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Q.

Electric charge 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

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a

1.8N

b

2.7N

c

3.6N

d

0.9N

answer is B.

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Detailed Solution

$\begin{aligned} F_{AC} = \frac{1}{4 π ε_{0}} \frac{1 \times 10^{- 6} \times 2 \times 10^{- 6}}{{(10 \times 10^{- 2})}^{2}} = 1.8 N \\ F_{BC} = \frac{1}{4 π ε_{0}} \frac{1 \times 10^{- 6} \times 2 \times 10^{- 6}}{{(10 \times 10^{- 2})}^{2}} = 1.8 N \end{aligned}$
Question Image
$F_{AC} = F_{BC}$ and angle between then $θ = 120^{0}$
Then the resultant also either $F_{AC} (Or) F_{BC}$
$∴ Net force 2 μ c is = 1.8 N$

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