If anyone could help me out with this physics problem I would greatly appreciate

Question

If anyone could help me out with this physics problem I would greatly appreciate it. Here is the problem: A charge 4.98 is placed at the origin of an xy-coordinate system, and a charge -1.99 is placed on the positive x-axis at = 4.02 . A third particle, of charge 5.99 is now placed at the point = 4.02 , = 2.96 . A. Find the x-component of the total force exerted on the third charge by the other two. B. Find the y-component of the total force exerted on the third charge by the other two. C. Find the magnitude of the total force acting on the third charge. D. Find the direction of the total force acting on the third charge.

Explanation / Answer

1) A charge=q1= 5.01*10^-9 C is placed at the origin Another charge =q2= - 2.03 *10^-9 C is placed s at x= 4.01*10^-2 m . A third charge = q3 = 5.98*10^-9 C is placed at the point x = 4.01*10^-2 m . , Y = 3.02 *10^-2 m . Distance of q3 from q1 = r31= sq rt [ (4.01)^2+(3.02)^2 ]=5.02*10^-2 m Distance of q3 from q2 = r32=3.02*10^-2 m a) The force exerted on q3 by q1 =F31 F31 =kq3q1/(r31)^2 F31 =9*10^9*5.98*10^-9 [5.01*10^-9 /(5.02*10^-2)^2 F31 =53.82 [5.01*10^-9 /(5.02*10^-2)^2 ] F31 =53.82 *10^-5[0.19881 ] F31 =10.6999*10^-5 AngleO is given by tanO=3.02/4.01 Angle O =36.984 degree cos36.984 =0.7988 component along x axis =Fcos36.984 =8.5471*10^-5N(i) sin36.984 =0.6016 component along y axis =Fsin36.984 =6.4370 *10^-5N(j) F31=8.5471*10^-5N(i) + 6.4370*10^-5 N(j) _______________________ F32=kq3q2/(r32)^2 F32 =9*10^9*5.98*10^-9* 2.03 *10^-9 / (3.02*10^-2)^2 F32 =53.82 *10^-5[0.2226] F32 =11.98*10^-5 N along negativey axis In vector notation,F32=11.98*10^-5 N (-j) ______________________ Total force =F =F31+F32 F =8.5471*10^-5N(i) + 6.4370*10^-5 N(j) +11.98*10^-5 N (-j) F=8.5471*10^-5N(i) -5.543*10^-5N(j) a) The x-component of the total force exerted on the third charge by the other two=8.5471*10^-5N in positive x axis direction b) The y-component of the total force exerted on the third charge by the other two=5.543*10^-5 N in negative y axis direction

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