# Three point charges of each of charge q are placed on three vertices of a square of side "l" meter . Find the magnitude of electric field at the centre of the square .

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

Toolbox:
• 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 $q_{2}\;.$
$Distance\;of\;q_{2}\;from\;O (r)=\large\frac{l}{\sqrt{2}}$
$E=\large\frac{kq}{r^2}$
$E=\large\frac{2kq}{l^2}\;.$
edited Jan 30, 2014 by yamini.v