2
$\begingroup$

I am having trouble with this example : enter image description here

As it says that the contact angle of the glass water interface is $0^{\circ}$ , so the force due to surface tension should act downward on the plates (rather than along the horizontal which would've given a clear idea of attractive interactions).

However, I also notice that the tendency of water would to decrease the area of its free surface. Therefore, I tried to approach this by calculating the surface potential energy as a function of the separation between the plates, $$U = T \times A$$ And thereby calculate force, $F$ as $F = -\frac{dU}{dx}$ where $x$ is the separation between the plates.

I am, unfortunately, still unable to solve it, because I am failing to find the expression for $U$. May I get help as to how to approach the problem and hence derive the expression for the force between the plates?

$\endgroup$
  • $\begingroup$ I think that when the question says the contact angle is 0˚ it means that the water surface is perfectly horizontal and normal to the glass plates, as shown in the figure. As for the energy, remember that you need to use the total energy in order to calculate the force. The constants $g$ and $\rho$ that you see in all of the possible answers are a hint. $\endgroup$ – Samuel Weir Apr 27 at 18:06
1
$\begingroup$

It is not directly,attraction, as you would say . What actually happens is as explained below: The water rises up in the plates due to capillary action. The water forms a concave surface, and due to this concavity, the pressure inside water is lesser than the pressure outside(the excess pressure). Now, if you consider a plate, there is force from inside due to pressure of water, and an external force due to atmospheric pressure which is greater. This basically causes an inward force , what you call an attractive force between the plates. Hope it helps you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.