Particle A of charge 4.00 10-4 C is at the origin, particle B of charge -5.00 10
ID: 1918688 • Letter: P
Question
Particle A of charge 4.00 10-4 C is at the origin, particle B of charge -5.00 10-4 C is at (3.00 m, 0) and particle C of charge 2.00 10-4 C is at (0, 4.00 m). (a) What is the x-component of the electric force exerted by A on C? N= (b) What is the y-component of the force exerted by A on C? N= (c) Find the magnitude of the force exerted by B on C. N= (d) Calculate the x-component of the force exerted by B on C. N= (e) Calculate the y-component of the force exerted by B on C. N= (f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C. N= (g) Repeat part (f) for the y-component. N= (h) Find the magnitude and direction of the resultant electric force acting on C. magnitude N= direction (Explanation / Answer
qa=4*10^-4C qb=-5*10^-4C qc=2*10^-4C r1=3m r2=4m r3 =5m a) zero b) E= Kqa/r2^2 =36*10^5/16 =2.25*10^5 N/C c) F=kq2q3/r3^2 =9*10^9*5*2*10^-8/5^2 =900/25=36N d)x component of force =Fsin@ =36*sin(tan^-1(3/4)) =36*3/5=21.6N e) y component of F=fcos@ =36*4/5 =28.8N f)resultant x component=21.6 +0=21.6 N g)resultant y component=- 28.8 +36=7.2N h)resulting F on C= F/q3=(21.6^2+7.2^2)^1/2=22.768N direction: tan@=7.2/21.6 @=0.321 this angle is counterclockwise from +ve x axis