Static Electricity Homework: Positive, Negative, or No Charge?

[harimakenji]

Homework Statement

1.a ball is attracted to both positive and negative charge. state whether the ball has negative, positive, or no charge

2.why does electroscope use gold for the "leaf"?can others metal be used?

3.http://www.sciencebyjones.com/induction.gif
i do not understand that event

Homework Equations

The Attempt at a Solution

1.i do not know.same charge = repel, different charge = attract, no charge = nothing happen.how can a ball attract both positive and negative

2.do not know why it uses gold.maybe another metal can be used

3.i understand the right rod with positive charge will attract negative charge on middle rod.why left rod with positive charge attracts positive charge on middle rod? and at second figure,how can the it be like that? shouldn't the positive charge and negative charge on the middle rod mix again so that middle rod neutral?

 

[collinsmark]

1.i do not know.same charge = repel, different charge = attract, no charge = nothing happen.how can a ball attract both positive and negative

Does it make a difference if the ball is a conductor? (Hint: you might want to think about this before you answer.)

2.do not know why it uses gold.maybe another metal can be used

Interesting question. You might wish to consider some mechanical properties of gold, such as its flexibility, brittleness, etc. Also considering its electrical properties couldn't hurt either.

3.i understand the right rod with positive charge will attract negative charge on middle rod.why left rod with positive charge attracts positive charge on middle rod? and at second figure,how can the it be like that? shouldn't the positive charge and negative charge on the middle rod mix again so that middle rod neutral?

I'm not sure what the question is. The link just shows a picture, with a little explanation involving effect of a charged rod. What exactly are you supposed to find or explain?

 

[harimakenji]

Does it make a difference if the ball is a conductor? (Hint: you might want to think about this before you answer.)

ok,the positive charge attracts negative charge on the ball and negative charge attracts positive charge on the ball,so the ball = no charge = neutral?

Interesting question. You might wish to consider some mechanical properties of gold, such as its flexibility, brittleness, etc. Also considering its electrical properties couldn't hurt either.

i'll try to find it later.about another metal,is it right gold can be replaced by other metals?

I'm not sure what the question is. The link just shows a picture, with a little explanation involving effect of a charged rod. What exactly are you supposed to find or explain?

the link doesn't have question.as i said,i don't understand the event
a.'2 metal rods in contact with positively charge rod nearby'.the two rods are the left and one and middle one,right?from the picture,to begin with,the left rod is positively charged and middle is neutral?

b.the positively charged rod is placed near the middle rod so it attracts the negative charge on the middle rod and repel the positive charge to the left.why can the positive charge on middle rod stay on the left side while there is positively charge rod on the left?shouldn't the left rod repels positive charge on the middle one?

c.'result of taking charged rod away.charges stay intact'.taking charged rod away means that the right rod is taken away,right?how can middle rod become negative?i think after right rod is taken away,the positive charge and negative charge on middle rod will mix again and middle rod becomes neutral, not negatively charged.how can this happen?where has the positive charge on the middle rod gone?

thank you

 

[collinsmark]

Does it make a difference if the ball is a conductor? (Hint: you might want to think about this before you answer.)

ok,the positive charge attracts negative charge on the ball and negative charge attracts positive charge on the ball,so the ball = no charge = neutral?

You might have the right answer in the end, but you may wish consider the reasons more carefully.

Suppose you do have a electrically neutral, conducting ball. What happens to the charge distribution of the conducting ball when you bring some other charge near it? (Hint: the ball will remain electrically neutral as a whole, but its charge distribution will no longer be spherically symmetric.) Given its new charge distribution, will the ball be attracted to the other charge?

i'll try to find it later.about another metal,is it right gold can be replaced by other metals?

I would think so, yes. A quick google search on "electroscope" in the shopping category, shows some electroscope products that do not use gold. But you still might want to consider some reasons why gold leaf (as opposed to some sort of non-gold) is a good choice.

the link doesn't have question.as i said,i don't understand the event
a.'2 metal rods in contact with positively charge rod nearby'.the two rods are the left and one and middle one,right?from the picture,to begin with,the left rod is positively charged and middle is neutral?

I interpret the figure such that before the positively charged (rightmost) rod is brought into proximity, both the middle and leftmost rods are neutral --and both are touching.

Only after the rightmost rod is brought into proximity with the middle rod, does the leftmost rod gain a net positive charge and the middle rod gain a net negative charge.

c.'result of taking charged rod away.charges stay intact'.taking charged rod away means that the right rod is taken away,right?how can middle rod become negative?i think after right rod is taken away,the positive charge and negative charge on middle rod will mix again and middle rod becomes neutral, not negatively charged.how can this happen?where has the positive charge on the middle rod gone?

I think what's happening is that in the beginning, the middle and left rod are touching (and both neutral). Then the positive (rightmost) rod comes into play, giving the middle rod a net negative charge and the left rod a net positive charge (remember both rods are touching). But the middle rod is attracted to rightmost rod more than it is the leftmost rod, given the new charge distribution (take a look at "+" and "-" signs in the middle rod in the first part of the picture). So the middle rod accelerates to the right. This causes the left and middle rod to disconnect (now they are no longer touching).

(One final note. You may find that many people on this forum do not tolerate text-message like wording. It is also part of the forum rules to make a good effort to write in proper English. We all make grammar mistakes, spelling mistakes, and typos. But those should be honest mistakes. Totally ignoring capitalization and punctuation sends the message that the writer doesn't care enough about his or her readers to make one's posts easy to read.)

 

[harimakenji]

You might have the right answer in the end, but you may wish consider the reasons more carefully.

Suppose you do have a electrically neutral, conducting ball. What happens to the charge distribution of the conducting ball when you bring some other charge near it? (Hint: the ball will remain electrically neutral as a whole, but its charge distribution will no longer be spherically symmetric.) Given its new charge distribution, will the ball be attracted to the other charge?

The charge distribution will be divided, one side of the ball will be negative and the other side will be positive. The negative side will be attracted to outside positive charge and positive side will be attracted to outside negative charge. That's the correct reasoning?

I would think so, yes. A quick google search on "electroscope" in the shopping category, shows some electroscope products that do not use gold. But you still might want to consider some reasons why gold leaf (as opposed to some sort of non-gold) is a good choice.

OK, I'll take a look at electrical properties of gold

I think what's happening is that in the beginning, the middle and left rod are touching (and both neutral). Then the positive (rightmost) rod comes into play, giving the middle rod a net negative charge and the left rod a net positive charge (remember both rods are touching). But the middle rod is attracted to rightmost rod more than it is the leftmost rod, given the new charge distribution (take a look at "+" and "-" signs in the middle rod in the first part of the picture). So the middle rod accelerates to the right. This causes the left and middle rod to disconnect (now they are no longer touching).

Because the two rods are connected, the negative charge of the leftmost rod is also attracted by the rightmost rod? And that's why after they are disconnected, the leftmost is positive and the middle is negative?

(One final note. You may find that many people on this forum do not tolerate text-message like wording. It is also part of the forum rules to make a good effort to write in proper English. We all make grammar mistakes, spelling mistakes, and typos. But those should be honest mistakes. Totally ignoring capitalization and punctuation sends the message that the writer doesn't care enough about his or her readers to make one's posts easy to read.)

Thanks for your advice

 

[collinsmark]

The charge distribution will be divided, one side of the ball will be negative and the other side will be positive. The negative side will be attracted to outside positive charge and positive side will be attracted to outside negative charge. That's the correct reasoning?

Yes, that's the right reasoning.

Now consider the distances involved. Imagine dividing up the charge distribution on the ball into a bunch of very small blocks of charge (some positive on one side, and some negative on the other side). Imagine that one of these blocks has charge q. The electric force between this charge q, and the outside charge Q is governed by the equation

$$F = k_e \frac{qQ}{r^2},$$

where k_e is the electric constant, and r is the distance from the outside charge Q to the particular small block of charge q in question, on the ball. The magnitude of this force (regardless if it is attractive or repulsive) is much greater at closer distances. In other words, smaller distances mean bigger force.

With this in mind, will a conducting ball (with overall neutral net-charge) be attracted to an outside charge?

OK, I'll take a look at electrical properties of gold

I'll help you out.

Gold is very good electrical conductor. It's not as conductive as silver or copper, but its still very good. That being said, the electrical properties of other metals would be just fine for use in an electroscope. So yes, gold is a good conductor, but don't dwell on that. Gold's electrical properties are not the main thing that makes it such a good choice, as opposed to other metals. (See below for some additional hints, but here is something to keep in mind: generally, metals are good conductors but corroded metals [like rust] are usually not.)

Here are the more important things to concentrate on:

Gold's chemical properties.
How easily does gold corrode as compared to other metals? (Hint: this is probably the most important of all the properties. The leaf is very thin, so if the surface corrodes, it pretty much means the whole leaf corrodes.)

Gold's physical (mechanical) properties.
How easy is it to form a thin leaf of gold compared to other metals? Once formed to a given thickness, how flexible would this leaf be? How brittle?

Because the two rods are connected, the negative charge of the leftmost rod is also attracted by the rightmost rod? And that's why after they are disconnected, the leftmost is positive and the middle is negative?

Yes, that's one way of looking at it. Charge can flow from one rod to another only if the rods are connected. Once unconnected, any net charge on a given rod stays on that rod.

 

[harimakenji]

Yes, that's the right reasoning.

Now consider the distances involved. Imagine dividing up the charge distribution on the ball into a bunch of very small blocks of charge (some positive on one side, and some negative on the other side). Imagine that one of these blocks has charge q. The electric force between this charge q, and the outside charge Q is governed by the equation

$$F = k_e \frac{qQ}{r^2},$$

where k_e is the electric constant, and r is the distance from the outside charge Q to the particular small block of charge q in question, on the ball. The magnitude of this force (regardless if it is attractive or repulsive) is much greater at closer distances. In other words, smaller distances mean bigger force.

With this in mind, will a conducting ball (with overall neutral net-charge) be attracted to an outside charge?

Yes it will if the distance is close enough?

I'll help you out.

Gold is very good electrical conductor. It's not as conductive as silver or copper, but its still very good. That being said, the electrical properties of other metals would be just fine for use in an electroscope. So yes, gold is a good conductor, but don't dwell on that. Gold's electrical properties are not the main thing that makes it such a good choice, as opposed to other metals. (See below for some additional hints, but here is something to keep in mind: generally, metals are good conductors but corroded metals [like rust] are usually not.)

Here are the more important things to concentrate on:

Gold's chemical properties.
How easily does gold corrode as compared to other metals? (Hint: this is probably the most important of all the properties. The leaf is very thin, so if the surface corrodes, it pretty much means the whole leaf corrodes.)

Gold's physical (mechanical) properties.
How easy is it to form a thin leaf of gold compared to other metals? Once formed to a given thickness, how flexible would this leaf be? How brittle?

Gold is not easily corroded so that the "leaf" can function properly, it is also ductile and it has good flexibility or elastic enough so that the "leaf" can move apart if induced. I think that's the answer.I have another question : http://dev.physicslab.org/img/7198daf9-e4f6-40c1-8542-b95188baa848.gif
I do not understand the 3rd figure. Where does the electrons go? Are they flow to Earth through the outside cap?

Thanks again

 

[collinsmark]

Yes it will if the distance is close enough?

Sounds good to me.

Gold is not easily corroded so that the "leaf" can function properly, it is also ductile and it has good flexibility or elastic enough so that the "leaf" can move apart if induced. I think that's the answer.

I'd buy that.

I have another question : http://dev.physicslab.org/img/7198daf9-e4f6-40c1-8542-b95188baa848.gif
I do not understand the 3rd figure. Where does the electrons go? Are they flow to Earth through the outside cap?

I do not interpret the figure as having any connection to ground/earth at all. The electrons don't flow to earth. However, some do flow into the positively charged rod, at the moment the rod is touches the electroscope's electrode (the metal ball on top). The end result is a somewhat net positive charge for both the rod and the electroscope (although the rod doesn't have quite as much positive charge as it did before -- some of the rod's previous positive charge flowed into the electroscope).

 

[harimakenji]

I do not interpret the figure as having any connection to ground/earth at all. The electrons don't flow to earth. However, some do flow into the positively charged rod, at the moment the rod is touches the electroscope's electrode (the metal ball on top). The end result is a somewhat net positive charge for both the rod and the electroscope (although the rod doesn't have quite as much positive charge as it did before -- some of the rod's previous positive charge flowed into the electroscope).

I didn't think about it before. I agree that the electrons flow to the rod

I think that's my question for now. Thanks a bunch for your generous help collinsmark. Hope to see you again