The dipole question

Conversation started Jul 18, 2019 at 7:34.

Jasmine

Jul 18, 2019 07:34

@JohnRennie I thought of the pair of charges are Q,-Q on the diagonals

John Rennie

Maybe I've misinterpreted the question, but it says the charges are Q, Q, -Q, -Q in cyclic order. I assumed that meant if we start at one corner and go round the square the charges are +Q then +Q then -Q then -Q i.e. the two +Q charges are next to each other, and likewise for the two -Q charges.

Like that

That seems to fit because the answer they give is what you would expect for the two dipoles.

Jasmine

@JohnRennie okay I understand now because of the given condition y>>L you have used the pair Q and -Q as a dipole

John Rennie

@Jasmine yes, exactly!

Jasmine

I had tried this question with too much use of trigonometry in it and used binomial theorem which even gave the wrong answer

John Rennie

Like that

You should be able to do it by doing the detailed calculation, but it's going to be a tortuous calculation.

Jasmine

Jul 18, 2019 07:41

@JohnRennie Thank you I got it !! :-)

John Rennie

We can go through it if you want.

Jasmine

@JohnRennie yes I actually want to go through it as I dont know what mistake I am making

John Rennie

Let me update my diagram ...

Jasmine

There two pairs of Q,-Q on the diagonals the net electric field will be in the sirection of -Q parallel to the diagonal for one pair

And for the other pair Q,-Q on the other diagonal the net electric will be again in the direction towards -Q parallel to the diagonal

And diagonals of square bisect at 90° and since both electric field are equal in magnitude so net will be $\sqrt{2}E_{net}$

John Rennie

I would do the calculation like this ...

The square base and the point P form a square based pyramid. Start by considering one of the faces as I've drawn on the right.

The distance $r$ will be a function of $y$ and $L$, but we don't need to worry exactly what the function is for now.

The field at P can be split into a component parallel to the line $Q$ to $-Q$ and a component perpendicular.

But the perpendicular component will be zero because the $+Q$ and $-Q$ charges will create equal and opposite perpendicular components and they will cancel.

Jasmine

Jul 18, 2019 07:54

@JohnRennie yes

John Rennie

So we only need consider the component parallel to the $Q$ to $-Q$ line.

OK so far?

Jasmine

@JohnRennie yes

John Rennie

And when we do the other two charges, the back face of the pyramid, they will also cancel the perpendicular component and they will produce a parallel component in the same direction. So the total field will be parallel to the two $Q$ to $-Q$ lines.

So we can do the calculation for just one face, as drawn, then double it to get the final result.

Is this all OK? If so we can start actually calculating something!

Jasmine

@JohnRennie yes

John Rennie

OK. Start with the left diagonal. The field is along the diagonal with magnitude $kQ/r^2$.

Jasmine

Jul 18, 2019 07:59

@JohnRennie yes

John Rennie

Call the half angle at the top $\theta$ (so the angle between the two diagonals is $2\theta$) then the horizontal component of the field is $E\sin\theta = kQ\sin\theta/r^2$

Where $\sin\theta = (L/2)/r$

OK so far?

Jasmine

@JohnRennie ok

John Rennie

Substitute for $\sin\theta$ in our expression and we get:

$$ E = \frac{kQL}{2r^3} $$

Jasmine

@JohnRennie yes

John Rennie

That's the field for the $+Q$ charge. From the symmetry the field due to the $-Q$ charge is going to be the same, so the total field from both the charges is:

$$ E = \frac{kQL}{r^3} $$

Jasmine

Jul 18, 2019 08:04

@JohnRennie yes

John Rennie

Note that $QL$ is just the dipole moment $p$, so we have got the standard equation for the dipole field $E = kp/r^3$. That's the equation I started with when I was discussing the problem with Aladdin.

And the other two charges give the same result, so the total field for all four charges is:

$$ E = \frac{2kQL}{r^3} $$

Now we need to express $r$ as a function of $y$ and $L$.

But it should be obvious that if $y \gg L$ then $r \approx y$ and we don't really need to bother. But I will go through it if you want.

Jasmine

@JohnRennie okay

John Rennie

So our final result is:

$$ E \approx \frac{2kQL}{y^3} $$

Jasmine

@JohnRennie yes....

I got it :-)

John Rennie

Note that I haven't cheated anywhere in the calculation, but I have taken advantage of the symmetry to simplify it as much as possible.

With complicated calculations it's always worth looking hard to see if they can be simplified.

Or of course you could do what I did first, wave your hands in the air and say "oh look it's just two dipoles" :-)

Jasmine

Jul 18, 2019 08:14

@JohnRennie yes,, I have asked for the other way because the the other question just asks me to find Electric field on the centroid of an equilateral triangle at a distance y above it without any condition given

I have got the correct answer for both the questions :-)

Conversation ended Jul 18, 2019 at 8:16.

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