If the earth's surface is treated as a conducting surface with some charge , then the order of magnitude of the charge per unit area `sigma "in" C//m^(2)` , so that a proton remains suspended in space near the earh's surface will be

Options

A. `sigma=(e)/(epsi_(0)mg)C//m^(2)`
B. `sigma=(epsi_(0)mg)/(e)C//m^(2)`
C. `sigma=(mg)/(epsi_(0)e)C//m^(2)`
D. `sigma=(epsi_(0)e)/(mg)C//m^(2)`

Answer

B

Explanation

Let `sigma` be the surface density of cahrge on the earth's surface. Then the electric field intensity just near the earth's surface
is given by `E=(sigma)/(epsi_(0))`
Let e and m be the charge and mass of the proton.
`therefore`The upward force on the proton`=eE=(esigma)/(epsi_(0))`
The downward force acting on the proton=mg
For equilibrium,`mg=(esigma)/(epsi_(0))`
`therefore sigma=(epsi_(0)mg)/(e)`