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