Electrostatics - finding the work done

[Saitama]

How did you get $n = 1$? Remember, $\sigma$ is a charge density, not a charge.

Is $n=3$? Am I right in saying that the charge density doubles after the first fold?

 

[voko]

Are you sure we you can't solve this with just dimensional analysis?

To perform a fold, it seems to me that the only quantities that the work can depend on are the initial charge density $\sigma$ and the initial leg length $a$ (as well as Coulomb's constant). That should be enough to get the answer.

You are probably right. There may be a methodological difficulty in justifying the constant, though, unless a deeper analysis is performed.

The potential energy approach should reveal that constant.

 

[TSny]

Is $n=3$? Am I right in saying that the charge density doubles after the first fold?

I would say yes to both questions.

 

[Saitama]

I would say yes to both questions.

Thanks a lot TSny and voko!

I get $W_2=\sqrt{2}\,\, J$.

But how do I find the expression for potential energy? I still don't see how to set up the integrals.

And where one can find such problems? Are they present in any text?

 

[voko]

For the integrals, let $\Delta_1$ be the first triangle, then $\Delta_2$, etc.

The potential due to the first triangle is $$ \phi_1(\xi, \eta) = \int\limits_{(x,y) \in \Delta_1} k \sigma_1 \frac {dx dy} {\sqrt{(x - \xi)^2 + (y - \eta)^2}} $$ *edit: clarified the potential integral.

The potential energy is $$ U_1 = \frac 1 2 \int\limits_{(\xi,\eta) \in \Delta_1} \sigma_1 \phi_1(\xi, \eta) {d\xi d\eta}$$ Now, if we replace $\sigma_1$ with $\sigma_2$ and $\Delta_1$ with $\Delta_2$, we will get $U_2$. It is obvious how to replace the charge density. To replace the area of integration, we can map linearly $\Delta_1 \to \Delta_2$, so the integral for $U_2$ will now be taken over $\Delta_1$, but the change of variables will yield another scaling factor $s$, so $$U_2 = \frac {\sigma_2^2} {\sigma_1^2} s U_1 $$

 

[TSny]

voko's explanation is nice. I was thinking of a similar but cruder way to look at it by using $U = \frac{k}{2}\displaystyle\sum\limits_{i\neq j} \frac{q_i q_j}{r_{ij}}$ for a system of particles.

Thus, break the triangle into small patches of area $\Delta A_i$ so you have $U = \frac{k\sigma^2}{2}\sum\limits_{i\neq j} \frac{\Delta A_i \Delta A_j}{r_{ij}}$.

Then scale the entire figure (including the grid) by a linear scaling factor so that you still have the same number of patches. See attached figure. You have a new energy

$U = \frac{k\sigma\,'\,^2}{2}\sum\limits_{i\neq j} \frac{\Delta A_i' \Delta A_j'}{r_{ij}'}$ Then think about how the new primed quantities compare to the original unprimed quantities. But for the new $\sigma\,'$ you'll want to use the charge density that results from folding, not the charge density that would result from the shrinking of the figure.

[EDIT: You can see that this is just a "poor man's version" of voko's elegant derivation.]

 

[TSny]

$$U_2 = \frac {\sigma_2^2} {\sigma_1^2} s U_1 $$

Note that the area $\Delta_1$ gets integrated over twice: once with the $dxdy$ integration and then again with the $d\xi d\eta$ integration. So, I think s should be replaced by s³ in your result.

 

[voko]

Note that the area $\Delta_1$ gets integrated over twice: once with the $dxdy$ integration and then again with the $d\xi d\eta$ integration. So, I think s should be replaced by s³ in your result.

I did not actually mean that $s$ was the scaling factor in the $\Delta_1 \to \Delta_2$ map. I let Pranav figure out what it really is :)

 

[TSny]

I did not actually mean that $s$ was the scaling factor in the $\Delta_1 \to \Delta_2$ map. I let Pranav figure out what it really is :)

Ah, I see. My apologies.

Pranav..., better not trust me, think it out.

 

[Saitama]

The potential due to the first triangle is $$ \phi_1(\xi, \eta) = \int\limits_{(x,y) \in \Delta_1} k \sigma_1 \frac {dx dy} {\sqrt{(x - \xi)^2 + (y - \eta)^2}} $$ *edit: clarified the potential integral.

Sorry but I have no idea what that integral is supposed to represent. What does "the potential due to the first triangle" means? We calculate the potential at some point, right? Potential due to the triangle but where?

 

[voko]

At some point with coordinates $(\xi, \eta)$. That integral is pretty much the definition of potential in 2D.

 

[Saitama]

Hi voko! I can make sense of the potential integral but I can't comprehend the next integral.

The potential energy is $$ U_1 = \frac 1 2 \int\limits_{(\xi,\eta) \in \Delta_1} \sigma_1 \phi_1(\xi, \eta) {d\xi d\eta}$$

Why there is factor of 1/2?

It is obvious how to replace the charge density.

Not obvious to me, do you mean I simply write $\sigma_2=2\sigma_1$?

...we can map linearly $\Delta_1 \to \Delta_2$,

What does this mean?

I am sorry for asking stupid questions but I have never dealt with these kind of integrals before. :(

 

[voko]

Hi voko! I can make sense of the potential integral but I can't comprehend the next integral.

Why there is factor of 1/2?

There is, of course, a reason for that, but it is unimportant now, because it disappears in the ratio of potential energies. But if you must know, look up the potential energy of a system of charges. If still unclear, come back.

Not obvious to me, do you mean I simply write $\sigma_2=2\sigma_1$?

Well, yes, but. The purpose here is to obtain $U_2$ as $fU_1$, where $f$ is some factor. So we want to keep $\sigma_1$ inside the integral. That implies that $f$ will contain $\sigma_1$ and $\sigma_2$, as shown in the message you quoted.

What does this mean?

We have a region $\Delta_1$ and a region $\Delta_2$. We can have a function $g: \Delta_1 \to \Delta_2$. Now, because the regions are very similar, we can find a $g$ that is linear.

 

[Saitama]

There is, of course, a reason for that, but it is unimportant now, because it disappears in the ratio of potential energies. But if you must know, look up the potential energy of a system of charges. If still unclear, come back.



Well, yes, but. The purpose here is to obtain $U_2$ as $fU_1$, where $f$ is some factor. So we want to keep $\sigma_1$ inside the integral. That implies that $f$ will contain $\sigma_1$ and $\sigma_2$, as shown in the message you quoted.



We have a region $\Delta_1$ and a region $\Delta_2$. We can have a function $g: \Delta_1 \to \Delta_2$. Now, because the regions are very similar, we can find a $g$ that is linear.

Thanks voko for the patience and time you have spent in explaining the alternative method but to be honest, I am unable to comprehend the integrals.

I am not sure but do I have to express $\phi_1 (\xi,\eta)$ in terms of $\phi_2 (\xi,\eta)$? The integral for $\phi_1$ is taken over $\Delta_1$ and for $\phi_2$, the integral is taken over $\Delta_1/2$ but how to write $\phi_2$ in terms of $\phi_1$?

 

[voko]

It probably makes sense to express $U_1$ and $U_2$ directly, without $\phi_1$ and $\phi_2$. So $$ U_1 = \frac 1 2 \int\limits_{(x,y) \in \Delta_1, (\xi,\eta) \in \Delta_1} k \sigma^2_1 \frac {dx dy d\xi d\eta} {\sqrt{(x - \xi)^2 + (y - \eta)^2}} $$ and $$ U_2 = \frac 1 2 \int\limits_{(x,y) \in \Delta_2, (\xi,\eta) \in \Delta_2} k \sigma^2_2 \frac {dx dy d\xi d\eta} {\sqrt{(x - \xi)^2 + (y - \eta)^2}} $$ In the second integral, change the variables: $$ x = mx', y = my', \xi = m \xi', \eta = m \eta' $$ so that $(x', y') \in \Delta_1$ is mapped to $(x, y) \in \Delta_2$ and $(\xi', \eta') \in \Delta_1$ mapped to $(\xi, \eta) \in \Delta_2$. Then $$ U_2 = \frac 1 2 \int\limits_{(x',y') \in \Delta_1, (\xi',\eta') \in \Delta_1} k \sigma^2_2 \frac {m^4 dx' dy' d\xi' d\eta'} {\sqrt{(mx' - m\xi')^2 + (my' - m\eta')^2}} = \frac 1 2 \int\limits_{(x',y') \in \Delta_1, (\xi',\eta') \in \Delta_1} k m^3 \sigma^2_2 \frac {dx' dy' d\xi' d\eta'} {\sqrt{(x' - \xi')^2 + (y' - \eta')^2}} $$ Now compare that with $U_1$.

 

[ehild]

With those integrals, first you determine the potential at a point inside the triangle, then multiply it with the charge of a small are around that point and integrate to the whole triangle to get the potential energy. And off course, you take the half.

Switch to the dimensionless "lengths" x/a=p, y/a=q, $\xi/a=s$, $\eta/a$=t.

$$U=1/2 k \int(σ \int (σ\frac{1}{\sqrt{(ap-as)^2+(aq-at)^2}}adp adq )ads adt=$$,both integrals over the same range of the variables. σ is constant, you can drawn it out from the integral, and do the same with the scale factor a.

$$U=1/2 k σ^2 \frac{a^4}{a}\int(\int (\frac{1}{\sqrt{(p-s)^2+(q-t)^2}}dp dq )ds dt)=1/2 k σ^2 a^3 I$$

I is the dimensionless integral, it depends only on the shape not on the size of the domain. When you fold the isosceles right triangle you get an isosceles right triangle again. The integral is the same, only a and σ are different.

Now you only need to follow how the surface density and the size of the triangle changes when folding.

ehild

 

[Saitama]

With those integrals, first you determine the potential at a point inside the triangle, then multiply it with the charge of a small are around that point and integrate to the whole triangle to get the potential energy. And off course, you take the half.

Switch to the dimensionless "lengths" x/a=p, y/a=q, $\xi/a=s$, $\eta/a$=t.

$$U=1/2 k \int(σ \int (σ\frac{1}{\sqrt{(ap-as)^2+(aq-at)^2}}adp adq )ads adt=$$,both integrals over the same range of the variables. σ is constant, you can drawn it out from the integral, and do the same with the scale factor a.

$$U=1/2 k σ^2 \frac{a^4}{a}\int(\int (\frac{1}{\sqrt{(p-s)^2+(q-t)^2}}dp dq )ds dt)=1/2 k σ^2 a^3 I$$

I is the dimensionless integral, it depends only on the shape not on the size of the domain. When you fold the isosceles right triangle you get an isosceles right triangle again. The integral is the same, only a and σ are different.

Now you only need to follow how the surface density and the size of the triangle changes when folding.

ehild

Thanks a lot ehild! That was very nice and detailed. I can make sense of the integral method now.