the −y axis. Thus the resultant vector should be in the4thquadrant.3.2. b. We start with the usual equation for electric potential and write:(10) V =XVi=kq1r1+kq2r2= 9 × 103(−4 + 3) = −9 × 103V,which is simply the sum of electric potential from each of the charges at the origin.3.3. c. Since the test charge is negative, the force should be in the opposite di-rection as the field vector you drew in part a. It should be a vector in the 2ndquadrant.3.4. d. Note that the electric potential at a point is defined to be the work doneon a unit positive test charge in bringing it from infinity to the point in question.That is to say:(11) ∆U = q∆V,so ∆U equals ∆V when q = 1 C. Using our electric potential from part b, wesimply use (11) with the charge q3= −4 µC, obtaining ∆U = 3.6 ×
View Full Document
Unlocking...