1. ## tiny spheres conducting

Two tiny conducting spheres are identical and carry charges of -21.6 mC and +51.4 mC. They are separated by a distance of 2.43 cm. What is the magnitude of the force that each sphere experiences?

2. Originally Posted by lovinhockey26
Two tiny conducting spheres are identical and carry charges of -21.6 mC and +51.4 mC. They are separated by a distance of 2.43 cm. What is the magnitude of the force that each sphere experiences?
Either
$\displaystyle F = \frac{kQ_1Q_2}{r^2}$
or
$\displaystyle F = \frac{Q_1Q_2}{4 \pi \epsilon _0 r^2}$
depending on how your book treats it.

(Note: r will be the distance between the centers of the two spheres. The problem did not specify this.)

-Dan

3. ## still lost

okay...
so using that formula i did the following calculations....

-.0000216 x .0000514/5.9049

which gave me a final answer of -1.88E-10 N

which is apparently the wrong answer... what am I missing here?

4. Originally Posted by lovinhockey26
Two tiny conducting spheres are identical and carry charges of -21.6 mC and +51.4 mC. They are separated by a distance of 2.43 cm. What is the magnitude of the force that each sphere experiences?
Originally Posted by lovinhockey26
okay...
so using that formula i did the following calculations....

-.0000216 x .0000514/5.9049

which gave me a final answer of -1.88E-10 N

which is apparently the wrong answer... what am I missing here?
The "m" in -21.6 mC is "milli" or "micro?" (It reads as "milli" so I don't know if you have that right or not.)

Also the radius is in cm, not m, so the radius squared is $\displaystyle (2.43 \times 10^{-2}~m)^2 = 0.00059049~m^2$

Finally you forgot the constant out at the front: $\displaystyle k = 9 \times 10^9 Nm^2/C^2$.

-Dan

5. it is supposed to be micro coloumbs
in my problem it displays the symbol for mu, but when i copy/paste the mu is lost and simply becomes the letter m... hence..its supposed to be 10^-6, or at least from my knowledge that should be right....

6. ## hmm

when i multiplied the constant by the remaining number (after fixing my cm to m) I came up with 15246...
which I'm fairly confident is incorrect....

I don't know why I keep getting the wrong numbers...I am also unsure of how to label my answer..that is if I ever figure it out...

7. Originally Posted by lovinhockey26
when i multiplied the constant by the remaining number (after fixing my cm to m) I came up with 15246...
which I'm fairly confident is incorrect....

I don't know why I keep getting the wrong numbers...I am also unsure of how to label my answer..that is if I ever figure it out...
I don't know why your calculator isn't working either. This is just "plug-n-chug" stuff.

$\displaystyle F = \frac{(9 \times 10^9)(-21.6 \times 10^{-6})(51.4 \times 10^{-6})}{(2.43 \times 10^{-2})^2} = -194400$

As to the units:
k has a unit of $\displaystyle \frac{Nm^2}{C^2}$

each charge has a unit of C

and r has a unit of m.

Thus F has units of
$\displaystyle \frac{\frac{Nm^2}{C^2} \cdot C \cdot C}{m^2} = N$
as we expect, since F is a force.

-Dan

8. ## :(

I tried the exact way that you have it posted here...

I'm down to my last try and I don't understand what could possibly be wrong...

9. ## maybe

okay,
so I tried it yet again... just calculating numbers... I used your setup that you have posted here....
I came up with -16968.81 N

Is it possible that this could be the answer?? Or am I still missing something somewhere with my calculations?

I'm not going to plug in the answer to the site until I'm more confident, because I really can't afford to lose the points...

10. Originally Posted by lovinhockey26
okay,
so I tried it yet again... just calculating numbers... I used your setup that you have posted here....
I came up with -16968.81 N

Is it possible that this could be the answer?? Or am I still missing something somewhere with my calculations?

I'm not going to plug in the answer to the site until I'm more confident, because I really can't afford to lose the points...
I'm getting -16921.8 N. I'm not sure why there is a difference between our answers.

-Dan