Electrostatic force experienced by `-3 muC` charge placed at point 'P' due to a system 'S' of fixed point charges as shown in figure is `vec(F)=(21 hat(i)+9hat(j)) muN`.
(i) Find out electric field intensity at point P due to S.
(ii) If now, `2 muC` charge is placed and `-3 muC` is removed at point P then force experienced by it will be.

Explanation

(i) `vec(F)=qvec(E) implies (21 hat(i)+9hat(j))muN=-3 muC (vec(E))`
`implies vec(E)=-7 hat(i) -3 hat(j) N/C`
(ii) Since the source charges are not disturbed the electric field intensity at 'P' will remain same.
`vec(F)_(2 muC)=+2(vec(E))=2(-7 hat(i)-3hat(j))=(-14 hat(i)-6 hat(j)) muN`