Electromagnetic forces, finding alpha.

[sphys4]

Homework Statement

When a third charge is placed at the origin, it is found that the force on it is zero. Find alpha?

Two charges Q and 5Q are located as shown in the figure below.
https://chip.physics.purdue.edu/protected/GiordanoEMimg/chapter17/17prob16.jpg



If possible, please show work so i can learn how to do it. Thanks!

Homework Equations

The Attempt at a Solution

I know that force varies as a function of charge and inversely proportional to square of distance.
So i need to use this equation somehow:
F = k Qa Qb / d^2

 

[ideasrule]

That link can only be accessed by Purdue University students. Can you attach the image instead?

 

[sphys4]

 

[gneill]

So suppose that some charge, q, was placed at the origin. What would be the formula for the force on it due to the charge Q? (And for the sake of argument, suppose for now that all the charges, q, Q, and 5Q are positive).

 

[rwisz]

First, we need to find the distance between the two charges Q and 5Q. We can use the Pythagorean theorem to find the distance d:

d = √(10^2 + 10^2)
d = √200
d = 14.14 cm

Now, we can plug in the values into the equation for force:

F = k Qa Qb / d^2
F = (8.99x10^9)(Q)(5Q) / (14.14 cm)^2
F = 4.05x10^11 Q^2 / 200 cm^2
F = 2.03x10^9 Q^2 N

Since we know that the force on the third charge at the origin is zero, we can set the force equal to zero and solve for Q:

2.03x10^9 Q^2 = 0
Q^2 = 0
Q = 0

Therefore, the charge Q must be zero for the force on the third charge to be zero. This means that alpha must also be zero since it is the ratio of Q to 5Q.

alpha = Q / 5Q
alpha = 0 / 5(0)
alpha = 0

In conclusion, the value of alpha is zero.