Electric field at point (30,30,0) due to a charge of 0.008 C placed at origin will be, (coordinates are in cm )

Step-by-step solution

E= K q r^2 r = K q r^3 r Here, aligned r & = (x₂-x₁ )^2+ (y₂-y₁ )^2+ (z₂-z₁ )^2 & = (30)^2+(30)^2+0^2 & =30 2 ~cm & =30 2 10⁻² ~m aligned and q=8 10⁻³ 10⁻⁶ C Also, r =(30 i +30 j ) 10⁻² ~m aligned So, E & = 9 10^9 8 10⁻³ 10⁻⁶ (30 2 10⁻² )^3 (30 i +30 j ) 10⁻² & = 9 8 10^9 10⁻¹¹ 27 2 2 10⁻⁶ 10^3 30( i + j ) & = 9 8 10^2 3 27 2 2 ( i + j ) & =200 2 ( i + j ) NC ⁻¹ aligned