XANPLE 15 e Triangle Goal Apply Coulomb\'s law in two dimensions. Problem Consid
ID: 1865532 • Letter: X
Question
XANPLE 15 e Triangle Goal Apply Coulomb's law in two dimensions. Problem Consider three point charges at the corners of a triangle, as shown in the figure, where 6.00 X 10-9 C, ???" "2.0OX 10.9 C, and ?, s 5.00x 10.9 C. (a) Find the components of the force exerted by q2 on es. (b) Find the components of the force F13 exerted by e2 on q. (e) Find the resultant force on q in terms of components and also in terms of magnitude and direction 00 m Strategy Coulomb's law gives the ma?nitude-tumf,.Fa of each force, which can be split with right triangle trigonometry into x and y-components. Sum the vectors componentváse and then find the magnitude and direction of the resultant vector SOLUTION (a) Find the components of the force exerted by az on g Find the magnitude of Fs with Coulomb's law ych (2.00 × 10.9 C)(5.00 4.00 m)2 10-9 C) F23-5.62% 10.9 N Because F23 is horizontal and points in the negative r-direction. Faax5.62 X 109N the negative of the magnitude gives the x component, and the y-component is zero. (b) Find the components of the force exerted by 91 o0 93 Find the magntude of Fatlealel 10-9 ?)(5.00 (5.00 m)a (6.00 10-9 ? - (8.99 x 10% N m2/c2) F13 1,08 x 108N fur-Fi, cos ??( 1.08 × 10'S N) cos(370)- 8.63 x 10-9 N F13,-Ft3 sin ?-(1.08 x 10-8 N) sin(379) .50 xi 10.9 N Uze the given triangle to find the components of F1 (e) Find the components of the resultant vector. Sum the x-components to find the resultant F Surm the y-components to find the resultant Fy Find the magnitude of the resultant forre.no.the.rharon .?.A+11 T 111 F.--5.62 x 10.9 N + 8.63 × 10-9 N- 01 x 10.9 01 x 109 N)2+(6.50 x 10 0 Type here to search l IExplanation / Answer
q1=6.00*10^-9 C, q2=-2.00*10^-9C, q3=5.00*10^-9C
a)
F23 = kq2q3/r23^2 = (9*10^9*2.00*10^-9*5*10^-9) /(4.00^2) = 5.625*10^-9 N
F23x= - 5.625*10^-9 N
F23y = 0.00N
b)
F13 = kq1q3/r13^2 = (9*10^9*1.00*10^-9*5*10^-9) /(5.00^2) = 1.8*10^-9 N
F13x= F13cos37 = 1.8*10^-9cos37 = 1.437*10^-9N
F13y= F13sin37 = 1.8*10^-sin37 = 1.083*10^-9N
c)
Fnetx= F23x+F13x= - 5.625*10^-9 + 1.437*10^-9 = -4.188*10^-9 N
Fnety= F23y+F13y= 0.00+1.083*10^-9 = 1.083*10^-9N
Fnet = sqrt(Fnetx^2+Fnety^2) = sqrt[(4.188*10^-9)^2+(1.083*10^-9)^2)] = 4.3*10^-9 N
?=tan^-1[(1.083*10^-9)/( -4.188*10^-9)] = -14.5 deg …..Meassured from –x axis
?=180-14.5 = 165.5 deg…..anticlockwise from +x axis
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