# Capillary action, pressure at differents places

I figured out that I really have no intuition about pressure in capillary action. I understand the math, especially forces (tension force and gravitational force), but the pressure is foreign to me.

So can you explain in which points the pressure is higher and where it is lower?

My reasoning tells me that the pressure at point A is higher than in point B, but I also know that the pressure at points C and D are equal so that means that pressure at points A, C and D is roughly 100 kPa. So is it right to assume that the pressure at point C is $$P_C = P_B - \rho gh$$. Is this correct or am I completely wrong here?

• Please show your attempt to answer the question, including the relevant reasoning. Otherwise this reads—regardless of your intentions—like a request to do homework. Aug 9, 2021 at 17:27
• If you read through the answers to this question physics.stackexchange.com/questions/648141/… it might help Aug 9, 2021 at 17:37
• @Chemomechanics thanks for helpful comment, I added some personal thought process to the question to avoid being closed down. Aug 9, 2021 at 18:03

In any case, the pressures on either side of a flat interface are equal because the interface exerts no forces, and the pressures on equal heights of the same fluid are equal when the fluid is able to shift. So I agree with the rest of your reasoning that $$P_\mathrm{C}=P_\mathrm{B}+\rho_\mathrm{water}gh$$. (I believe you had an inadvertent sign error.)