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

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Correct answer: C. 200 2 ( i + j ) NC ^ -1

E= K q r^2 r = K q r^3 r Here, aligned r & = (x_2-x_1 )^2+ (y_2-y_1 )^2+ (z_2-z_1 )^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

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