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princejhaizee · Jan 19, 2011, 01:51 AM

Two gilded pith balls, spherical in shape and having a mass of 1 gram are supported from a common point thread 1 meter ling. Determine the magnitude of a positive charge that must be given to each ball to cause them to be supported by 10 cm. Please answer it in Detailed Solution!

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Unknown008 · Jan 19, 2011, 02:55 AM

Something like that?

Use the formula:

$F = \frac{QQ'}{4\pi \epsilon_o r^2}$

and

$W = mg$

The angle with which the thread makes with the vertical is given by $\sin^{-1}$\frac{0.05}{1}$$

The tension in one thread is then given by:

$T\cos\theta = W$

$T\sin\theta = F$

You can get Q from those. What do you get?

EDIT: Changed sin^-1(0.1/1) by sin^-1(0.05/1). Thanks ebaines!

ebaines · Jan 19, 2011, 06:40 AM

One correction: I think the angle to the verical would actually be $\sin^{-1}$\frac{0.05}{1}$$ . The balls are separated by 10 cm, and half of that is the distance form each ball to the midpoint between them.

Unknown008 · Jan 19, 2011, 06:48 AM

Yes, I missed that :o

Thanks for pointing it out ebaines :)

princejhaizee · Jan 21, 2011, 02:06 PM

I know that the answer in this problem is 70StatC. Now can you show me the correct or let say complete solution on how to get this 70StatC. Thanks in Advance.

Honestly, even you show me the two formula, I don't know how to use that. Please...

ebaines · Jan 21, 2011, 03:30 PM

Prince - starting with :

$ F = \frac {Q^2} {4 \pi \epsilon_0 r^2}\\ F = T \sin \theta\\ \\ $

(If that first equation looks unfamilar to you, your text may show it this way: $F = k_e \frac {Q^2} {r^2}$ . It's the same thing, just using a different value for the constant.)

So:

$ T \sin \theta = \frac {Q Q'} {4 \pi \epsilon_0 r^2} \\ $

You also know that

$ W = T \cos \theta $

So

$ T = \frac W {\cos \theta} $

Therefore:

$ W \frac {\sin \theta}{\cos \theta} = W \tan \theta = \frac {Q^2} {4 \pi \epsilon_0 r^2} $

Remember that $W = mg$ , r = 0.1 m, and m = 0.001 Kg, and $\tan \theta$ is 0.05 (at such a small angle $\tan \theta$ is essentially the same as $\sin \theta$ .) Now rearrange to solve for Q. Good luck - post back what you get for an answer.

princejhaizee · Jan 23, 2011, 06:59 PM

Hello, this is what I get from your formula:

[(0.001 kg.) (9.8 m/s^2) (0.05)] = [(Q^2 / 9 x 10^9 N * m^2/s^2) (0.01 m)^2]
[0.00049 N] = Q^2 / [0.09 x 10^9 N * s^2]
Q^2 = 0.00049 / 0.09 x 10^9 N * s^2
Q^2 = 0.0054 x 10^9 s^2

Is my solution application correct? It looks like not, because the answer must be 70StatC.

Please do show the complete solution where the answer will be 70StatC.

Thanks

Unknown008 · Jan 23, 2011, 09:54 PM

No, if you see ebaines post, you are using the constant ke, and as such, this constant is in the numerator, not in the denominator of the formula.

Next, you didn't actually use 0.01^2, but you only used 0.01 (for r)

Otherwise, everything is okay, you can go on looking for the value of Q.

And I don't have the least idea of what 70StatC means...