Two charges $10^{- 8}$ and $- 10^{- 8} C$ are placed at two corners of an equilateral triangle of side 20cm. Electric intensity at the third corner is:

\sqrt{3} \times 2250 N C^{- 1}

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\sqrt{2} \times 2250 N C^{- 1}

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450 N C^{- 1}

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2250 N C^{- 1}

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Solution

The correct option is C $2250 N C^{- 1}$
Electricle field $=$ $(E - E C o s 60^{0})^x + (E S i n 60^{0})^y$
$= \frac{E}{2}^x + \frac{E \sqrt{3}}{2}^y$

$= E (\frac{1}{2}^x + \frac{\sqrt{3}}{2}^y)$

$= \frac{k (10^{- 8})}{(20 \times 10^{- 2})^{2}} (\frac{1}{2}^x + \frac{\sqrt{3}}{2}^y)$

$E = \frac{9 \times 10^{9} \times 10^{- 8}}{4 \times 10^{- 2}}$

$E = 2250 (\frac{1}{2}^x + \frac{\sqrt{3}}{2}^y)$

$| E | = 2250 \sqrt{\frac{1}{4} + \frac{3}{4}}$

$| E | = 2250 N / C$

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Two charges10−8and −10−8C are placed at two corners of an equilateral triangle of side 20cm. Electric intensity at the third corner is:

The correct option is C 2250NC−1Electricle field = (E−ECos600)^x+(ESin600)^y=E2^x+E√32^y=E(12^x+√32^y)=k(10−8)(20×10−2)2(12^x+√32^y)E=9×109×10−84×10−2E=2250(12^x+√32^y)|E|=2250√14+34|E|=2250 N/C

Two charges $10^{- 8}$ and $- 10^{- 8} C$ are placed at two corners of an equilateral triangle of side 20cm. Electric intensity at the third corner is: