2.25 N force is acting on `15 xx 10^(-4)`C charge placed at a point in a uniform electric field. So, intensity of this electric field is......

Options

A. 150 N
B. 15 N
C. `1500 N//C`
D. `0.15 N//C`

Answer

C

Explanation

`q = 15 xx 10^(-4) C, F= 2.25 N, E=?`
`therefore E = F/q`
`=(2.25)/(15 xx 10^(-4))`
`=1500 V//m`