To find the magnitude and direction of the net forces on Q3, we need to calculate the individual forces acting on Q3 from Q1 and Q2 and then take the vector sum of these forces.
The formula to calculate the magnitude of the force between two point charges is given by Coulomb's Law:
F = k * |q1 * q2| / r^2
where:
F is the magnitude of the force
k is Coulomb's constant (8.9875 x 10^9 N m^2/C^2)
q1 and q2 are the magnitudes of the charges
r is the distance between the charges
Let's first calculate the force acting on Q3 from Q1:
Distance between Q1 and Q3 = sqrt((0.4 - 0)^2 + (0 - 0.3)^2) = sqrt(0.16 + 0.09) = sqrt(0.25) = 0.5 m
F1 = k * |2 * 5| / 0.5^2
F1 = 8.9875 x 10^9 N m^2/C^2 * 10 C / 0.25 m^2
F1 = 359500 N
Now, let's calculate the force acting on Q3 from Q2:
Distance between Q2 and Q3 = sqrt((0.4 - 0)^2 + (0 + 0.3)^2) = sqrt(0.16 + 0.09) = sqrt(0.25) = 0.5 m
F2 = k * |4 * 5| / 0.5^2
F2 = 8.9875 x 10^9 N m^2/C^2 * 20 C / 0.25 m^2
F2 = 719000 N
Next, we calculate the net force acting on Q3 by taking the vector sum of the forces F1 and F2:
Fnet = sqrt(F1^2 + F2^2 + 2 * F1 * F2 * cos(180°))
Fnet = sqrt(359500^2 + 719000^2 + 2 * 359500 * 719000 * cos(180°))
Fnet = sqrt(129015120500 + 517764100000 + 516407000000)
Fnet = sqrt(1166181010500)
Fnet ≈ 1.08 x 10^6 N
Therefore, the magnitude of the net force acting on Q3 is approximately 1.08 x 10^6 N.
To find the direction of the net force, we need to calculate the angle it makes with the positive x-axis:
tanθ = (ΣFy) / (ΣFx)
tanθ = (F1 - F2) / 0
tanθ = (359500 - 719000) / 0 = undefined
Therefore, the direction of the net force is directly between the forces acting on Q3 from Q1 and Q2, but the exact angle is undefined.
Three charges are placed on the xy_plane Q1=2uc at (0,0.3) ,Q2=4uc at (0,-0.3) and Q3= 5uc at (0.4,0),what's the magnitude and direction of the net forces on Q3
1 answer