A charge 4.98 nC is placed at the origin of an xy -coordinate system, and a char

Question

A charge 4.98 nC is placed at the origin of an xy-coordinate system, and a charge -1.99 nC is placed on the positive x-axis at x = 4.00 cm . A third particle, of charge 6.04 nC is now placed at the point x = 4.00 cm , y = 3.00 cm .

Part A

Find the x-component of the total force exerted on the third charge by the other two.

Part B

Find the y-component of the total force exerted on the third charge by the other two.

Part C

Find the magnitude of the total force acting on the third charge.

Part D

Find the direction of the total force acting on the third charge.

Explanation / Answer

q1 = 4.98 nc

q2 = -1.99 nc

q3 = 6.04 nc

distance between q1 and q3 is sqrt(3^2+4^2) = 5 cm = 0.05 m

a) f1 = kq1q3/r^2 = 9*10^9*4.98*10^-9*6.04*10^-9/0.05^2 = 1.082*10^-4 N

angle = theta = tan^-1(3/4) = 36.9 degres

f1x = f1cos36.9 = 0.87*10^-4 N

f1y = f1sin36.9 = 0.65*10^-4 N

f2x = 0

f2y = -f2

f2 = 9*10^9*1.99*10^-9*6.04*10^-9/0.03^2 = 1.2*10^-4 N

f2y = -1.2*10^-4 N

fx = f1x+f2x = 0.87*10^-4 N

b) fy = f1y+f2y = -0.55*10^-4 N

c) magnitude f = sqrt(fx^2+fy^2) = 1.03 *10^-4 N

d) direction = tan^-1(0.55/0.87) = 32.3 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.