A current density of 8.50 10-13 A/m2 exists in the atmosphere at a location where the electric field is 143 V/m. Calculate the electrical conductivity of the Earth's atmosphere in this region.

The electrical conductivity is 5.94×10^-15 ohm^-1 m^-1

Explanation:

Electrical conductivity = current density ÷ electric field

Current density = 8.5×10^-13 A/m^2

Electric field = 143 V/m

Electrical conductivity = 8.5×10^-13 A/m^2 ÷ 143 V/m = 5.94×10^-15 ohm^-1 m^-1

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach