Explanation

(b) Two equal and opposite charges are placed at a distance d. Electric field at centre(B) due to +$\mathcal{Q}$charge
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$\left( E_{1} \right) = \frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}}{\left( \frac{d}{2} \right)^{2}}$
Similarly, electric field due to $- \mathcal{Q}$charge
$\left( E_{2} \right) = \frac{1}{4\pi\varepsilon_{0}}\frac{\left( \mathcal{Q} \right)}{\left( \frac{d}{2} \right)^{2}}$
Therefore, net electric field at point
$E = E_{1} + E_{2}$
$= \frac{1}{4\pi\varepsilon_{0}}\frac{4\mathcal{Q}}{d^{2}} + \frac{1}{4\pi\varepsilon_{0}}\frac{4\mathcal{Q}}{d^{2}} = \frac{1}{4\pi\varepsilon_{0}}\frac{8\mathcal{Q}}{d^{2}}\ $