In: Physics
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 = 4.01 cm . A third particle, of charge 6.05 nC is now placed at the point x = 4.01 cm , y = 2.98 cm .
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. (THETA)
The Coulomb's law is an equation that relates the magnitude of the force between two point charges, q1 and q2, seperated by a distance d as follows;
F = k *q1* q2/d2
In our case, let charge 1 be charge q1 = 4.98nC, charge 2 be q2 = − 1.99nC and charge 3 be q3 = 6.05 nC.
The distance between charges 1 and 3 is
r = √(4.01 cm)2 + (2.98 cm)2 =4.996 cm=0.04996 m
Therefore, the magnitude of the electric force exerted by charge 1 on charge 3 is
F13 = (8.9875 × 109 N⋅m2/C2) × |4.98 × 10−9C| × |6.05 × 10−9 C|(0.04996 m)2
= 1.0848124150 × 10−4 N
and this force makes an angle
α = arctan (2.98/4.01) = 36.62604402∘
with the positive x-axis since this force is repulsive (between two positive charges). Therefore, the x-component of the electric force exerted by charge 1 on charge 3 is
F13,x = F13 cosα = 1.084124150 × 10−4 N × cos36.6204402∘
=8.708510047 × 10−5 N
in the positive x-direction, and the y-component is
F13,y = F13 sinα
= 1.084124150 × 10−4 N × sin36.6204402
= 6.460680187 × 10−5 N∘
in the positive y-direction.
Similarly, the magnitude of the electric force exerted by charge 2 on charge 3 is
F23 = (8.9875 × 109 N⋅m2/C2) × |− 1.99 × 10−9 C| × |6.05 × 10−9 C|/(0.0298 m )2= 1.21825289 × 10−4 N
in the negative y-direction since this force is attractive.( no x component)
a: The x-component of the total force exerted on charge 3 by the other two charges is
Fx = F13,x + F23,x= (8.708510047 × 10−5 +0 ) N =8.708510047 × 10−5 N
b: The y-component of the total force exerted on charge 3 by the other two charges is
Fy = F13,y − F23,y= 6.460680187 × 10−5 N∘ − 1.21825289 × 10−4 N = − 5.245572705 × 10−5 N
c: The magnitude of the total force on charge 3 is
F = √F2x + F2y= √(8.708510047 × 10−5 N)2+(− 5.245572705 × 10−5N)2 =1.015565066 × 10−4 N
d.
force makes an angle
θ = − arctan(5.245572705× 10−5N)/(8.708510047 × 10−5 N) = − 31.0635842∘
with the positive x-axis, measured counterclockwise.