The electric field at point (30, 30, 0) due to a charge `0.008muC` at origin will be (coordinates are in cm)

Options

A. `8000NC^(-1)`
B. `4000(hati+hatj)NC^(-1)`
C. `200sqrt2(hati +hatj)NC^(-1)`
D. `400sqrt2(hati+hatj)NC^(-1)`

Answer

C

Explanation

Electric field `E=(kq)/(r^(2)),`
`r=sqrt(x^(2)+y^(2)+z^(2))=sqrt(30^(2)+30^(2)+0^(2))`
= Distance of point from origin `=30sqrt2cm`
`"So, "|E|=(9xx10^(9)xx0.008xx10^(-6))/((30sqrt2xx10^(-2))^(2))`
`=400N//C`
Also, unit vector in direction of (30, 30, 0) point from origin
`=(hati+hatj)/(sqrt2)`
`"So, E = Magnitude"xx"direction"`
`=400xx((hati+hatj)/(sqrt2))`
`=200sqrt2(hati+hatj)NC^(-1)`