Answer:
[tex]\rho=48.3 \ohm-m[/tex]
Explanation:
Given
resistance of the plant R =2.3 MΩ
separation between the electrode L= 20 cm
width of the leaf B= 2.1 cm
thickness t=0.20 mm
Area of the leaf= t×B
= 0.20 mm×2.1 cm=[tex]0.42\times10^{-5} m^2[/tex]
Resistance of the leaf related to length and area
[tex]R= \frac{\rho L}{A}[/tex]
[tex]\rho= \frac{R A}{L}[/tex]
plugging values we get
[tex]\rho= \frac{2.3\times10^{6}\times.42\times10^{-5}}{20\times10^{-2}}[/tex]
[tex]\rho=48.3\ohm-m[/tex]
