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In summary, the conversation includes a discussion on solving a problem involving an integral with a factor of r3 in the denominator, the confusion on the approach used for part c, a clarification on the meaning of potential and its physical significance, and a suggestion for converting the potential into cylindrical form.

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Your approach for part c looks confusing. We'll talk about part c, let's get through part b first.

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Sourabh N said:^{3}in the denominator outside the integral (in each of the three integrals). Remember r = x + y + z, so you cannot take it outside the integral.

Your approach for part c looks confusing. We'll talk about part c, let's get through part b first.

It seems [itex]r=\sqrt{x^{2}+y^{2}+z^{2}}[/itex], isn't it?

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Ah, ofcourse! My bad.

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Which is difficult to integrate...

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Not really. For example, for x integral take xathrun200 said:

Which is difficult to integrate...

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Sourabh N said:Not really. For example, for x integral take x^{2}+ y^{2}+ z^{2}= t, then 2xdx = dt and your integral is simply dt/2*(t^1.5)

Oh. I forget that I can use substitution.

But how about the new interval?

From 0 to x^2+y^2+z^2?

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Common convention is to express potential relative to potential at infinity, so limits would be[itex]\infty[/itex] to xathrun200 said:Oh. I forget that I can use substitution.

But how about the new interval?

From 0 to x^2+y^2+z^2?

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athrun200 said:Now, I find that the integral for x, y and z are the same.

Is that correct now?

2 things. First, although you wrote 0 as lower limit, you realize it's actually infinity (as I mentioned in previous post). Having the correct upper and lower limit will help you fix the sign of final value for x integral you get.

Second, note it's the x integral; it is accompanied with a [itex]\hat{x}[/itex] and similarly y and z integral are accompanied with respective unit vectors. Together they form [itex]\vec{E}[/itex] which has to look same as 8.11 (this is part c!)

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Do you mean I should put infinty as the upper limit and x^+y^+z^2 as lower limit?Sourabh N said:2 things. First, although you wrote 0 as lower limit, you realize it's actually infinity (as I mentioned in previous post). Having the correct upper and lower limit will help you fix the sign of final value for x integral you get.

What is the physical meaning if I put infinty as the lower limit and x^+y^+z^2 as the upper limit?

Sourabh N said:Second, note it's the x integral; it is accompanied with a [itex]\hat{x}[/itex] and similarly y and z integral are accompanied with respective unit vectors. Together they form [itex]\vec{E}[/itex] which has to look same as 8.11 (this is part c!)

I don't understand this point.

Do you mean that I miss out the vector?

However, inside the integral is a dot product. i.e.[itex]F\bullet dr[/itex]

Since dot product produce scalar only, there should be not any vector left.

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Look at it this way - xathrun200 said:Do you mean I should put infinty as the upper limit and x^+y^+z^2 as lower limit?

What is the physical meaning if I put infinty as the lower limit and x^+y^+z^2 as the upper limit?

On the meaning of upper lower limits - Griffith defines potential like

I don't understand this point.

Do you mean that I miss out the vector?

However, inside the integral is a dot product. i.e.[itex]F\bullet dr[/itex]

Since dot product produce scalar only, there should be not any vector left.

Ignore this. I jumped on part c in my head.

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Sourabh N said:Look at it this way - x^{2}+ y^{2}+ z^{2}= t (where y and z and not constant anymore). So, 2xdx + 2ydy + 2zdz = dt. This substitution makes more sense than taking x^{2}+ y^{2}+ z^{2}= t with y and z constant as I had suggested previously. Because now the limit of integral is infinity and x^{2}+ y^{2}+ z^{2}. I am not sure about limits of the integral if y and z are kept constant.

On the meaning of upper lower limits - Griffith defines potential like View attachment 37100 I cannot possibly give a better explanation. I highly recommend you have a look (Sec 2.3 in Griffith - Intro to electrodynamics)

Ignore this. I jumped on part c in my head.

So [itex]\phi=\frac{q}{r}[/itex]?

If it is correct, let's move to part c.

I would like to know if I get the correct cylindrical form.

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Yes that is correct. You can check that by using [itex]\vec{E}[/itex] = -[itex]\nabla[/itex] [itex]\phi[/itex] and comparing it with 8.11athrun200 said:So [itex]\phi=\frac{q}{r}[/itex]?

If it is correct, let's move to part c.

To get [itex]\phi[/itex] in cylindrical form, you have to write [itex]\vec{E}[/itex] in cylindrical coordinates and follow the procedure as above. (To know how to go to cylindrical from spherical, look at http://is.gd/JFitBh)I would like to know if I get the correct cylindrical form.

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Sourabh N said:To get [itex]\phi[/itex] in cylindrical form, you have to write [itex]\vec{E}[/itex] in cylindrical coordinates and follow the procedure as above. (To know how to go to cylindrical from spherical, look at http://is.gd/JFitBh)

Do you mean that my approach above can't get the answer?

If so, why?

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