Electrodynamics: Conducting sphere of radius R cut in half

In summary, we have a conducting sphere that has been cut in half to form a gap with a small distance between the two halves. One half is charged with q and the other half is uncharged. We are asked to find the charge density σ(r) on the planar face of the first half-sphere, where s << R. We can assume that all four σ's are uniform and neglect edge effects. We can express σ2, σ3, and σ4 in terms of σ1, and using Gauss' law, we can show that the electric field in the gap is equal to σ/εo.
  • #1
sdefresco
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Homework Statement
Electrodynamics: Conducting Sphere cut in half to form a gap, and a charge q is placed on the first half-sphere. Find all four σ, as well as E in the and the potential difference between the hemispheres.
Relevant Equations
Anything Gauss' Law, boundary conditions, known features of a conductor, and any E-field derivations that come out of Coulomb's law.
Summary: Electrodynamics: Conducting Sphere cut in half to form a gap, and a charge q is placed on the first half-sphere. Find all four σ.
A sphere of radius R is cut in half to form a gap of s << R (ignore edge effects) - the first hemisphere is charged with q, and the second hemisphere is left uncharged.

My first question is to determine σ(r) on the plane face of the first hemisphere.

My personal question is as to whether or not this σ depends on r (ie non-constant). If it does, it is due to geometry. My question, then, is how does σ depend on r for the plane-face of the first half-sphere? Does the sphere being locally flat not constitute a constant σ of σ1(r)=1/3(Q/(3πR^2)) to account for 3Sface=Stot

The question later on asks for the electric field between the gap, so this answer should be obtainable before making that conclusion. I know that E must point straight from σ1 to σ2, as E must be normal to both of the surfaces since they are conductors.
 
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  • #2
Interesting problem. See if you can get a reasonable answer by assuming that all four ##\sigma##'s are uniform (neglecting edge effects). So you could let ##\sigma_1## represent the unknown charge density on the flat surface of the upper sphere. Can you find expressions for the other three ##\sigma##'s in terms of ##\sigma_1##?
 
  • #3
TSny said:
Interesting problem. See if you can get a reasonable answer by assuming that all four ##\sigma##'s are uniform (neglecting edge effects). So you could let ##\sigma_1## represent the unknown charge density on the flat surface of the upper sphere. Can you find expressions for the other three ##\sigma##'s in terms of ##\sigma_1##?
Current, not-yet-debunked assumptions: Q is distributed uniformly about each hemisphere, so σ1,2(r) and σ3,4(θ) are constants. Thus, |σ|=Q/A for each case, simply

I feel as if the way the question is structured, it suggests σ1 through σ4 can all be obtaining as functions of r (possibly constants, as we've both questioned), without having to define using each other.

But, if σ1(r) is constant (Q is uniformly distributed amongst planar face), then since the planar face contains (1/3)Q, I think σ1=(1/3)Q/(4π R2).
So, Is it fair to claim that 1/3 of Q lies on the planar face due to S1=(1/3)Stotal? If so, my solution should be correct.

From there, σ1(r) =-σ2(r) due to inducted charge, and σ3,4(r) would come from the surface area relationship once again.

I've also thought about using the boundary condition that E over σ should be σ/εo, but am unsure if I can use that alone to say Egap=σ/εo

.
 
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  • #4
Tyler DeFrancesco said:
Current, not-yet-debunked assumptions: Q is distributed uniformly about each hemisphere, so σ1,2(r) and σ3,4(θ) are constants. Thus, |σ|=Q/A for each case, simply
First thing to do is to define which surfaces the subscripts 3 and 4 refer to. From your first post, I think ##\sigma_2## is the charge density on the flat surface of the lower hemisphere.

I feel as if the way the question is structured, it suggests σ1 through σ4 can all be obtaining as functions of r (possibly constants, as we've both questioned), without having to define using each other.
If you can express ##\sigma_2##, ##\sigma_3##, and ##\sigma_4## in terms of just ##\sigma_1## (and I think you can), then you will have reduced 4 unknowns to just 1 unknown. For example, later in your post you relate ##\sigma_2## to ##\sigma_1##. See below.

But, if σ1(r) is constant (Q is uniformly distributed amongst planar face), then since the planar face contains (1/3)Q, I think σ1=(1/3)Q/(4π R2).
So, Is it fair to claim that 1/3 of Q lies on the planar face due to S1=(1/3)Stotal? If so, my solution should be correct.
In the original post, the total charge on the upper hemisphere is denoted ##q##. Is this the same as ##Q##, or does ##Q## denote some other charge? You should be able to determine the four constant ##\sigma##'s without assuming that the flat surface has 1/3 of the total charge of the upper hemisphere. Then, you will see whether or not that assumption is true.

From there, σ1(r) =-σ2(r) due to inducted charge, and σ3,4(r) would come from the surface area relationship once again.
Can you explain in a little more detail how you arrive at ##\sigma_2 = -\sigma_1##? Note that you have succeeded in expressing ##\sigma_2## in terms of ##\sigma_1##. That just leaves ##\sigma_3## and ##\sigma_4## to be expressed in terms of ##\sigma_1##.

I've also thought about using the boundary condition that E over σ should be σ/εo
Do you mean that E at any point just outside the surface of one of the conductors is |σ|/εo? If so, then, yes. You can use Gauss' law to show this.
but am unsure if I can use that alone to say Egap=σ/εo.
Yes, the field at any point in the gap (away from the edges) is uniform.
 
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  • #5
o
TSny said:
First thing to do is to define which surfaces the subscripts 2, 3, and 4 refer to.

If you can express ##\sigma_2##, ##\sigma_3##, and ##\sigma_4## in terms of just ##\sigma_1## (and I think you can), then you will have reduced 4 unknowns to just 1 unknown. For example, using Gauss' law, can you express the ##\sigma## for the flat surface of the lower hemisphere in terms of ##\sigma_1##? (I think you actually do this later in your post. See below)

In the original post, the total charge on the upper hemisphere is denoted ##q##. Is this the same as ##Q##, or does ##Q## denote some other charge? You should be able to determine the four constant ##\sigma##'s without assuming that the flat surface has 1/3 of the total charge of the upper hemisphere. Then, you will see whether or not that assumption is true.

OK, here I think you are using the subscript 2 for the flat surface of the lower hemisphere. Can you explain in a little more detail how you arrive at ##\sigma_2 = -\sigma_1##?

Do you mean that E at any point just outside the surface of one of the conductors is σ/εo? If so, then, yes. You can use Gauss' law to show this.
Yes, the field at any point in the gap (away from the edges) is uniform.

σ1(r) -> planar face of the first hemisphere
σ1(r) -> planar face of the second hemisphere
σ3/4(θ) -> curved, "cap" surface of the first/second hemisphere

I say σ1(r) = -σ2(r) because σ1(r) induces σ2(r), as the second hemisphere is [also] a conductor.

Q=q, yes.

By the boundary condition, I mean that the difference in E across a surface charge is σ/εo.

As for your point about finding σ1(r) in the first place, I am most stuck on this - I do not know how to determine this without following the original logic to obtain σ1(r)=⅓Q/4πR2.
 
  • #6
Tyler DeFrancesco said:
σ1(r) -> planar face of the first hemisphere
σ1(r) -> planar face of the second hemisphere
σ3/4(θ) -> curved, "cap" surface of the first/second hemisphere
OK, thanks. The second line should be ##\sigma_2##, but I'm sure that's what you meant.

I say σ1(r) = -σ2(r) because σ1(r) induces σ2(r), as the second hemisphere is [also] a conductor.
You can make this more rigorous using Gauss' law.

By the boundary condition, I mean that the difference in E across a surface charge is σ/εo.
OK. Since you know the value of E inside the conducting material, this tells you what E is just outside the conductor.

As for your point about finding σ1(r) in the first place, I am most stuck on this - I do not know how to determine this without following the original logic to obtain σ1(r)=⅓Q/4πR2.
You'll need to bring in another concept to get the value for ##\sigma_1##. But this can wait. First, it will be helpful to have ##\sigma_2##, ##\sigma_3## and ##\sigma_4## expressed in terms of ##\sigma_1##. For example, can you express ##\sigma_3## in terms of ##\sigma_1## using the fact that ##q## is assumed given?
 
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  • #7
TSny said:
You'll need to bring in another concept to get the value for ##\sigma_1##.

The only think I could think of is to say that σ1=(1/3)q/A=(1/3)q/(πR2) under the assumption that σ1 is constant (i.e., (1/3)q is distributed uniformly across S1.

Then,
TSny said:
But this can wait. First, it will be helpful to have σ2, σ3 and σ4 expressed in terms of σ1σ1. For example, can you express σ3 in terms of σ1σ1 using the fact that q is assumed given?
The only thing I can think of is using the geometry argument,
S1,2=(1/2)S3,4
and so |σ3,4|=(2/3)q/A=(2/3)q/(2πR2)=|σ1,2|

Not my original result, so I'm curious if this is correct reasoning. This is obviously the same result as if I assumed σ=q/Stot=q/3πR2, meaning q is distributed completely uniformly about S, I believe.
 
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  • #8
Tyler DeFrancesco said:
The only think I could think of is to say that σ1=(1/3)q/A=(1/3)q/(πR2) under the assumption that σ1 is constant (i.e., (1/3)q is distributed uniformly across S1.
To say that 1/3 of q is on the flat surface of the upper hemisphere is the same as saying that ##\sigma_3= \sigma_1##. But that is not the correct relation between ##\sigma_3## and ##\sigma_1##.

Think about how you can express q in terms of ##\sigma_1##, ##\sigma_3##, and R.
 
  • #9
TSny said:
To say that 1/3 of q is on the flat surface of the upper hemisphere is the same as saying that ##\sigma_3= \sigma_1##. But that is not the correct relation between ##\sigma_3## and ##\sigma_1##.

Think about how you can express q in terms of ##\sigma_1##, ##\sigma_3##, and R.
By definition, q=∫∫σdS=∫∫σ3R2Sin(θ)dθd∅+∫∫σ1rdrdθ, which will come out to give some relationship of the two sigmas.

Could this reasoning yield their relationship?

-> q=2πσ3R2+πσ1R2
 
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  • #10
Tyler DeFrancesco said:
q=2πσ1R2+πσ3R2
Looks pretty good, but did you accidently switch ##\sigma_1## and ##\sigma_3##?
 
  • #11
TSny said:
Looks pretty good, but did you accidentally switch ##\sigma_1## and ##\sigma_3##?
I did, thank you for pointing that out.
Now, finding σ1 to complete the solution is still a bump for me.
 
  • #12
TSny said:
Looks pretty good, but did you accidently switch ##\sigma_1## and ##\sigma_3##?
I don't understand the 3D-symmetry involved that allow us to infer that ##\sigma_1## and ##\sigma_3## are constant throughout their domains.
 
  • #13
Tyler DeFrancesco said:
Now, finding σ1 to complete the solution is still a bump for me.
Before getting to that, what about ##\sigma_4##?
 
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  • #14
Delta2 said:
I don't understand the 3D-symmetry involved that allow us to infer that ##\sigma_1## and ##\sigma_3## are constant throughout their domains.
I think you can get an answer with constant σ's that is a good approximation when s/R can be neglected compared to 1. This would be a "zeroth-order" approximation in s/R. But I might be overlooking something.

At first, I thought there was an answer that was accurate to first order in s/R with constant σ's. But I no longer think so.
 
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  • #15
TSny said:
At first, I thought there was an answer that was accurate to first order in s/R with constant σ's. But I no longer think so.
Delta2 said:
I don't understand the 3D-symmetry involved that allow us to infer that σ1σ1 and σ3σ3 are constant throughout their domains.
This is one of the main things I’ve been questioning. The problem is structured such that I feel as if I’m expected to give the σ’s as functions of r or theta, depending on if it’s the planar or cap face, respectively.
Note: r is defined as the distant from the sphere, but I believe it refers to the radial distance from the center of the planar face. Theta would be measured from the third-axis down from the cap (equator at θ=π/2).

If I have to abandon the assumption of constant surface charge, where do I go from there? Obviously the geometry of the hemisphere is key, but how does it dictate charge distribution aside from E=0 inside?

Clearly, the charges on the cap and face need to rearrange themselves to cancel E within the hemisphere: this shouldn't really give me any details on how to answer this. If the E field between the gaps really is
Egap(r)=σ1(r) /εo, due to boundary conditions for a conductor (since E=0 inside, E=σ/εo outside),
then this field too should be non-constant. Does that make sense for s<<R^2 and ignoring edge effects? Why would these hemispheres not behave like parallel plate capacitors in-between the gap?

If I assume σcap depends on θ, then I am not sure what the limits should even logically be at θ=0 and θ=Pi/2 in order to determine what σcap is. Should σcap(0)=σ1(0) due to these points only being separated by R and being located on the third axis?
 
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  • #16
It's not clear from the statement of the problem how accurate your answer is expected to be. I think I can help you get the lowest-order approximate solution in which the σ's constants. (This is the solution in the limit where the gap distance ##s## approaches zero.) If you are expected to go beyond this level of approximation, then the σ's will not be constants and I don't see a manageable way to solve it. May I ask at what level you are studying this material? For example, are you expected to be familiar with solving Laplace's equation as a boundary value problem? If this is a problem from a textbook, can you tell us the title and author? If this is from a course, what is the level of the course?

If you want to continue with finding the lowest-order approximate solution we can carry on with it. You have succeeded in expressing ##\sigma_2## and ##\sigma_3## in terms of ##\sigma_1##. If you can also express ##\sigma_4## in terms of ##\sigma_1##, then you'll almost be home.
 
  • #17
TSny said:
It's not clear from the statement of the problem how accurate your answer is expected to be. I think I can help you get the lowest-order approximate solution in which the σ's constants. (This is the solution in the limit where the gap distance ##s## approaches zero.) If you are expected to go beyond this level of approximation, then the σ's will not be constants and I don't see a manageable way to solve it. May I ask at what level you are studying this material? For example, are you expected to be familiar with solving Laplace's equation as a boundary value problem? If this is a problem from a textbook, can you tell us the title and author? If this is from a course, what is the level of the course?

If you want to continue with finding the lowest-order approximate solution we can carry on with it. You have succeeded in expressing ##\sigma_2## and ##\sigma_3## in terms of ##\sigma_1##. If you can also express ##\sigma_4## in terms of ##\sigma_1##, then you'll almost be home.
I'm currently an undergrad taking Electrodynamics I (junior/senior level class depending on the major). We're following Griffith's 4e text. Currently, using Laplace's equation to solve boundary value problems has not yet been broached, although I'm ahead in the textbook so I know of the general method. The problem itself was created by my professor, I believe (or is at least not from Griffith's or any other indexed source I can find).

My prof. suggests that the entire problem set should be solvable using knowledge in Griffith's from Ch.1 up to Conductors in Ch.2 (Electrostatics).
 
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  • #18
OK. So, I think we just want an approximation that is good for very small ##s## where we can take the ##\sigma##'s as constant.

Let's try this:

What does Enet have to be at any point inside the conducting material?
What's the total E inside the material that's produced by the total charge of ##\sigma_1## and ##\sigma_2## taken together (neglecting edge effects)?

So, what can you say about the total E inside the material produced by the total charge of ##\sigma_3## and ##\sigma_4## taken together?
 
  • #19
TSny said:
OK. So, I think we just want an approximation that is good for very small ##s## where we can take the ##\sigma##'s as constant.

Let's try this:

What does Enet have to be at any point inside the conducting material?
What's the total E inside the material that's produced by the total charge of ##\sigma_1## and ##\sigma_2## taken together (neglecting edge effects)?

So, what can you say about the total E inside the material produced by the total charge of ##\sigma_3## and ##\sigma_4## taken together?
Enet=0 inside the material at every point,
So E inside produced by sigma 1 and 2 should be equal in magnitude and opposite in sign.
Therefore, 3 and 4 should also be equal in magnitude and opposite in sign?
 
  • #20
Tyler DeFrancesco said:
Enet=0 inside the material at every point,
So E inside produced by sigma 1 and 2 should be equal in magnitude and opposite in sign.
Therefore, 3 and 4 should also be equal in magnitude and opposite in sign?
When you say "3 and 4 should also be equal in magnitude and opposite in sign", are you saying that the field produced by the charge on surface 3 should cancel the field produced by the charge on surface 4 at each point inside the material? If so, I agree.

From this, can you deduce any relationship between ##\sigma_3## and ##\sigma_4##?

Also, can you deduce the total amount of charge on surfaces 3 and 4 together?
 
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  • #21
TSny said:
When you say "3 and 4 should also be equal in magnitude and opposite in sign", are you saying that the fields produced by ##\sigma_3## and ##\sigma_4## should cancel at each point inside the material? If so, I agree.

From this, can you deduce any relationship between ##\sigma_3## and ##\sigma_4##?

Also, can you deduce the total amount of charge on surfaces 3 and 4 together?
Aside from using my original geometry argument, no. If by 3 and 4 you mean the cap surfaces, then I honestly don’t know. I feel like overanalyzing this problem without anymore guidance is just further confusing me. I can make many arguments that appear logical for any answer and arrive at something that I cannot check.
 
  • #22
Tyler DeFrancesco said:
Aside from using my original geometry argument, no.
I'm not sure which of my questions your "no" refers to.

You know that the net electric field inside the material must be zero. You also correctly stated that the two flat surfaces together don't produce any electric field inside the material. So, does it make sense that the net field produced by the two spherical caps must be zero at any point inside the material?
 
  • #23
TSny said:
I'm not sure which of my questions your "no" refers to.

You know that the net electric field inside the material must be zero. You also correctly stated that the two flat surfaces together don't produce any electric field inside the material. So, does it make sense that the net field produced by the two spherical caps must be zero at any point inside the material?
In total, yes, the E field from each of the caps should also net to zero. That seems to suggest sigma1,2=sigma3,4, but you’ve already stated that’s not a correct relationship.
Likewise, |sigma3|=|sigma4|, but this doesn’t give a relationship between the four or with q. This set is due in a couple of hours and I’ve not made much progress at all on this final problem.
Specifically, sigma3=-sigma4.
 
  • #24
Tyler DeFrancesco said:
In total, yes, the E field from each of the caps should also net to zero.
Yes, at any point inside the conducting material, the field from the top cap cancels the field from the bottom cap.

That seems to suggest sigma1,2=sigma3,4, but you’ve already stated that’s not a correct relationship.
It's not a correct relationship. You'll deduce the correct relationship later.

Let's continue with the fact that the charges on the two spherical caps must be such that they produce zero net field anywhere inside the sphere.

Suppose I handed you a basketball and asked you to spread some charge on the surface of the basketball such that the net E field at every point inside the basketball is zero. How would you distribute the charge?
 
  • #25
TSny said:
Yes, at any point inside the conducting material, the field from the top cap cancels the field from the bottom cap.

It's not a correct relationship. You'll deduce the correct relationship later.

Let's continue with the fact that the charges on the two spherical caps must be such that they produce zero net field anywhere inside the sphere.

Suppose I handed you a basketball and asked you to spread some charge on the surface of the basketball such that the net E field at every point inside the basketball is zero. How would you distribute the charge?
The obvious answer of Q distribution amongst a conducting sphere is uniformly about the surface, which would be the equivalent of both of the caps’ surface charges.
It’s when the sphere is cut that my logic starts to become second-guessed.
 
  • #26
Tyler DeFrancesco said:
The obvious answer of Q distribution amongst a conducting sphere is uniformly about the surface, which would be the equivalent of both of the caps’ surface charges.
Yes. The charge must be uniformly spread over the entire spherical surface.

It’s when the sphere is cut that my logic starts to become second-guessed.
The gap is so narrow that the two caps together essentially make a spherical surface. Neglecting the gap means that we are only getting an approximate answer. But the approximation should be good for s << R.

So, as you said ##\sigma_3 = \sigma_4##.
Can you see what the total charge on the surface of the sphere must be in terms of ##q##?
 
  • #27
TSny said:
Yes. The charge must be uniformly spread over the entire spherical surface.

The gap is so narrow that the two caps together essentially make a spherical surface. Neglecting the gap means that we are only getting an approximate answer. But the approximation should be good for s << R.

So, as you said ##\sigma_3 = \sigma_4##
By that logic, E=0 in the gap and sigma1,2 should be zero if they effectively don’t exist. That doesn’t seem right to me either.
 
  • #28
Why would ##\sigma_1## and ##\sigma_2## be zero?
 
  • #29
There will be a nonzero E in the gap. But E will be zero everywhere inside the conducting material. This requires uniform distribution of charge over the surface of the sphere.
 
  • #30
TSny said:
There will be a nonzero E in the gap. But E will be zero everywhere inside the conducting material. This requires uniform distribution of charge over the surface of the sphere.
I’m not understanding how that reveals information about the planar faces of the hemispheres. They don’t exist in a non-cut sphere analogue, so what happens to the charge on that?
 
  • #31
Tyler DeFrancesco said:
I’m not understanding how that reveals information about the planar faces of the hemispheres. They don’t exist in a non-cut sphere analogue, so what happens to the charge on that?
You'll be able to deduce the charge on the planar faces after you see how much charge is on each hemispherical cap. But you do already know that the net charge of both planar surfaces is zero since they carry equal but opposite charge.

What's the total charge of the entire system?

What's the total charge on the two hemispherical caps?

{supper time}
 
  • #32
TSny said:
You'll be able to deduce the charge on the planar faces after you see how much charge is on each hemispherical cap. But you do already know that the net charge of both planar surfaces is zero since they carry equal but opposite charge.

What's the total charge of the entire system?

What's the total charge on the two hemispherical caps?

{supper time}
Total charge of the system is +Q.
I don’t know what the total charge on the caps is. +Q or some fraction of Q dictated by geometry, I suppose.
They need to net to Q, so 1/2 Q each sounds right. I feel the caps’ charge need to be opposite in order for E=0 to be maintained within each of the hemispheres, however.
 
  • #33
Tyler DeFrancesco said:
Total charge of the system is +Q.
I don’t know what the total charge on the caps is. +Q or some fraction of Q dictated by geometry, I suppose.
They need to net to Q, so 1/2 Q each sounds right.
Right. Each cap has a net charge of Q/2, where Q is the amount of charge that was initially placed on the upper hemispere.

I feel the caps’ charge need to be opposite in order for E=0 to be maintained within each of the hemispheres, however.
Remember the basketball. You had to place a uniform charge density over the entire surface in order to get E = 0 everywhere inside. So, the charge density for each hemispherical cap must be the same (including sign).
 
  • #34
TSny said:
Right. Each cap has a net charge of Q/2, where Q is the amount of charge that was initially placed on the upper hemispere.

Remember the basketball. You had to place a uniform charge density over the entire surface in order to get E = 0 everywhere inside. So, the charge density for each hemispherical cap must be the same (including sign).
It seems a bit counter intuitive that this could be true for electrically isolated objects. This would mean that σ1 and σ3 have the same sign, but that's impossible since E=0 would not be true in this case for the hemisphere. I feel like calling the objects "electrically isolated" (as specified in the prompt), and then saying s<<R, doesn't actually give me anything to go off of, but rather only further confuses me due to these questions. I could say I know how everything will re-arrange themselves by concluding E=0 in a conductor, but that's not exactly true because clearly it's been days and I still don't have a solid answer. If I'm at a constant loss to really justify that my physical system follows my train of logic, I won't gain anything out of this.

Frankly, I have to turn this problem set in within a couple of hours, and still have nothing worth writing on the page.
 
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  • #35
Tyler DeFrancesco said:
It seems a bit counter intuitive that this could be true for electrically isolated objects. This would mean that σ1 and σ3 have the same sign, but that's impossible since E=0 would not be true in this case for the hemisphere.
Given that s<<R we can make the approximation that all ##\sigma_i## will be constant throughout their respective domains. Also that the two spherical caps each carry charge +Q/2 since the whole system is very similar to an uncut sphere (because s<<R). This leaves that the flat face of the upper hemisphere (to which we put total charge +Q) has charge +Q/2 and the flat face of the lower hemisphere has charge -Q/2. Again because s<<R the electric field due to the charges on the flat surfaces will be zero in the space that is not in-between the flat surfaces, because the two flat surface each with equal and opposite charge are like a parallel plate capacitor. The electric field in-between the flat surfaces will be that of a parallel plate capacitor with charge Q/2 and -Q/2 in each plate.
Frankly, I have to turn this problem set in within a couple of hours, and still have nothing worth writing on the page.
I think with the guide of @TSny the problem has been solved, you just haven't realized it :D.
 

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