$(a)\;\large\frac{kq}{l^2}\qquad(b)\;\large\frac{kq}{2l^2}\qquad(c)\;\large\frac{2kq}{l^2}\qquad(d)\;none$

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- The SI unit of quantity of electric charge is the coulomb, which is equivalent to about $6.242 \times 10^{18}$ e (e is the charge of a proton). Hence, the charge of an electron is approximately $?1.602 \times 10^{?19}$ C.

Answer : (c) $\;\large\frac{2kq}{l^2}$

Explanation : Electric field due to $q_{1}\; and \; q_{3}\;$ is equal and opposite and therefore cancel each other . Net electric field will be due to only $q_{2}\;.$

$Distance\;of\;q_{2}\;from\;O (r)=\large\frac{l}{\sqrt{2}}$

$E=\large\frac{kq}{r^2}$

$E=\large\frac{2kq}{l^2}\;.$

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