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3 •• Determine the Concept During this sequence of events, negative charges are attracted from ground to the rectangular metal plate B. When S is opened, these charges are trapped on B and remain there when the charged body is removed. Hence B is negatively charged and (c) is correct. 6 • Determine the Concept The forces acting on +q are shown in the diagram. The force acting on q due to Q is along the line joining them and directed toward Q. The force acting on q due to Q is along the line joining them and directed away from Q. *8 • Determine the Concept E is zero wherever the net force acting on a test charge is zero. At the center of the square the two positive charges alone would produce a net electric field of zero, and the two negative charges alone would also produce a net electric field of zero. Thus, the net force acting on a test charge at the midpoint of the square will be zero. (b) is correct. 30 •• Picture the Problem The configuration of the charges and the forces on the fourth charge are shown in the figure … as is a coordinate system. From the figure it is evident that the net force on q4 is along the diagonal of the square and directed away from q3. We can apply Coulomb’s law to express F1, 4 , F2, 4 and F3, 4 and then add them to find the net force on q4. Express the net force acting on q4: F4 F1, 4 F2, 4 F3, 4 Express the force that q1 exerts on q4: kq q F1, 4 12 4 ˆj r1, 4 Substitute numerical values and evaluate F1, 4 : 3 nC ˆ j 3.24 105 N ˆj F1, 4 8.99 109 N m 2 /C 2 3 nC 2 0 . 05 m kq q F2, 4 22 4 iˆ r2, 4 Express the force that q2 exerts on q4: Substitute numerical values and evaluate F2, 4 : 3 nC ˆ i 3.24 105 N iˆ F2, 4 8.99 109 N m 2 /C 2 3 nC 2 0.05 m kq q F3, 4 32 4 r̂3, 4 , where r̂3, 4 is a unit vector r3, 4 Express the force that q3 exerts on q4: pointing from q3 to q4. r3, 4 r3,1 r1, 4 Express r3, 4 in terms of r3,1 and r1, 4 : 0.05 m iˆ 0.05 m ˆj r 0.05 m iˆ 0.05 m ˆj ˆr3, 4 3, 4 r3, 4 0.05 m2 0.05 m2 Convert r3, 4 to r̂3, 4 : 0.707 iˆ 0.707 ˆj Substitute numerical values and evaluate F3, 4 : 3 nC 0.707 iˆ 0.707 ˆj F3, 4 8.99 109 N m 2 /C 2 3 nC 0.05 2 m 2 5 5 ˆ ˆ 1.14 10 N i 1.14 10 N j Substitute and simplify to find F4 : F4 3.24 10 5 N ˆj 3.24 10 5 N iˆ 1.14 10 5 N iˆ 1.14 10 5 N ˆj 2.10 10 5 N iˆ 2.10 10 N ˆj 5 35 •• Picture the Problem By considering the symmetry of the array of charges we can see that the y component of the force on q is zero. We can apply Coulomb’s law and the principle of superposition of forces to find the net force acting on q. Express the net force acting on q: Fq FQ on x axis, q 2FQ at 45, q Express the force on q due to the charge Q on the x axis: kqQ FQ on x axis, q 2 iˆ R Express the net force on q due to the charges at 45: Substitute to obtain: kqQ 2 FQ at 45, q 2 2 cos 45iˆ R 2 kqQ ˆ i 2 2 R kqQ 2 kqQ ˆ Fq 2 iˆ i 2 R 2 R kqQ 2 ˆ 1 i 2 R 2 kqQ Note: I think the right answer of this problem should be 2 1 2 iˆ . Since the answer R on the textbook is also wrong, this problem will not be graded. 40 • Picture the Problem We can compare the electric and gravitational forces acting on an electron by expressing their ratio. We can equate these forces to find the charge that would have to be placed on a penny in order to balance the earth’s gravitational force on it. (a) Express the magnitude of the electric force acting on the electron: Fe eE Express the magnitude of the gravitational force acting on the electron: Fg me g Express the ratio of these forces to obtain: Fe eE Fg mg Substitute numerical values and evaluate Fe/Fg: Fe 1.6 10 19 C 150 N/C Fg 9.11 10 31 kg 9.81 m/s 2 2.69 1012 or Fe 2.69 1012 Fg , i.e., the electric force is greater by a factor of 2.691012. (b) Equate the electric and gravitational forces acting on the penny and solve for q to obtain: q Substitute numerical values and evaluate q: q mg E 3 10 3 kg 9.81m/s 2 150 N/C 1.96 10 4 C 44 •• Picture the Problem The diagram shows the locations of the charges q1 and q2 and the point on the x axis at which we are to find E . From symmetry considerations we can conclude that the y component of E at any point on the x axis is zero. We can use Coulomb’s law for the electric field due to point charges to find the field at any point on the x axis. We can establish the results called for in parts (b) and (c) by factoring the radicand and using the approximation 1 1 whenever << 1. (a) Express the x-component of the electric field due to the charges at y = a and y = a as a function of the distance r from either charge to point P: kq E x 2 2 cos iˆ r Substitute for cos and r to obtain: kq x 2kqx 2kqx ˆ E x 2 2 iˆ 3 iˆ i r r r x 2 a 2 3 2 2kqx iˆ 2 2 3 2 x a and (b) For x a, x2 + a2 a2, so: For x a, x2 + a2 x2, so: 2kqx Ex x Ex 2kqx Ex 2 a2 32 a 2kqx a3 2kqx 2kq x2 2 32 x 2 32 For x a, the charges separated by a would appear to be a single charge (c) 2kq of magnitude 2q. Its field would be given by E x 2 . x Factor the radicand to obtain: For a << x: a 2 Ex 2kqx x 2 1 2 x a2 1 2 1 x 3 2 and Ex 2kqx x 2 3 2 2kq x2 55 •• Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the electron in terms of the parameter t and Newton’s 2nd law to express the constant acceleration in terms of the electric field. Eliminating the parameter will yield an equation for y as a function of x, q, and m. We can decide whether the electron will strike the upper plate by finding the maximum value of its y coordinate. Should we find that it does not strike the upper plate, we can determine where it strikes the lower plate by setting y(x) = 0. Express the x and y coordinates of the electron as functions of time: x v0 cos t and y v0 sin t 12 ayt 2 F eE a y net, y y me me Apply Newton’s 2nd law to relate the acceleration of the electron to the net force acting on it: eE y Substitute in the y-coordinate equation to obtain: y v0 sin t Eliminate the parameter t between the two equations to obtain: yx tan x To find ymax, set dy/dx = 0 for extrema: eE y dy tan x' dx mev02 cos 2 2me t2 eE y 2m v cos 2 e 0 2 x 2 (1) 0 for extrema me v02 sin 2 (See remark below.) 2eE y Solve for x to obtain: x' Substitute x in y(x) and simplify to obtain ymax: ymax me v02 sin 2 2eE y Substitute numerical values and evaluate ymax: 9.1110 kg5 10 m/s sin 45 1.02 cm 21.6 10 C3.5 10 N/C 31 ymax 2 6 19 2 3 and, because the plates are separated by 2 cm, the electron does not strike the upper plate. To determine where the electron will strike the lower plate, set y = 0 in equation (1) and solve for x to obtain: x me v02 sin 2 eE y Substitute numerical values and evaluate x: 9.1110 kg5 10 m/s sin 90 x 1.6 10 C3.5 10 N/C 31 2 6 19 3 4.07 cm Remarks: x is an extremum, i.e., either a maximum or a minimum. To show that it is a maximum we need to show that d2y/dx2, evaluated at x, is negative. A simple alternative is to use your graphing calculator to show that the graph of y(x) is a maximum at x. Yet another alternative is to recognize that, because equation (1) is quadratic and the coefficient of x2 is negative, its graph is a parabola that opens downward. 72 •• Picture the Problem Choose the coordinate system shown in the diagram and let Ug = 0 where y = 0. We’ll let our system include the ball and the earth. Then the work done on the ball by the electric field will change the energy of the system. The diagram summarizes what we know about the motion of the ball. We can use the work-energy theorem to our system to relate the work done by the electric field to the change in its energy. Using the work-energy theorem, relate the work done by the electric field to the change in the energy of the system: Welectricfield K U g K 2 K1 U g,2 U g,1 or, because K1 = Ug,2 = 0, Welectricfield K 2 U g,1 Substitute for Welectric field, K2, and Ug,0 and simplify: qEh 12 mv12 mgh Solve for m: m *78 •• 12 m 2 gh qE g 2 mgh mgh Picture the Problem Each sphere is in static equilibrium under the influence of the tension T , the gravitational force Fg , and the electric force FE . We can use Coulomb’s law to relate the electric force to the charge on each sphere and their separation and the conditions for static equilibrium to relate these forces to the charge on each sphere. (a) Apply the conditions for static equilibrium to the charged sphere: Fx FE T sin kq2 T sin 0 r2 and F y T cos mg 0 Eliminate T between these equations to obtain: tan Solve for q: qr kq2 mgr 2 mg tan k r 2L sin Referring to the figure, relate the separation of the spheres r to the length of the pendulum L: Substitute to obtain: q 2 L sin mg tan k (b) Evaluate q for m = 10 g, L = 50 cm, and = 10: q 20.5 m sin 10 0.01kg 9.81m/s 2 tan 10 8.99 109 N m 2 /C 2 80 •• Picture the Problem Let the origin be at the lower left-hand corner and designate the charges as shown in the diagram. We can apply Coulomb’s law for point charges to find the forces exerted on q1 by q2, q3, 0.241 C and q4 and superimpose these forces to find the net force exerted on q1. In part (b), we’ll use Coulomb’s law for the electric field due to a point charge and the superposition of fields to find the electric field at point P(0, L/2). (a) Using the superposition of forces, express the net force exerted on q1: Apply Coulomb’s law to express F2 ,1 : F1 F2,1 F3,1 F4,1 kq q kq q F2,1 22 1 rˆ2,1 23 1 r2,1 r2,1 r2,1 Apply Coulomb’s law to express F4 ,1 : kq q kq q F4,1 42 1 rˆ4,1 43 1 r4,1 r4,1 r4,1 Apply Coulomb’s law to express F3,1 : 2 k q q ˆj kq ˆj L L3 L2 2 k q q ˆ kq iˆ L i L3 L2 kq q kq q F3,1 32 1 rˆ3,1 33 1 r3,1 r3,1 r3,1 kq2 L iˆ L ˆj 23 2 L3 kq2 3 2 2 iˆ ˆj 2 L Substitute and simplify to obtain: kq2 kq2 F1 2 ˆj 3 2 2 iˆ L 2 L 2 kq ˆ ˆ kq2 2 i j 32 2 L 2 L (b) Using superposition of fields, express the resultant field at point P: kq2 L2 iˆ ˆj 1 ˆ ˆ 1 i j 2 2 E P E1 E2 E3 E4 2 ˆj kq iˆ L2 (1) Use Coulomb’s law to express E1 : kq kq L E1 2 1 rˆ1, P 3 r1, P r1, P 2 Use Coulomb’s law to express E 2 : kq L 3 L 2 2 ˆj ˆj 4kq ˆj 2 L kq kq L E3 2 3 rˆ3, P 3 Liˆ r3, P r3, P 2 Use Coulomb’s law to express E 4 : ˆj 4kq ˆj 2 L kq k q L E 2 2 2 rˆ2, P 3 r2, P r2, P 2 Use Coulomb’s law to express E3 : kq L 3 L 2 2 ˆj 8kq ˆ 1 i 53 2 L2 2 ˆj ˆj kq k q L E 4 2 4 rˆ3, P 3 Liˆ r4, P r4, P 2 8kq ˆ 1 i 53 2 L2 2 ˆj ˆj Substitute in equation (1) and simplify to obtain: 4kg 4kg 8kg 1 E P 2 ˆj 2 ˆj 3 / 2 2 -iˆL L 5 L 2 ˆj 8kg iˆ- 1 3/ 2 2 5 L 2 ˆj 8kg 1 5 ˆj 2 25 L 85 ••• Picture the Problem We can use Coulomb’s force law for point masses and the condition for translational equilibrium to express the equilibrium position as a function of k, q, Q, m, and g. In part (b) we’ll need to show that the displaced point charge experiences a linear restoring force and, hence, will exhibit simple harmonic motion. (a) Apply the condition for translational equilibrium to the point mass: kqQ mg 0 y02 Solve for y0 to obtain: (b) Express the restoring force that acts on the point mass when it is displaced a distance y from its equilibrium position: kqQ mg y0 F kqQ kqQ 2 2 y0 y y0 kqQ kqQ 2 y 2 y0 y y0 2 0 because y << y0. Simplify this expression further by writing it with a common denominator: F 2 y0 ykqQ y04 2 y03 y 2 y0 ykqQ y y04 1 2 y0 2ykqQ y03 again, because y << y0. From the 1st step of our solution: kqQ mg y 02 Substitute to obtain: F Apply Newton’s 2nd law to the displaced point charge to obtain: m 2mg y y0 d 2 y 2mg y 2 dt y0 or d 2 y 2 g y 0 dt 2 y0 the differential equation of simple harmonic motion with 2 g y0 .