Interpret the figure

The wording suggests a right-angled arrangement with equal perpendicular sides of length $l$, and charges placed at the three corners other than point $O$.

So take point $O$ as one corner of a square-like right-angle figure, with three charges $q$ located at:

• distance $l$ along the $x$-axis,
• distance $l$ along the $y$-axis,
• distance $\sqrt{l^2+l^2}=l\sqrt2$ along the diagonal.

We now find the electric field at $O$ due to these three equal charges.

Field due to charge at distance $l$ on the horizontal side

Magnitude: $E_1=\frac{1}{4\pi\varepsilon_0}\frac{q}{l^2}$

Direction: along the negative $x$-axis.

Field due to charge at distance $l$ on the vertical side

Magnitude: $E_2=\frac{1}{4\pi\varepsilon_0}\frac{q}{l^2}$

Direction: along the negative $y$-axis.

Field due to charge at the diagonal corner

Distance from $O$ is: $r=l\sqrt2$

Hence magnitude:

Add components

Total $x$-component: $E_x=\frac{1}{4\pi\varepsilon_0}\left(\frac{q}{l^2}+\frac{q}{2\sqrt2\,l^2}\right)$

Total $y$-component: $E_y=\frac{1}{4\pi\varepsilon_0}\left(\frac{q}{l^2}+\frac{q}{2\sqrt2\,l^2}\right)$

Since both are equal, magnitude is: $E=\sqrt{E_x^2+E_y^2}=\sqrt2\,E_x$

So, $E=\sqrt2\cdot \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{l^2}+\frac{q}{2\sqrt2\,l^2}\right)$

$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{l^2}\left(\sqrt2+\frac12\right)$

Simplify

$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{2l^2}(2\sqrt2+1)$

Compare with options

• Option A: $\frac{1}{4\pi\varepsilon_0}\frac{q}{l^2}$ — not correct.
• Option B: $\frac{1}{4\pi\varepsilon_0}\frac{q}{2l^2}(2\sqrt2-1)$ — does not match.
• Option C: $\frac{q}{4\pi\varepsilon_0(2l)^2}$ — not correct.
• Option D: $\frac{1}{4\pi\varepsilon_0}\frac{2q}{2l^2}(\sqrt2)=\frac{1}{4\pi\varepsilon_0}\frac{q\sqrt2}{l^2}$ — not correct.

So the derived result is: $\boxed{E=\frac{1}{4\pi\varepsilon_0}\frac{q}{2l^2}(2\sqrt2+1)}$

This does not match any option, including the stored answer B.

Why stored answer B may arise

Option B, $\frac{1}{4\pi\varepsilon_0}\frac{q}{2l^2}(2\sqrt2-1),$ would occur if the diagonal charge had opposite sign (so its field subtracts from the two side contributions), or if the figure differs from the usual three-like-charges corner arrangement. Since the actual figure is not visible here, the stored answer likely assumes such a configuration.

But for the standard interpretation of three equal charges at the other three corners, the field is $\frac{1}{4\pi\varepsilon_0}\frac{q}{2l^2}(2\sqrt2+1).$