Question 93 of 311

The three charges $\dfrac{q}{2},q$ and $\dfrac{q}{2}$ are placed at the corners $A,D$ and $C$ of a square of side $a$ as shown in figure. The magnitude of electric field $\left(E\right)$ at the corner $B$ of the square, is

A$\dfrac{q}{4\pi {ϵ}_{0}{a}^{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\right)$B$\dfrac{q}{4\pi {ϵ}_{0}{a}^{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)$C$\dfrac{q}{4\pi {ϵ}_{0}{a}^{2}}\left(1-\dfrac{1}{\sqrt{2}}\right)$D$\dfrac{q}{4\pi {ϵ}_{0}{a}^{2}}\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right)$

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Answer: (A) $\dfrac{q}{4\pi {ϵ}_{0}{a}^{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\right)$

${E}_{A}=\dfrac{k\left(\dfrac{q}{2}\right)}{{a}^{2}}=\dfrac{kq}{2{a}^{2}}$, ${E}_{C}=\dfrac{k\left(\dfrac{q}{2}\right)}{{a}^{2}}=\dfrac{kq}{2{a}^{2}}$ and ${E}_{D}=\dfrac{k\left(q\right)}{{\left(\sqrt{2}a\right)}^{2}}=\dfrac{kq}{2{a}^{2}}$
As we can see from diagram, {E}_{C}&{E}_{A} are perpendicular.
Therefore, net electric field will be,
$=\dfrac{kq}{2{a}^{2}}+\sqrt{{\left(\dfrac{kq}{2{a}^{2}}\right)}^{2}+{\left(\dfrac{kq}{2{a}^{2}}\right)}^{2}}=\dfrac{kq}{2{a}^{2}}+\dfrac{kq}{\sqrt{2}{a}^{2}} =\dfrac{q}{4\pi {ϵ}_{0}{a}^{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\right)$