Calculating Electrostatic Force: q1 and q2

[Laxman2974]

Homework Statement

The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +3.5 µC, x1 = 1.5 cm, y1 = 0.50 cm, and q2 = -4.0 µC, x2 = -2.0 cm, y2 = 1.5 cm.
(a) Find the magnitude of the electrostatic force on q2. Done I got this part =
94.98868N
(b) Find the direction of this force. Got this also =
-15.9453959° (counterclockwise from the +x axis)
(c) At what coordinates should a third charge q3 = +5.5 µC be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero?
Can't get this - saw problem solving methods for questions that are very similar in previous postings, but I can't get the math to work out

Homework Equations

Got as far as q1/x^2 = q2/(x+r)^2
r being the distance between q1 and q2 since q3 must lie to the right of q1

The Attempt at a Solution

many attempts at a solution - but its been 15 years since I have done significant math and I can't get anywhere now.

 

[cryptoguy]

the equation you gave in part 2 is may not be right since q3 may not be located on the same Y value as q1. The equation would look more like:
[tex]\frac{q_1q_3}{x^2+y^2} = \frac{q_2q_3}{(r+\sqrt{x^2+y^2})^2}[/tex] (where x is distance from q1 x-coordinate to q3 x-coordinate and y is dist from q1 y-coord to q3 x-coord)

Now I'm not sure if this is the only way to solve this but I would start with equating the x-components of the forces aka [tex]F_{1-3x} = F_{2-3x}[/tex] and then do the same for y.

 

[Moetasim]

I understand your struggle and I am here to help. Let's work through this problem together.

First, let's review the information we have been given. We have two charged particles, q1 and q2, with fixed coordinates in the xy plane. We also know their respective charges: q1 = +3.5 µC and q2 = -4.0 µC.

For part (a), you correctly calculated the magnitude of the electrostatic force on q2 to be 94.98868N. This is the force that q1 exerts on q2 due to their charges and the distance between them.

For part (b), you also correctly found the direction of this force to be -15.9453959° counterclockwise from the +x axis. This is the direction in which q2 will be pushed by the electrostatic force from q1.

Now, for part (c), we need to find the coordinates where a third charge, q3 = +5.5 µC, can be placed such that the net electrostatic force on q3 due to q1 and q2 is zero. In other words, we need to find the point where the forces from q1 and q2 balance each other out.

To do this, we can use Coulomb's Law, which states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this can be written as:

F = k(q1q2)/r^2

Where F is the electrostatic force, k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.

Since we want the net force on q3 to be zero, we can set up the following equation:

F1 + F2 = 0

Where F1 is the force from q1 and F2 is the force from q2. Substituting in the values we know, we get:

k(q1q3)/r1^2 + k(q2q3)/r2^2 = 0

Where r1 and r2 are the distances between q1 and q3, and q2 and q3, respectively.

Now, we can solve for r1 and r2 by setting the two equations equal to each other and solving for r. This will give us