At each corner of an equilateral triangle, there is a charge, as diagrammed below. The length of each side of the triangle is .4 meters. Find the net force acting upon q1, given the following values for each of the charges: q1 = −2.00 μC, q2 = 3.00 μC, and q3 = 4.00 μC.
forces on q1 attractive
k =10^-12* usual k to account for μ
force in direction of Fq3=k(8)/.4^2
force in direction of Fq2=k(6)/.4^2
force perpendicular to opposite side
Fy= Fq3 cos 30 + Fq2 cos 30
force parallel to opposite side
Fx= Fq3 cos 60 - Fq2 cos 60
F = sqrt(Fx^2+Fy^2) as usual
To find the net force acting on q1, we need to calculate the force exerted by each charge on q1 and then combine them vectorially. The force between charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The equation for the force between two charges q1 and q2 separated by a distance r is:
F = k * (|q1 * q2|)/r^2
Where k is the electrostatic constant, approximately 9.0 x 10^9 Nm^2/C^2.
Let's calculate the forces first:
Force between q1 and q2:
F12 = k * (|q1 * q2|)/r^2
= (9.0 x 10^9 Nm^2/C^2) * (|(-2.0 x 10^-6 C) * (3.0 x 10^-6 C)|)/ (0.4 m)^2
Force between q1 and q3:
F13 = k * (|q1 * q3|)/r^2
= (9.0 x 10^9 Nm^2/C^2) * (|(-2.0 x 10^-6 C) * (4.0 x 10^-6 C)|)/ (0.4 m)^2
Now, let's calculate the magnitudes of the forces using the given values:
F12 = (9.0 x 10^9 Nm^2/C^2) * (2.0 x 10^-6 C * 3.0 x 10^-6 C)/ (0.4 m)^2
F13 = (9.0 x 10^9 Nm^2/C^2) * (2.0 x 10^-6 C * 4.0 x 10^-6 C)/ (0.4 m)^2
Next, we need to consider the direction of each force:
The force between q1 and q2 will be repulsive since they have opposite signs.
The force between q1 and q3 will be attractive since they have the same sign.
Since the charges are arranged symmetrically, the forces will cancel each other out in the x-direction, leaving only the y-component of the net force.
Finally, we need to find the net force acting on q1. We can do this by adding the y-component of the forces acting on q1:
Net force on q1 = F12(sin 60°) + F13(sin 60°).
Now, we can substitute the known values and calculate the net force on q1.