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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?

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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?

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  • $\begingroup$ 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. $\endgroup$ Aug 9, 2021 at 17:27
  • $\begingroup$ If you read through the answers to this question physics.stackexchange.com/questions/648141/… it might help $\endgroup$ Aug 9, 2021 at 17:37
  • $\begingroup$ @Chemomechanics thanks for helpful comment, I added some personal thought process to the question to avoid being closed down. $\endgroup$ Aug 9, 2021 at 18:03

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My reasoning tells me that the pressure in point A is higher than in point B, but I also know thath the pressure in point C and D is equal so that means that pressure at point A,C and D is rougly 100 kPa. So is it right to assume that the pressure in point C is PC=PB−ρgh. Is this correct or am I completely wrong here? [emph. added]

Your reasoning would be completely correct if the oil were replaced by air; the problematic part is shown in bold. Considering the density of air to be negligible relative to that of water, the air pressure everywhere for an air–water system would be essentially identical (nominally 1 atm). Thus, the pressures at A and D would be essentially identical. However, we can't make the assumption of negligible density for oil.

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.)

(We deduce that the inside of the capillary has a higher energy when in contact with oil than with water. This energy compensates for the increased gravitational potential of the water as it spontaneously rises.)

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