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ABC is an equilateral triangle of each side 5 cm . Two charges of $\;\pm\large\frac{50}{3}\times10^{-3}\;\mu C\;$ are placed at B and A respectively . Find the magnitude of resultant field at C .

$(a)\;6\times10^{4}\;NC^{-1}\qquad(b)\;6\times10^{-4}\;NC^{-1}\qquad(c)\;6\times10^{0}\;NC^{-1}\qquad(d)\;None$

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Answer : (a) $\;6\times10^{4}\;NC^{-1}$
Explanation : E due to A = $\;\large\frac{9\times10^9\times50\times10^{-9}\times10^{4}}{3\times5^2}$
$=6\times10^4\;NC^{-1}$
Similarly E due to charge at B = $\;6\times10^{4}\;NC^{-1}$
Both are at $\;120^{0}\;$ to each other and of equal magnitude
Therefore
resultant $\;|E|_{net}=6\times10^{4}\;NC^{-1}\;.$
answered Feb 3, 2014 by yamini.v
edited Feb 3, 2014 by yamini.v
 

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