Force between Two Conducting Spheres

[EmilyfromOH]

Homework Statement

Two tiny conducting spheres are identical and carry charges of -19.4 μC and +49.3 μC. They are separated by a distance of 2.55 cm. What is the magnitude of the force that each sphere experiences?

Homework Equations

Coulombs Law: F=k_elq₁q₂l/r²
K_e = 8.99E9

The Attempt at a Solution

The way I write down the numbers is how I put them into my calculator.
F=((8.99E9)(-19.4E-6)(49.3E-6)) / ((2.55E-2)²)
F= -8.5982158 / 6.5025E-4
F=-13222.9385
but magnitude can't be negative so it would be
F= 13222.9385 N

I am not sure what I am doing wrong. I have tried multiple tries at reworking/entering the problem into the calculator to see if I'm somehow messing it up, however I keep getting the same answer.
F=[((-19.4E-6)(49.3E-6)) / ((2.55E-2)²)] * (8.99E9)= 1.47084967E-6 * (8.99E9) = 13222.9385 N
F=(8.99E9)(-19.4E-6)(49.3E-6) = -8.5982158 ; (2.55E-2)² =6.5025E.4 ; F= -8.5982158 / 6.5025E.4 = 13222.9385 N


Help would be greatly appriciated!

 

[TSny]

Hello, EmilyfromOH. Welcome to PF.

Homework Statement

Homework Equations

Coulombs Law: F=k_elq₁q₂l/r²
K_e = 8.99E9

Note the absolute value signs in the formula. When you want the magnitude of the force, use the absolute values of the charges.

 

[EmilyfromOH]

Hello, EmilyfromOH. Welcome to PF.



Note the absolute value signs in the formula. When you want the magnitude of the force, use the absolute values of the charges.

When I did the calculations with the abs value, the number tuned out the same.

 

[TSny]

Note |(-2)(+3)| = |-6| = 6 (a positive result).

Likewise |(-2)(-3)| = |6| = 6 (a positive result again)

So, no matter what the sign of the charges, |q₁q₂| will yield a positive number. All of the other numbers in Coulomb's law are also positive. So, F must come out positive.

Realizing this, there is no need to include any negative signs with the charges when plugging into your calculator while calculating F. Just plug in their absolute values.

(But be careful: There are formulas for other quantities in electricity where you will need to include the signs of the charges and the answers for those quantities can be negative.)

 

[EmilyfromOH]

Note |(-2)(+3)| = |-6| = 6 (a positive result).

Likewise |(-2)(-3)| = |6| = 6 (a positive result again)

So, no matter what the sign of the charges, |q₁q₂| will yield a positive number. All of the other numbers in Coulomb's law are also positive. So, F must come out positive.

Realizing this, there is no need to include any negative signs with the charges when plugging into your calculator while calculating F. Just plug in their absolute values.

(But be careful: There are formulas for other quantities in electricity where you will need to include the signs of the charges and the answers for those quantities can be negative.)

Ok, thanks a lot, that was very helpful!

 

[lightgrav]

wtf? Force is a vector, so it has direction ... a positive direction is NOT the same as a negative direction.
Your calculator entry is flawless, the correct answer IS -13.22 kN away from the other charge ...
"negative away" means "toward" the other charge; each attracts the other.
This is very different than a +19μC charge and a +24μC charge: they would repel, not attract.

probably you're entering the numerical value wrong into the on-line text-box, or messing the units entry.

 

[TSny]

Hi, lightgrav. The question asks for the magnitude of the force on each sphere. So, a positive answer would be required.

 

[lightgrav]

the OP knew that in the very first post; but thought that her computed positive value was wrong anyway.
(probably because it didn't match the mistaken "answer key" - that was her real issue!)
Your comments (and her textbook) could mislead her into thinking that Coulomb Forces are always positive.
Sorry, but that approach ignores the most important feature of a Force, so we should carefully avoid it.

 

[TSny]

Yes, the OP knew she had to find the magnitude. I thought maybe you had overlooked that point, since you said the the correct answer IS negative. I thought by "answer", you were referring to the answer to the OP question since you suggested that she was not entering the answer correctly into the online text-box.

Sorry for any misinterpretation.