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Question: Electric charges of \(1\mu C, - 1\mu C\) and \(2\mu C\) are placed in air at the corners A, B and C ...

Electric charges of 1μC,−1μC1\mu C, - 1\mu C and 2μC2\mu C 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

B

1.8 N

C

2.7 N

D

3.6 N

Answer

1.8 N

Explanation

Solution

FA = force on C due to charge placed at A

=9×109×10−6×2×10−6(10×10−2)2=1.8N= 9 \times 10^{9} \times \frac{10^{- 6} \times 2 \times 10^{- 6}}{(10 \times 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= 9 \times 10^{9} \times \frac{10^{- 6} \times 2 \times 10^{- 6}}{(0.1)^{2}} = 1.8N

Net force on C

Fnet=(FA)2+(FB)2+2FAFBcos⁡120o=1.8NF_{net} = \sqrt{(F_{A})^{2} + (F_{B})^{2} + 2F_{A}F_{B}\cos 120^{o}} = 1.8N