An electron experiences a force equal to its weight, when placed in an electric field. The intensity of the field will be

Options

A. `10xx10^(-31)N//C`
B. `5.5xx10^(-11)N//C`
C. `15xx10^(11)N//C`
D. `2xx10^(-11)N//C`

Answer

B

Explanation

`F=mg and F=qE`
`therefore qE=mg`
`therefore E=(mg)/(q)=(9xx10^(-31)xx9.8)/(1.6xx10^(-19))`
`=5.5xx10^(-11)N//C`