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

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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 $\;|E|_{net}=6\times10^{4}\;NC^{-1}\;.$

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