The magnitude of point charge due to which the electric field 30 c away has the magnitude `2"N C"^(-1)` will be

Options

A. `2 xx10^(-11)C`
B. `3 xx10^(-11)C`
C. `5 xx10^(-11)C`
D. `9 xx10^(-11)C`

Answer

A

Explanation

The magnitude of electric field due to point charge q at a distance r form the charge is
`E=1/(4piepsilon_0)q/r^2`
Here, `E=2 N C^(-1) , r = 30 cm = 0.3 m, `
`1/(4piepsilon_0)=9xx10^9N m^2C^(-2),q = ` ?
`:. 2N C^(-1) = ((9xx10^9N m^2C^(-2))q)/((0.3m)^2)`
or `q=((2NC^(-1))(0.09m^2))/ ((9xx10^9N m^2C^(-2)))=2xx10^(-11)C`
Thus, the magnitude of point charge will be `2xx10^(-11)C`.