Understanding Electric Charge and Potential: Exploring Concepts and Questions

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In summary: Remember you're trying to figure out the number of electrons that make up 1.602 micro C. Therefore it is Total charge/charge on one electron = 10^13 electrons. Try this equation: E = -grad(V) This equation tells you the electric field (in volts) as a function of the potential (in volts). So if you want to find the y-component of the electric field, just do the following: y = E^(-grad(V))
  • #1
Bradracer18
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A few questions for my homework that I'm having trouble with. I'd like to really grasp these concepts, so any help is appreciated(especially good explanations).

1. When a previously uncharged object is charged to 1.602 microcoulombs...

A. It has an excess of electrons
B. one million electrons have been removed from it
C. 10^13 electrons have been added to it
D. None


---I'm thinking D right now...as I can't find a way to make the others make sense. I know a microcoulomb is 10^-6 C. The sign isn't negative either...which means charged by protons??

2. If the potential is given by V = xy - 3z^-2, then the electric field has a y-component...

A. x
B. y
C. x + y
D. x + y - 6z^-3

I have no clue on this one...need some guidance in getting started. I guess I don't understand how to come up with the individual components, using that equation.

Thank you for the help!
Brad
 
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  • #2
Bradracer18 said:
1. When a previously uncharged object is charged to 1.602 microcoulombs...

A. It has an excess of electrons
B. one million electrons have been removed from it
C. 10^13 electrons have been added to it
D. None


---I'm thinking D right now...as I can't find a way to make the others make sense. I know a microcoulomb is 10^-6 C. The sign isn't negative either...which means charged by protons??
Realize that for most things the net charge is changed by adding or subtracting electrons, not protons. (The protons are usually fixed in a lattice, while some of the electrons can be easily removed.) In any case, since the net charge is positive, electrons must have been removed. How many?

2. If the potential is given by V = xy - 3z^-2, then the electric field has a y-component...

A. x
B. y
C. x + y
D. x + y - 6z^-3

I have no clue on this one...need some guidance in getting started. I guess I don't understand how to come up with the individual components, using that equation.
What's the relationship between electric field and potential? Hint: There's a derivative involved. (Look it up!)
 
  • #3
Ok, so I've found this much anyways...not sure if it is helpful to me or not. But I think V(potential) = Intergal(Electric Field). Is this helpful to me at all...if so, I don't know how to do that intergual to find the y-component.
 
  • #4
Doc, thanks for the good explanation on the electrons. So basically, they are easier to remove...and my 1.602 is positive, then the answer has to be B or D. I don't understand how it can be B...but assuming it is...how do I come up with 1 million?

I'm still not finding that equation(derivitive)...I've even looked on the net for it...any help?
 
  • #5
The charge on one electron is -1.602 x 10-19 C (coulombs). By convention the charge is negative, and the charge on the proton is + while being of the same magnitude.

Figure out how many electrons make 1.602 microcoulombs, and compare that to 1 million.
 
  • #6
yeah ok, that is what I did beforehand. I took the 1.602 x 10^-19C/1.602 x 10^-6 C = 1 x 10 ^-13. This is not even close to a million, but now makes me wonder about answer C. I don't think it is answer C, due to the electrons need to be removed. So, I'm going to say it is answer D.

Thanks Astronuc! Any help with that equation I need for the above question?(that is, if I have the correct answer for this question first).
 
  • #7
Bradracer18 said:
yeah ok, that is what I did beforehand. I took the 1.602 x 10^-19C/1.602 x 10^-6 C = 1 x 10 ^-13. This is not even close to a million, but now makes me wonder about answer C. I don't think it is answer C, due to the electrons need to be removed. So, I'm going to say it is answer D.

Remember you're trying to figure out the number of electrons that make up 1.602 micro C. Therefore it is Total charge/charge on one electron = 10^13 electrons.

Any help with that equation I need for the above question?(that is, if I have the correct answer for this question first).

It's E = -grad(V)
 
  • #9
Ok, well my math is a little rusty on this one(actually, not sure I've been fully taught this stuff)...but from researching it a little more, I found that I just need to kinda break it up into coordinates(and take their derivative). So with that, do I just differentiate xy with respect to y?? If so, I'm not real sure how to do that, as I can't come up with the answer(but if that is what I do, I'll work on it for a bit longer).
 
  • #10
Yes, you'll need to take the (negative of the) partial derivative of V with respect to y to find the y-component of the field.
 
  • #11
So the partial derivative of V, with respect to y. That eliminates the z portion of the equation(I think). So I'm left with dy/dx=xy.

I'm going to say the answer is B, or Y.
 
  • #12
Careful. It's true that (partial) dV/dy = x, but Ey = - dV/dy.
 
  • #13
So...does that make it a -x(which I doubt). My math on these is terrible. If you don't want to explain it any more, that is fine Doc...I'm not really holding up my end of the deal real well...as my math seems to be failing me.
 
  • #14
Bradracer18 said:
So...does that make it a -x(which I doubt).
That's it.

My math on these is terrible.
Your math is OK. What I don't understand is why your homework includes stuff that you haven't covered. (For example, it seems like you haven't covered the mathematical relationship between field and potential.)

Another thing that bugs me: The correct answer for question 2 is not given as a choice. (And for question 1, why didn't they give the actual answer as a choice instead of "NONE"?)
 
  • #15
No kidding Doc. I hate none, as it makes you explore every single angle of the question to make sure you aren't forgetting something(because you have an answer that isn't listed). Also, on question 2...I don't understand why they put the wrong answer either, but I will write something in.

I don't know why he asks questions like these...sometimes I don't think he knows what he's taught us, and what he hasn't.

Thanks again Doc...I'm sure I'll have a few more as the night goes on...

Brad
 

FAQ: Understanding Electric Charge and Potential: Exploring Concepts and Questions

What is the difference between a charged object and a neutral object?

A charged object has an imbalance of positive or negative charges, while a neutral object has an equal number of both positive and negative charges. This imbalance in charges is what creates an electric field around a charged object.

How does the distance between two charged objects affect their potential energy?

The potential energy between two charged objects is directly proportional to the distance between them. As the distance increases, the potential energy decreases, and vice versa. This is known as Coulomb's Law.

Can charged objects have both positive and negative charges?

Yes, charged objects can have both positive and negative charges. This is known as a dipole, where one end of the object has a positive charge and the other end has a negative charge. An example of this is a water molecule.

How does the shape of a charged object affect its potential?

The shape of a charged object can affect its potential by changing the distribution of its charges. For example, a long, thin rod will have a different potential than a spherical object with the same amount of charge. This is because the electric field is stronger at the edges of a charged object, and the shape can affect the strength and direction of the field.

Can the potential of a charged object be changed?

Yes, the potential of a charged object can be changed by either adding or removing charges. Bringing a charged object closer to or farther away from a second charged object can also change the potential of both objects. Additionally, changing the shape or size of a charged object can also affect its potential.

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