Where Should the Third Charge Be Placed to Achieve Zero Net Electric Force?

[Gummibare610]

Homework Statement

Three point charges lie in a straight line along the y-axis. A charge of q1 is -9 microcouloumbs at y = 6 m and a charge of q2 is -8 microcoulombs at y = -4m. The net electric force on the third point charge is zero. Where is this charge located.

Homework Equations

Fe = k (q1*q2/r^2)

The Attempt at a Solution

Fe = k (q1*q2/r^2)
= (9*10^9)((9*10^-6)(8*10^-6)/(10m)^2))
= 6.5*10^-3 N

Then i attempted to plug in this value for Fe to receive the r, and the location, but
i don't know which two charges i had to use.

Thanks,

SLM

 

[ideasrule]

What's the equation for the electric field? Do you know how to add electric fields induced by different charges?

 

[Gummibare610]

I know how to add different electrical forces, but the problem i have is finding the location of it, instead of the force.

 

[collinsmark]

I know how to add different electrical forces, but the problem i have is finding the location of it, instead of the force.

So you've added the two equations for the electric fields of the first two charges together, giving you an equation for the overall electric field E as a function of y (considering both charges). What location along y produces a zero electric field? (Hint: you need to set something to zero and solve for y.)

If a charge is in a zero electric field, what can you say about the net electric force acting on that charge?

 

[rdl]

I would first clarify the problem statement and make sure that all the given information is accurate and complete. From the given information, it seems that there are only two point charges (q1 and q2) and the third charge's location is unknown. The net electric force on the third charge is zero, which means that the third charge must be located in a position where the electric forces from q1 and q2 are equal in magnitude but opposite in direction.

To determine the location of the third charge, we can use the equation for electrostatic force between two point charges, Fe = k (q1*q2/r^2). Since the net force on the third charge is zero, we can set the forces from q1 and q2 equal to each other and solve for r.

Fe from q1 = k (q1*q3/r^2) = (9*10^9) * (-9*10^-6) * (q3/r^2)
Fe from q2 = k (q2*q3/r^2) = (9*10^9) * (-8*10^-6) * (q3/r^2)

Since these two forces must be equal, we can set them equal to each other and solve for r:

(9*10^9) * (-9*10^-6) * (q3/r^2) = (9*10^9) * (-8*10^-6) * (q3/r^2)
-9*10^-6 * q3 = -8*10^-6 * q3
q3 cancels out, leaving us with:

r = 8/9 * (6m + 4m) = 6.67m

This means that the third charge must be located at a distance of 6.67 meters from q1 and 4 meters from q2, on the y-axis. We can also calculate the magnitude of this third charge by plugging in the value of r into the equation:

Fe = k (q1*q3/r^2)
6.5*10^-3 N = (9*10^9) * (-9*10^-6) * (q3/6.67m)^2
q3 = -5.36*10^-6 C

Therefore, the third charge must be located at y = -5.36*10^-6 C