Position vector of q1, r1=2i^+3j^+3k^
Position vector of q2, r2=i^+j^+k^
Vector from q1 to q2 is r21=r2−r1=(i^+j^+k^)−(2i^+3j^+3k^)=−i^−2j^−2k^
Magnitude ∣r21∣=(−1)2+(−2)2+(−2)2=9=3
Force on q2 due to q1 is given by Coulomb's law in vector form:
F21=4πϵ01∣r21∣3q1q2r21
Substituting the given values:
F21=339×109×(3×10−6)×(−4×10−6)(−i^−2j^−2k^)
F21=27−108×10−3(−i^−2j^−2k^)
F21=−4×10−3(−i^−2j^−2k^)
F21=(4i^+8j^+8k^)×10−3 N
Answer: (4i^+8j^+8k^)×10−3