Four charges of +q, +q +q and +q are placed at the corners A, B, C and D of a square. The resultant force on the charge at D

Given data: Four charges with charge +q

Concept used: Electric fields superposition

Detailed solution:

Force between the charges at D and C

$F_{DC}=\frac{K{q}^2}{a^2}$

Force between the charges at D and A

$F_{DA}=\frac{K{q}^2}{a^2}$

Force between the charges at D and B

$$\begin{gathered} F_{DB}=\frac{K{q}^2}{{\left(\sqrt{2}a\right)}^2} \\ {F}_{DB}=\frac{K{q}^2}{2a^2} \end{gathered}$$

Net force on the charge at D is

$$\begin{gathered} F_{DC}=F_{DA}=F \\ F=\sqrt{F^2+F^2+2F\cos \theta }+F_{BD} \\ F=\sqrt{2}F+F_{DB} \\ F=\frac{K{q}^2}{a^2}\sqrt{2}+\frac{K{q}^2}{2a^2} \\ F=\frac{K{q}^2}{2a^2}\left(1+2\sqrt{2}\right) \\ F=\frac{q^2}{8\pi {\in }_0{a}^2}\left(1+2\sqrt{2}\right) \end{gathered}$$