How Do Sign Conventions Affect Calculating Net Force on Charges?

[JoeyBob]

Homework Statement

See attached

Relevant Equations

F=9E9 *(q1q2/r)

So 0=q1q3/r+q2q3/(13.6-r)

0=-8.5/r-3.63/13.6+3.63/r

0.2669=3.63/r-8.5/r

r=-18.2465, but the answer is supposed to be 8.24

 

[Charles Link]

I think you figured out that the minus charge must be placed between the two positive charges. Both forces are attractive. The first charge pulls charge 3 to the left, giving it a negative sign. The second charge pulls charge 3 to the right=that force is positive.

Meanwhile you need an exponent of 2 on the distances in the denominator.

Your algebra is also incorrect. e.g. $6/(4-2) \neq 6/4-6/2$.

 

[JoeyBob]

I think you figured out that the minus charge must be placed between the two positive charges. Both forces are attractive. The first charge pulls charge 3 to the left, giving it a negative sign. The second charge pulls charge 3 to the right=that force is positive.

Meanwhile you need an exponent of 2 on the distances in the denominator.

Your algebra is also incorrect. e.g. $6/(4-2) \neq 6/4-6/2$.

Are they not both negative though, because q1*q3 and q2*q3 are both negative?

 

[JoeyBob]

I think you figured out that the minus charge must be placed between the two positive charges. Both forces are attractive. The first charge pulls charge 3 to the left, giving it a negative sign. The second charge pulls charge 3 to the right=that force is positive.

Meanwhile you need an exponent of 2 on the distances in the denominator.

Your algebra is also incorrect. e.g. $6/(4-2) \neq 6/4-6/2$.

I get the right answer as you've said when I make one positive and fix my algebra, but could you explain further why its positive? I understand it pulls the charge to the right, but I was under the impression that, from the equation below, it was the charges that delineated a positive or negative sign.

 

[haruspex]

I get the right answer as you've said when I make one positive and fix my algebra, but could you explain further why its positive? I understand it pulls the charge to the right, but I was under the impression that, from the equation below, it was the charges that delineated a positive or negative sign.

That formula only gives the magnitude. The vector form includes r as a vector, e.g. $\frac{kq_1q_2}{r^3}\vec r$. For the two cases, the r vector is pointing opposite ways.

 

[JoeyBob]

That formula only gives the magnitude. The vector form includes r as a vector, e.g. $\frac{kq_1q_2}{r^3}\vec r$. For the two cases, the r vector is pointing opposite ways.

So the vector from the first charge interaction would be r/r i hat and from the second charge interaction (r-13.6)/sqrt(r-13.6)^2 *i hat. The first would remain positive and cancel, but the second would be negative and cancel because we know r is smaller than 13.6 and a negative divided by the absolute value of a positive is negative.

Is this correct?

 

[Charles Link]

@haruspex wrote it with an $\vec{r}$ in the numerator, and an $r^3$, in the denominator, but when you are first getting started, you are better to look at it as a unit vector divided by $r^2$. If you got the correct answer, I believe you now have the right idea.

As he pointed out, you only get the amplitude of the force from the formula. You then determine which way the force points, (left or right), and put the sign on there accordingly.

 

[Steve4Physics]

Are they not both negative though, because q1*q3 and q2*q3 are both negative?

May I add my two-penn’orth?

The confusion comes from mixing two different sign-conventions and is easily avoided.

When using $F = \frac{kq_1q_2}{r^2}$we are using a radial sign-convention. A positive value of F means outwards (repulsion); a negative value of F means inwards (attraction). Positive or negative values of F do not tell us if F acts in the +x or -x directions; that needs a different sign-convention.

When you have charges along a line it's often simplest to consider the magnitudes of the forces. In this question we have the arrangement:
[+q₁] [-q₃] [+q₂]
For zero force on q₃ we need the 2 attractive forces on it to balance:
|F₁₃| = |F₂₃|

Also worth noting is that answers should have units and be rounded to the appropriate number of significant figures.