Q.
Electric charges of 1μC,−1μC and 2μC are placed in air at the corners A,B and C, respectively, of an equilateral triangle ABC having length of each side 10cm. The resultant force on the charge at C is
FA= Force on C due to charge placed at A =9×109×(10×10−2)210−6×2×10−6=1.8N FB= Force on C due to charge placed at B =9×109×(0.1)210−6×2×10−6=1.8N
Net force on C, Fnet=(FA)2+(FB)2+2FAFBcos120∘=1.8N