Stumped by This Physics Problem? Find Out What's Wrong Here!

[RM86Z]

Homework Statement

Particles of charge Q and 3Q are placed on the x axis at 
x = -L and x = +L, respectively. A third particle of charge q is placed on the x axis, and it is found that the total electric force on this particle is zero. Where is the particle?

Relevant Equations

F = kq1q2/r^2; q1 and q2 are two charges, k is Coulomb's constant and r is the separation between the two charges.

I am stumped by this one! My attempt at solving this problem was to do a coordinate transform to put Q at the origin and 3Q at 2L on the x-axis. The distance then between q and Q is x and the distance between q and 3Q is (2L - x).

The electrical force between q and Q is F1 = kqQ/x^2.
The electrical force between q and 3Q is F2 = kq3Q/(2L - x)^2.

The net force on q is zero which implies that F1 = -F2.

F1 = -F2
kqQ/x^2 = -kq3Q/(2L-x)^2
(2L - x)^2 = -3x^2
4L^2 - 4Lx +4x^2 = 0
L^2 - Lx + x^2 = 0

This has no real solutions.

What am I doing wrong here?!

Thank you!

 

[etotheipi]

F1 = -F2
kqQ/x^2 = -kq3Q/(2L-x)^2
(2L - x)^2 = -3x^2
4L^2 - 4Lx +4x^2 = 0
L^2 - Lx + x^2 = 0

That relation is only true in vector form, $\vec{F_{1}} = -\vec{F_{2}}$, or a signed component form ${F_{x}}_{1} = -{F_{x}}_{2}$, not magnitudes. You want the net $x$ component of force to be zero, and this is the sum of the individual signed $x$ components of the forces. Alternatively, drawing a diagram should help you to see how the magnitudes must be related.

To avoid sign errors, it's sometimes good practice to resolve the forces to find the net force, and then set this to $0$ or $ma$.

 

[kuruman]

kqQ/x^2 = -kq3Q/(2L-x)^2

This is what you are doing wrong. You are setting a positive quantity on the left side equal to a negative quantity on the right side.

 

[jbriggs444]

The electrical force between q and Q is F1 = kqQ/x^2.
The electrical force between q and 3Q is F2 = kq3Q/(2L - x)^2.

The net force on q is zero which implies that F1 = -F2.

It seems you have a sign convention problem here. You expect the position to be somewhere between 0 and 2L, correct?

Per your formulae, that means that both forces are positive. Yet you are setting one to be the inverse of the other. Did you forget that the forces are exerted in opposite directions?

 

[RM86Z]

Thank you very much guys that was indeed a mistake that I should have picked up! I've solved the following given:

$$ F_{q_1q_2} = \frac {kq_1q_2} {r^2} $$

$$ F_1 - F_2 = 0 $$
$$ F_1 = F_2 $$
$$ \frac {kqQ}{x^2} = \frac {kq3Q}{(2L - x)^2} $$
$$ (2L - x)^2 = 3x^2 $$
$$ 2x^2 + 4Lx - 4L^2 = 0 $$
$$ x^2 + 2Lx - 2L^2 = 0 $$

Solutions are:

$$ x = ( \sqrt {3} - 1) L $$
$$ x = -(1 + \sqrt {3}) L $$

Answer in the book is:

$$ x = -0.27 L $$

Even transforming the problem back to the original locations on the x-axis doesn't give the correct answer?

 

[etotheipi]

Solutions are:

$$ x = ( \sqrt {3} - 1) L $$
$$ x = -(1 + \sqrt {3}) L $$

Answer in the book is:

$$ x = -0.27 L $$

Even transforming the problem back to the original locations on the x-axis doesn't give the correct answer?

You can eliminate one of the solutions right off the bat since you know $x$ in your transformed system has to be positive.

Then how do you transform back into the original coordinates?

 

[RM86Z]

I am a fool! Thanks guys, really struggled with this one.

 

[kuruman]

When you solve quadratics in physics usually only one solution makes sense in view of the situation described. However, the "other solution" could also make sense under a modified description of the physical situation that is nevertheless formulated by the same quadratic. Can you think of a situation in which you have charges of magnitudes Q and 3Q separated by 2L such that the force on a third charge would be zero not between the charges but at a point greater than 2L from one of the charges?