Equillibrium of electrostatic forces of 2 charges on a 3rd

[afulldevnull]

Homework Statement

Two particles lie on the x axis. The first particle is at the origin with a charge of +1.6uC and the other has a charge of -3.1uC and is +10.5cm away. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero. what is the coordinates of the third particle?

Homework Equations

Coloumbs Law: F = k (q1 q2) \ r^2

The Attempt at a Solution

Since the charges of the particles are opposite, and because particle 1 has a smaller magnitude, then the third particle must go to the left of particle 1 and must have a negative charge. This is because for the net to be 0, then the strong force at a distance must repel as strongly as the small force nearby attracts. If the net force is to be 0, then the forces must have the same magnitude. So: k(1.6q)/d^2 = k(3.1q) / (10.5+d)^2.
I get the 26 cm, but that can't be right. Any thoughts?

 

[PeterO]

Homework Statement

Two particles lie on the x axis. The first particle is at the origin with a charge of +1.6uC and the other has a charge of -3.1uC and is +10.5cm away. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero. what is the coordinates of the third particle?

Homework Equations

Coloumbs Law: F = k (q1 q2) \ r^2

The Attempt at a Solution

Since the charges of the particles are opposite, and because particle 1 has a smaller magnitude, then the third particle must go to the left of particle 1 and must have a negative charge. This is because for the net to be 0, then the strong force at a distance must repel as strongly as the small force nearby attracts. If the net force is to be 0, then the forces must have the same magnitude. So: k(1.6q)/d^2 = k(3.1q) / (10.5+d)^2.
I get the 26 cm, but that can't be right. Any thoughts?

You said that d was on the left of the origin - so is negative.
The solution of the quadratic equation yields approx 26 and also around -4.
Perhaps it is the "-4" you are after.

Also, the charge of q3 can be either positive or negative.

 

[afulldevnull]

You said that d was on the left of the origin - so is negative.
The solution of the quadratic equation yields approx 26 and also around -4.
Perhaps it is the "-4" you are after.

Also, the charge of q3 can be either positive or negative.

d is a distance, so it cannot be negative, and because we are solving for d then -4 cannot be an answer either

edit: 26 is the right answer

 

[PeterO]

d is a distance, so it cannot be negative, and because we are solving for d then -4 cannot be an answer either

Points taken.
Your answer of 26 actually sounds pretty good - if unexpected.

had the two given charges been 1.6 and 3.2, then the "charge contribution" to the attractive/repulsive forces would be "doubled"
That means the "distance contribution" has to be 1/2

The distance is squared in the formula, so the distance to the larger charge is approx 1.4 times the distance to the smaller charge. we don't really need sqrt(2) = 1.414213... since this is an approximation anyway.

so d + 10.5 is about 1.4 * d

When the extra 10.5 cm represents 0.4 of d, which gives a d value around 26?