Question:
The top of the atmosphere is at about $400\, kV$ with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about $100 \,Vm^{-1}$. Still, we do not get an electric shock as we step out of our house into the open because (assume the house to be a steel cage so that there is no field inside)
The top of the atmosphere is at about $400\, kV$ with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about $100 \,Vm^{-1}$. Still, we do not get an electric shock as we step out of our house into the open because (assume the house to be a steel cage so that there is no field inside)
Updated On: Aug 1, 2022
- there is a pd between our body and the ground
- $100\, Vm^{-1}$ is not a high electric field so that we do not feel the shock.
- our body and the ground forms an equipotential surface.
- the atmosphere is not a conductor
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The Correct Option is D
Solution and Explanation
Dry atmosphere is not a conductor. Only when top of the atmosphere increases to a very large value and there is moisture in the air the lightning strikes the ground. Since $400\, kV $ is not very high voltage so, it will not strike the ground.
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).