Sharpening Rod
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You’ve decided to protect your house by placing a 7.0-m-tall iron lightning rod next to the house. The top is sharpened to a point and the bottom is in good contact with the ground. From your research, you’ve learned that lightning bolts can carry up to 50kA of current and last up to 50 uS .
How much charge is delivered by a lightning bolt with these parameters?
You don’t want the potential difference between the top and bottom of the lightning rod to exceed 160 V. What is the minimum diameter, in cm, the rod can be?
Q=I∆t The current is given in kA so I=50,000A. Time is in µs so ∆t=5*10^-5
Q=(50000)(5.0*10^-5)=2.5C
R=pL/A
R=Resistance p=resistivity L=Length A=Cross-sectional area(πr^2)
The resistivity of iron is 9.7*10^-8 Ωm
The minimum resistance is equal to the minimum Potential difference divided by the current
∆V/I=Rmin -> 160V/50000A=.0032Ω
rearranging the Resistance equation to solve for r, r=(pL/Rmin*π)^(1/2)
[(9.7*10^-8)(7m)/(.0032Ω)π]^(1/2) = .0082
d=r*2 d=.0164m d=1.64cm
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