Why is a small nut attracted to a straw whereas it should be repulsed?

[Lotto]

Why do you need to be able to say that?
If the initial charges are q₀ and Q₀, what is the repulsive force at distance r, time t?

The repulsive force is $F_\mathrm e =k\frac {q_0 Q_0}{r^2}$ in this case, but I must be able to calculate $Q$ of the straw dependent on time, see post #12. I can do so by measuring the distance $r$, and then by using geometry, I can determine $F_\mathrm e$ thanks to the deflected nut/washer. Then I calculate $Q$ as $Q=2r\sqrt{\pi \epsilon F_\mathrm e}$.

 

[haruspex]

The repulsive force is $F_\mathrm e =k\frac {q_0 Q_0}{r^2}$ in this case, but I must be able to calculate $Q$ of the straw dependent on time, see post #12. I can do so by measuring the distance $r$, and then by using geometry, I can determine $F_\mathrm e$ thanks to the deflected nut/washer. Then I calculate $Q$ as $Q=2r\sqrt{\pi \epsilon F_\mathrm e}$.

You did not answer the question I asked. What is the force at time t? What is the ratio between that and the force at time 0?

 

[Lotto]

You did not answer the question I asked. What is the force at time t? What is the ratio between that and the force at time 0?

I am not sure whether I understand, but $$F_\mathrm e (t)=\frac{Q_0 k q}{r^2} {\mathrm e}^{-\frac{\sigma}{\epsilon} t}=F_\mathrm{e_0} \left(\frac{r_0 }{r }\right )^2 \frac{q}{q_0} {\mathrm e}^{-\frac{\sigma}{\epsilon} t}.$$ So
$$ \frac{F_\mathrm e (t)}{F_\mathrm {e_0}}=\left(\frac{r_0 }{r }\right )^2 \frac{q}{q_0} {\mathrm e}^{-\frac{\sigma}{\epsilon} t}$$
But why do we need it?

 

[nasu]

You don't need the actual values of Q and Qo. And it would be very difficult to find them. You just need to find how the ratio $Q/Q_o$ depend on some distance or angle in your specific setup. Then you will plot this (distance or angle) versus time.

 

[Lotto]

You don't need the actual values of Q and Qo. And it would be very difficult to find them. You just need to find how the ratio $Q/Q_o$ depend on some distance or angle in your specific setup. Then you will plot this (distance or angle) versus time.

Well, I am actually supposed to determine $Q$ dependent on time, see post #12.

 

[Lotto]

However, let's say I have only one nut and after rubbing the straw, the nut is attracted to it, so if I put the straw far enough so that the nut can't touch it, what would be their distance $r$, when I then want to use equation $Q=2r\sqrt{\pi \epsilon F_\mathrm e}$? The distance between the center of mass of the charged part of the straw and the center of mass of the nut? And will their charges have the same value? I think yes, but I am no sure.

 

[haruspex]

Well, I am actually supposed to determine $Q$ dependent on time, see post #12.

No, it says find the product of the charges, qQ. But if the purpose is to find $\sigma$ then you do not even need to find that.

 

[nasu]

Well, I am actually supposed to determine $Q$ dependent on time, see post #12.

There is no such requirement in the post #12. Using that formula does not mean that you need to know Q.

 

[haruspex]

I am not sure whether I understand, but $$F_\mathrm e (t)=\frac{Q_0 k q}{r^2} {\mathrm e}^{-\frac{\sigma}{\epsilon} t}=F_\mathrm{e_0} \left(\frac{r_0 }{r }\right )^2 \frac{q}{q_0} {\mathrm e}^{-\frac{\sigma}{\epsilon} t}.$$ So
$$ \frac{F_\mathrm e (t)}{F_\mathrm {e_0}}=\left(\frac{r_0 }{r }\right )^2 \frac{q}{q_0} {\mathrm e}^{-\frac{\sigma}{\epsilon} t}$$
But why do we need it?

Since you are working with two balloons, both discharging, the equations are $$F_\mathrm e (t)=k\frac{Q(t)q(t)}{r^2} =k\frac{Q_0q_0}{r^2} {\mathrm e}^{-2\frac{\sigma}{\epsilon} t} =
F_\mathrm{e_0} \left(\frac{r_0 }{r }\right )^2 {\mathrm e}^{-2\frac{\sigma}{\epsilon} t}$$.
Using the angle, you will measure F(t) and r(t), so you can plot.. what against what to find $\sigma$?

 

[Lotto]

Since you are working with two balloons, both discharging, the equations are $$F_\mathrm e (t)=k\frac{Q(t)q(t)}{r^2} =k\frac{Q_0q_0}{r^2} {\mathrm e}^{-2\frac{\sigma}{\epsilon} t} =
F_\mathrm{e_0} \left(\frac{r_0 }{r }\right )^2 {\mathrm e}^{-2\frac{\sigma}{\epsilon} t}$$.
Using the angle, you will measure F(t) and r(t), so you can plot.. what against what to find $\sigma$?

I understand, but I won't do it with balloons, I will do it this way: I will rub the straw and then made the washer deflect because it will be attracted to the straw (but the straw will be far enough so that it won't touch it). Then I will measure the distance between the washer and the straw. What can I say about their charges? If the straw has a charge $Q$, what charge has the washer? It is attracted, so it must have a charge, maybe only a partial one, but still some. Isn't its charge the same as $Q$ (in value)? Or are their charges different? I know it doesn't matter, but it interests me.

And I still think that when I am to measure the dependence of the straw's self-discharge on time, I am to measure its $Q$ dependent on time, but that is only my opinion.

 

[haruspex]

Or are their charges different? I know it doesn't matter, but it interests me.

And I still think that when I am to measure the dependence of the straw's self-discharge on time, I am to measure its $Q$ dependent on time, but that is only my opinion.

Since the problem statement only expects you to be able to find the product of the charges, the question setter clearly understood that you cannot find the individual charges in the asymmetric (washer and straw) model.
Different discharge rates is a problem, though. You can only find the sum of the $\sigma$ values.
With two balloons you can assume the discharge rates are the same, but I am not sure about washer and straw.

what charge has the washer? It is attracted, so it must have a charge,

No, even an uncharged conductor will be attracted because of the induced charge distribution.

 

[Lotto]

No, even an uncharged conductor will be attracted because of the induced charge distribution.

And can I say that the washer has a partial charge, that it has on one side a negative charge and on the other side a positive charge?

 

[haruspex]

And can I say that the washer has a partial charge, that it has on one side a negative charge and on the other side a positive charge?

If this is in the case where you touched the straw against the washer, both will have a negative charge overall, and the side of the washer away from the straw will be negative.
On cursory analysis, the side of the washer near the straw may be positive or negative. But maybe a more detailed view would resolve that.