Three charges, each of +4 $\mathrm{\mu C}$ are placed at the corners B, C and D of a square ABCD of side 1 m. The electric field at the center O of the square is
 

 

Here we shall use a logical trick. If a charge of 4$\mu$C had also been placed at A then
$\begin{array}{r}{\overrightarrow{\mathrm{E}}}_{\mathrm{A}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{B}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{C}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{D}}=\overrightarrow{0}\\ \Rightarrow {\overrightarrow{\mathrm{E}}}_{\mathrm{B}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{C}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{D}}=-{\overrightarrow{\mathrm{E}}}_{\mathrm{A}}\end{array}$
$\begin{array}{l}\Rightarrow \left|{\overrightarrow{\mathrm{E}}}_{\mathrm{B}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{C}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{D}}\right|=\frac{\mathrm{k}\left(4\times {10}^{-6}\right)}{(1/\sqrt{2}{)}^2}\\ \Rightarrow \left|{\overrightarrow{\mathrm{E}}}_{\mathrm{B}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{C}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{D}}\right|=\left(9\times {10}^9\right)\left(4\times {10}^{-6}\right)\left(2\right)\\ \Rightarrow \left|{\overrightarrow{\mathrm{E}}}_{\mathrm{B}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{C}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{D}}\right|=7.2\times {10}^4 {\mathrm{NC}}^{-1}\end{array}$

towards A