GOAL Apply Coulomb\'s law in two dimensions 36.9° His sin 36.9° 4.00 m F, cos 36
ID: 1572632 • Letter: G
Question
GOAL Apply Coulomb's law in two dimensions 36.9° His sin 36.9° 4.00 m F, cos 36.9 PROBLEM Consider three point charges at the corners of a triangle, as shown in the figure, where g1 6.00 x 10-0 C 3.00 m 5.00 m 92 2.00 x 10 C, and 5.00 x 10 C. (a) Find the components of the force F23 exerted by az on QT (b) Find the components of the force F 13 exerted by a on g3. (c) Fin The rotce ted by 1 on F13. turce Led by 92 on 43 SF zs. The ros F3 exerted on 43 the sum F13 F the resultant force on a3, n terms of components and also in terms of magnitude and direction. STRATEGY Coulomb's law gives the magnitude of each force, which can be split with right-triangle trigonometry into x- and y-components. Sum the vectors componentwise and then find the magnitude and direction of the resultant vector, SOLUTION (a) Find the components of the force exerted by a2 on g3 Find the magnitude of F23 with 92 23 Coulomb's law 10-9 c) 5.00 x 10-9 2.00 x 8.99 x 105 N m /CP 4.00 m 5.62 x 10-9 5.62 x 10-9 N Because 23 is horizontal and points in 23x the negative r-direction, the negative 23y of the magnitude gives the x component, and the y componen is t zero (b) Find the components of the force exerted by gi on a 3 Find the magnitude of F13 6.00 10 CO 00 x 10 8.99 x 109 N. m2/C2 5.00 m 1.08 x 10 Use the given triangle to find the F1 cos 0 N) cos (36.9°) 1.08 x 10 components of 13 8.64 x 10 N sin 1.08 10 N sin (36.9 3y 6.48 x 10 N (c) Find the components of the resultant vector. 5.62 x 10 9 N 8.64 x 10-9 N Sum the x-components to find the 3.02 x 10-2 esultant Sum the y-components to find the 0 6.48 x 10 6.48 x 10 resultant F. Find the magnitude of the resultant F.2 force on the charge 13, using the Pythagorean theorem. 3.01 x 10 6.50 x 10 7.15 x 10Explanation / Answer
a) F23 = kq2q3 / r^2 = 8.99*10^9*1.71*5.07*10^-18 / 4^2 = 4.87*10^-9 N
F23 x = -4.87*10^-9 N
F23 y = 0
b) F13 = kq1q3 / r^2 = 8.99*10^9*6.32*10^-9*5.07*10^-9 / 5^2 = 1.1522*10^-8 N
F13 x = F13CosQ = 1.1522*10^-8 * Cos36.9 = 9.21*10^-9 N
F13 y = F13SinQ = 1.1522*10^-8 * Sin36.9 = 6.92*10^-9 N
c) Fx = 9.21*10^-9 + (-4.87*10^-9) = 4.34*10^-9 N
Fy = 0 + 6.92*10^-9 = 6.92*10^-9 N
Fnet = sqrt (Fx^2 + Fy^2 )
Fnet = sqrt { (4.34*10^-9)^2 + (6.92*10^-9)^2} = 8.168*10^-9 N
TanQ = Fy / Fx = 6.92*10^-9 / 4.34*10^-9
Q = 57.9 degree
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