Deriving the electrostatic pressure of a water droplet

[etotheipi]

Homework Statement

Calculate the outward electrostatic pressure acting on a spherical, isolated water droplet of net charge q, radius r.

Relevant Equations

Gauss's Law, electric fields

I assumed a uniform distribution of charge within the droplet such that $E = \frac{q}{4\pi\epsilon_{0}r^{2}}$ at the outside surface. I then said that the pressure acting at the surface would be the force on a charge element $dq$ within an area $dA$ on the surface, divided by the area. This doesn't appear to be particularly useful:

$$P = \frac{q}{4\pi\epsilon_{0}r^{2}}\frac{dq}{dA}$$

The right term is the area density $\sigma$, yet I can't find a way to obtain this quantity (and if the area element is of zero width, it seems as if $\sigma = 0$ because it is the volume distribution which is uniform).

So how could we compute pressure?

 

[haruspex]

Homework Statement:: Calculate the outward electrostatic pressure acting on a spherical, isolated water droplet of net charge q, radius r.
Homework Equations:: Gauss's Law, electric fields

I assumed a uniform distribution of charge within the droplet

Water is a conductor, no?

 

[etotheipi]

Water is a conductor, no?

Ah, that was quite an oversight on my part! In that case, I suppose we'd have $\sigma = \frac{q}{4\pi r^{2}}$ on the surface of the droplet.

If I set my Gaussian surface to be the same surface as the layer of charge, does the charge enclosed include the charge actually on the surface? Because if it does, then the electric field in the vicinity of the charges is equal to the expression in my first post and it becomes just a case of substitution to calculate the pressure, however if it doesn't, then that would seem to imply that the electric field only becomes non-zero strictly above the layer of charge. I'm guessing it is the first option? Thanks a bunch!

 

[TSny]

Water is a conductor, no?

Yes, I think it's likely that this is the intended interpretation. However, it is interesting to note that very pure water is not very conductive.

https://en.wikipedia.org/wiki/Purified_water#Electrical_conductivity

But any drop of water that you are likely to meet in everyday life will have dissolved ions which will make it much more conductive.

 

[TSny]

If I set my Gaussian surface to be the same surface as the layer of charge, does the charge enclosed include the charge actually on the surface? Because if it does, then the electric field in the vicinity of the charges is equal to the expression in my first post and it becomes just a case of substitution to calculate the pressure, however if it doesn't, then that would seem to imply that the electric field only becomes non-zero strictly above the layer of charge. I'm guessing it is the first option? Thanks a bunch!

You have to be careful here. The external electric field acting on a small patch of surface area $dA$ is not equal to the electric field inside the sphere (zero) nor is it equal to the electric field just outside the surface of the sphere. This is tricky, and there are at least a couple of ways to construct an argument for determining the correct value of $E'$ to use for determining the force per unit area $\sigma E'$.

 

[etotheipi]

You have to be careful here. The external electric field acting on a small patch of surface area $dA$ is not equal to the electric field inside the sphere (zero) nor is it equal to the electric field just outside the surface of the sphere. This is tricky, and there are at least a couple of ways to construct an argument for determining the correct value of $E'$ to use for determining the force per unit area $\sigma E'$.

I wonder, suppose the sphere is centred on the origin and has radius 1, and that we want to calculate the electric field strength at $(x,y,z) = (1,0,0)$. By symmetry, there is no $y$ or $z$ component to the electric field, so I'm thinking that maybe we could integrate the contributions of successive rings of charge from $-1$ to $1$ to calculate the $x$ component of the field?

 

[haruspex]

Ah, that was quite an oversight on my part! In that case, I suppose we'd have $\sigma = \frac{q}{4\pi r^{2}}$ on the surface of the droplet.

If I set my Gaussian surface to be the same surface as the layer of charge, does the charge enclosed include the charge actually on the surface? Because if it does, then the electric field in the vicinity of the charges is equal to the expression in my first post and it becomes just a case of substitution to calculate the pressure, however if it doesn't, then that would seem to imply that the electric field only becomes non-zero strictly above the layer of charge. I'm guessing it is the first option? Thanks a bunch!

Yes, that's an interesting question. I seem to remember an argument that says it is half at the boundary. The scenario for that question was a spherical conducting shell with a small hole.

I just tried doing it by supposing the charge is uniformly distributed over a shell thickness $\Delta R$ at the surface, taking an element at radius r thickness dr and spanning a solid angle $d\Omega$ within that shell, finding the force on that, integrating over the shell thickness, etc. When I let $\Delta R$ tend to zero I did indeed get half.

 

[TSny]

I wonder, suppose the sphere is centred on the origin and has radius 1, and that we want to calculate the electric field strength at $(x,y,z) = (1,0,0)$. By symmetry, there is no $y$ or $z$ component to the electric field, so I'm thinking that maybe we could integrate the contributions of successive rings of charge from $-1$ to $1$ to calculate the $x$ component of the field?

You could do this. But when you integrate, you only want to integrate over rings that do not include any of the charge of the little patch of area $dA$ for which you are calculating the force.

The picture shows the spherical shell with the little patch of area $dA$ removed off to the side. This area would fill the hole shown in the shell. The force on $dA$ is due to the electric field $E'$ at the hole that is produced by all the charge on the shell (without the patch in place). Integrating over the rings of the shell (with the hole) would give you $E'$. But that doesn't seem like much fun.

Instead, see if you can use superposition ideas. The little patch $dA$ can be thought of as a little disk that produces its own electric field $E''$ as shown in blue. If you replace $dA$ back in the hole, the superposition of the red and blue fields must produce the total field of a complete spherical shell of charge. This idea can be used to find $E'$ without much calculation.

Another approach to the problem is to consider the potential energy $U$ of the (complete) shell as a function of its radius. Then think about how the pressure is related $U$.

 

[etotheipi]

Yes, that's an interesting question. I seem to remember an argument that says it is half at the boundary. The scenario for that question was a spherical conducting shell with a small hole.

I just tried doing it by supposing the charge is uniformly distributed over a shell thickness $\Delta R$ at the surface, taking an element at radius r thickness dr and spanning a solid angle $d\Omega$ within that shell, finding the force on that, integrating over the shell thickness, etc. When I let $\Delta R$ tend to zero I did indeed get half.

That's pretty awesome; I'm trying to work through your method now but I'm wondering how you determined the force on the element of the shell. I was thinking maybe since the electric field of a ring is $E = \frac{Qx}{4\pi\epsilon_{0}(x^{2}+r^{2})^{\frac{3}{2}}}$, if we split up the sphere into lots of cylindrical strips and integrate this quantity over $[-r,r]$, something along the lines of $$dE = \frac{2\pi y\sigma x}{4\pi\epsilon_{0}(x^{2}+r^{2})^{\frac{3}{2}}} dx$$ where $y = \sqrt{r^{2} - x^{2}}$.

 

[kuruman]

Another approach to the problem is to consider the potential energy $U$ of the (complete) shell as a function of its radius. Then think about how the pressure is related $U$.

My favorite approach to this type of problem.

 

[etotheipi]

View attachment 255884

Instead, see if you can use superposition ideas. The little patch $dA$ can be thought of as a little disk that produces its own electric field $E''$ as shown in blue. If you replace $dA$ back in the hole, the superposition of the red and blue fields must produce the total field of a complete spherical shell of charge. This idea can be used to find $E'$ without much calculation.

Ah, so if put $E''$ as $\frac{\sigma}{2\epsilon_{0}}$, then we must have

$\frac{\sigma}{2\epsilon_{0}} + E' = \frac{Q}{4\pi\epsilon_{0}r^{2}}$ just above the surface of $dA$?

 

[TSny]

Ah, so if put $E''$ as $\frac{\sigma}{2\epsilon_{0}}$, then we must have

$\frac{\sigma}{2\epsilon_{0}} + E' = \frac{Q}{4\pi\epsilon_{0}r^{2}}$ just above the surface of $dA$?

Yes

 

[etotheipi]

Another approach to the problem is to consider the potential energy $U$ of the (complete) shell as a function of its radius. Then think about how the pressure is related $U$.

I will admit I had to look up the equation for the self energy of the shell (I will try to derive it after!), however this turns out to be $U = \frac{q^{2}}{8\pi\epsilon_{0}R}$.

Then $P = \frac{U}{V} = \frac{3q^{2}}{32\pi^{2}\epsilon_{0}R^{4}}$. If that happens to be correct, it's an incredibly efficient method!

 

[kuruman]

I will admit I had to look up the equation for the self energy of the shell (I will try to derive it after!), however this turns out to be $U = \frac{q^{2}}{8\pi\epsilon_{0}R}$.

Then $P = \frac{U}{V} = \frac{3q^{2}}{32\pi^{2}\epsilon_{0}R^{4}}$. If that happens to be correct, it's an incredibly efficient method!

Actually, you should use $P =- \frac{dU}{dV}$. The negative sign says that when the volume expands by a virtual $dV$, the shell does work $dW=PdV$ on the environment which is accompanied by a decrease of $U$ by the same amount.

Yes, it is an efficient way of doing this and that's why it is my favorite approach.

 

[haruspex]

I had to look up the equation for the self energy of the shell

At a guess, deriving that involves something equivalent to my integration.

 

[kuruman]

At a guess, deriving that involves something equivalent to my integration.

It is usually done by integrating the electrostatic energy density $u=\frac{1}{2}\epsilon_0E^2$ over all space.

 

[gneill]

If we accept that the electric potential very close to the surface of a conducting spherical shell with radius R and holding charge Q is given by

$V = \frac{1}{4 \pi \epsilon_o} \cdot \frac{Q}{R }$

then a little rearrangement gives:

$\frac{Q}{V} = 4 \pi \epsilon_o R$

But $\frac{Q}{V}$ (Coulombs per Volt} is a capacitance.

The energy stored in a capacitor is given by:

$E = \frac{1}{2} C V^2$

We have expressions for both C and V, so:

$E = \frac{1}{2} \left(4 \pi \epsilon_o R \right) \left( \frac{1}{4 \pi \epsilon_o} \frac{Q}{R } \right)^2$

After simplifying:

$E = \frac{Q^2}{8 \pi \epsilon_o R}$

Edit:
I probably should have used the alternative formula for the energy stored in a capacitor:

$E = \frac{1}{2} \frac{Q^2}{C}$

or even better, skipping the (implied) capacitance equation step altogether:

$E = \frac{1}{2} Q V$

Less algebra that way! Really, I should have seen that simplification, but I so rarely use that last way of calculating the energy stored in a capacitor that it frankly slipped my mind (at least for a while).