Step 1: Understanding the Question:
The question asks for the specific pF value, which is a logarithmic measure of soil water tension (matric potential), that identifies the state of "Field Capacity."
Field Capacity is defined as the amount of soil moisture or water content held in the soil after excess water has drained away and the rate of downward movement has materially decreased.
Key Formula or Approach:
The concept of pF was introduced by Schofield to represent soil water tension on a logarithmic scale.
The formula for pF is:
\[ pF = \log_{10}(\text{suction in cm of water column}) \]
Step 2: Detailed Explanation:
• Definition of Field Capacity (FC):
Field capacity is usually reached $2$ to $3$ days after a heavy rain or irrigation in well-drained soils of uniform structure.
The tension at field capacity varies slightly with soil texture but is generally accepted as $-1/3$ bar or $0.33$ atmospheres for most agricultural soils.
• Calculation of pF for Field Capacity:
$1$ bar of pressure is approximately equivalent to $1023$ cm of water head.
Therefore, $0.33$ bars $\approx 0.33 \times 1023 \approx 337$ cm of water.
Applying the formula: \( pF = \log_{10}(337) \approx 2.52 \).
Hence, the standard pF value for Field Capacity is $2.5$.
• Analyzing Other pF Values:
pF 0: Represents a saturated soil where the suction is zero (water is at atmospheric pressure).
pF 1.0: Represents a very wet soil, near saturation ($10$ cm of suction).
pF 4.2: Represents the Permanent Wilting Point (PWP), which occurs at $15$ bars of tension ($\log_{10}(15000) \approx 4.18$).
• Agricultural Significance:
Between pF $2.5$ (FC) and pF $4.2$ (PWP) lies the "Available Water" range for plants.
Soil management aims to keep moisture content close to pF $2.5$ for optimal crop growth.
Step 3: Final Answer:
The pF value corresponding to the matric potential at field capacity is $2.5$.