Particle A of charge 2.91 10-4 C is at the origin, particle B of charge -5.46 10
ID: 1657092 • Letter: P
Question
Particle A of charge 2.91 10-4 C is at the origin, particle B of charge -5.46 10-4 C is at (4.00 m, 0), and particle C of charge 1.01 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C.
(a) What is the x component of the electric force exerted by A on C?
(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. 19.82 Correct
(d) Calculate the x component of the force exerted by B on C.
(e) Calculate the y component of the force exerted by B on C.
(f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. N (g) Similarly, find the y component of the resultant force vector acting on C. N
(h) Find the magnitude and direction of the resultant electric force acting on C. magnitude N direction ° counterclockwise from the +x-axis
Explanation / Answer
Given, Particle A, charge Qa = 2.91*10^-4 C is at (0,0)
Particle B, charge Qb = -5.46*10^-4 C at (4,0)
Particle C, charge Qc = 1.01*10^-4 C at (0,3)
a. x component of electric force exerted by A on C = 0 [ as particle is on y axis and the other one is at origin]
b. y compoennt of force exerted by A on C = kQaQc/3^2 = 8.98*2.91*1.01*10^-8*10^9/9 = 29.325 N ( +Y AXIS)
C. MAGNITUDE OF force exerted by B on C , F= kQbQc/(4^2 + 3^2) = 8.98*5.46*1.01*10^-8*10^9/25 = 19.8084432 N
d. x component of F = F*4/5 = 15.84675456 N ( + x axis )
e. y component of F = F*3/5 = 11.88506592 N ( - y axis)
f. Resulting x component = 15.8467 N
g. resulting y component = 19.8084 - 11.885 = 7.9234 N
h. Magnitude of force = sqroot(7.9234^2 + 15.8467^2) = 17.17 N
direction = arctan(7.92/15.84) = 26.565 deg CCW form +x axis