Question

Humid air breaks down (its molecules become ionized) in an electric field of 4.99 x 10^6 N/C. In that field, what is the magnitude of the electrostatic force on (a) an electron and (b) an ion with a single electron missing? (a) Number _____ Units _____ (b) Number _____ Units _______

Answer 1

Explanation:

It is given that,

Electric field, $E=4.99\times 10^6\ N/C$

The relation between the electric force and the electric field is given by :

$F=qE$

Where

q is the charge

(a) Charge on the electron, $q_e=-1.6\times 10^{-19}\ C$

Electrostatic force,

$F=-1.6\times 10^{-19}\times 4.99\times 10^6$

$F=-7.98\times 10^{-13}\ N$  

(b) If an electron is missing, the net charge remains the same only sign changes.

So, force becomes, $F=7.98\times 10^{-13}\ N$  

Hence, this is the required solution.