Two electric charges of $+ 10^{- 9} C$ and $- 10^{- 9} C$ are placed at the corners A and B of an equilateral triangle ABC side 5cm.The electric intensity at C is

1800 N / C

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3600 N / C

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900 N / C

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2700 N / C

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Solution

The correct option is B $3600 N / C$
Net $\to E$ in y direction gets cancelled
Field in x direction is given by:
$E_{x} = (\frac{k \times 10^{- 9}}{(5 \times 10^{- 2})^{2}} \times c o s 60^{0}) \times 2$

$= \frac{9 \times 10^{9} \times 10^{- 9}}{25 \times 10^{- 4}} \times \frac{1}{2} \times 2$

$E_{x} = 3600 N / C$

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