Answer: F= 19.09 ×10-3 N direction DB Explanation: given a square ABCD whose side 2 m corner A have charge 3×10-6 C B have charge 2×10-6 C C have charge 3×10-6 C find force on B force on B due to A F₁= k q₁q2/r² = k(3×10-6 Cx 2×10-6 C)/ 4 F₁= (3/2)kx10-12 AB direction force on B due to C F3=k q2q3/r²= k( 3×10-6 Cx 2×10-6 C)/4 F₁= (3/2)k×10-12 CB direction add F, and F2 angle between F, and F3 Ø=90° net force F=√ F₁² + F3²+2F₁F3 cosØ cos 90=0 F= √ F₁² + F32 F = √((3/2)k×10-12)²+((3/2)k×10-12)2 F = (3/2)k×10-12x√2 F = 13.5×10-2√2 F= 19.09 ×10-2 N