Electricity, force, and component form

[abeltyukov]

Homework Statement

Consider the four charges in Figure P25.46 http://i137.photobucket.com/albums/q208/infinitbelt/p25-46.gif , in which d = 3 cm, q = 4 nC, the +x-axis points to the right, and the +y-axis points up. What is the force on the 1 nC charge at the bottom? Give your answer in component form, using i and j for and , respectively.

Homework Equations

F 2on1 = (K|q1||q2|)/r^2
F 3on1 = (K|q1||q3|)/r^2

The Attempt at a Solution

I did for force 2 on 1 :[(9 x 10^9)(5 x 10^-9)(8 x 10^-9)]/(sqrt(0.03^2 + 0.048^2))^2
Then I did tan(theta) = 4.8/3.0 = 57.994 degrees

1.12 x 10^-4 N * sin(57.994) = 9.5 x 10^-5 N

Then for the other force 3 on 1: [(9 x 10^9)(5 x 10^-9)(5 x 10^-9)]/0.03^2 = 2.5 x 10^-4 N

I am confused as to what to do from here. Any ideas?


Thanks!

 

[BishopUser]

I'm not sure I follow your method. I know that the force on a charge due to another charge is $$F= \frac{kQq}{r^2}$$ where k is approximately $$9x10^9 N M^2 C^{-2}$$.

I'm not sure of your reasoning for plugging in 8nC or 5nC into the equation. You are also talking about q1 and q3; I don't know what those are because they aren't labeled on your diagram. Please explain your reasoning for reach step.

 

[m2003]

I would first like to commend you for attempting to solve the problem and showing your work. It is important to understand the concepts and equations involved in order to properly solve the problem.

To find the force on the 1 nC charge at the bottom, we can use the principle of superposition, which states that the net force on a charge is equal to the vector sum of the individual forces acting on it.

In this case, we have two individual forces acting on the 1 nC charge from the other two charges. Using the equation for force 2 on 1, we can calculate the x and y components of this force:

Fx = 9.5 x 10^-5 N * cos(57.994) = 4.8 x 10^-5 N

Fy = 9.5 x 10^-5 N * sin(57.994) = 7.3 x 10^-5 N

Similarly, for force 3 on 1, we can calculate the x and y components:

Fx = 2.5 x 10^-4 N * cos(45) = 1.8 x 10^-4 N

Fy = 2.5 x 10^-4 N * sin(45) = 1.8 x 10^-4 N

Now, to find the net force on the 1 nC charge, we simply need to add the x and y components of the two forces:

Fnetx = 4.8 x 10^-5 N + 1.8 x 10^-4 N = 2.3 x 10^-4 N

Fnety = 7.3 x 10^-5 N + 1.8 x 10^-4 N = 2.5 x 10^-4 N

Therefore, the net force on the 1 nC charge is given by Fnet = (2.3 x 10^-4 N)i + (2.5 x 10^-4 N)j.

I hope this helps clarify the solution for you. Remember, always break down the problem into smaller, manageable steps and use the appropriate equations to solve for the desired quantities.