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Question

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 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
=9×109×106×2×106(10×102)2=1.8N
FB = force on C due to charge placed at B
=9×109×106×2×106(0.1)2=1.8N

Net force on C
Fnet=(FA)2+(FB)2+2FAFBcos120=1.8N

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