Answer:
current through water layer is given as 0.0090051 A
Explanation:
Given data:
Radius of kite string is 2.10 mm
Thickness of water coating 0.517 mm
distance of kite in upward direction = 0.823 km
Resistivity = 168 ohm. m
Potential difference is 165 MV
Cross-sectional area of water
[tex]A = pi*((2.61*10^{-3}m)^2 - (2.10*10^{-3})^2[/tex]
[tex]= 7.546*10^{-6} m^2[/tex]
Resistance of water[tex]R = resistivity \times \frac{length\ L }{area\ A}[/tex]
[tex]= 168\times \frac{823}{7.5465*10^{-6}}[/tex]
[tex]= 1.832 \times 10^{10} Ohm[/tex]
Current through water [tex]I =\frac{ V }{ R}[/tex]
[tex]= \frac{165\times 10^6}{1.832\times 10^{10}} Ohm[/tex]
= 0.0090051 A
